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Want to crack the Accenture coding round but don’t know where to start from? Look no further because we’re here with Accenture coding questions with Solutions asked in interviews which will help you crack accenture coding tests.Accenture is an international services company which includes a coding round for their hiring process. In the coding round you need to solve the problems in order to get recruited. There are multiple elements included in this coding round like data structures, algorithms, and sometimes coding exercises related to real-world scenarios.The coding rounds are conducted to test the ability of the candidate and how effectively he can write logical code. Candidates are expected to write the right answer and solve the problem and it’s pretty crucial that you solve the problem to get hired at the company.If you’re preparing for coding rounds you can check our recent post on IBM coding round. To give it to you directly the coding questions asked in the Accenture aren’t that tough if you are prepared. There are multiple elements that you need to focus on in order to prepare for the test. Today we’ve covered Accenture Coding Questions with Solutions. There are Accenture Coding Questions in Python, Accenture Coding Questions in C and many other languages. you can download the accenture coding questions and answers pdf at the bottom of this post.Accenture Coding Questions Round Duration Writing the code from scratch the candidate will be given 20 questions and 45 minutes of time to solve the problem. The languages to write the codes are:CC++JavaPythonDot NetRules for Accenture Coding Questions for Freshers Round Section: Number of questions Type of Test Time Duration Negative Marking 20 Questions Non- Adaptive 45 minutes No Latest Accenture Coding Questions And Answers 2024 for placement 1. Chocolate Distribution (Asked in Accenture, On Campus 10 Aug 2022, Slot 2)Problem Description:The function accepts an integer array ‘arr’ of size ‘n’ as its argument. Each element of ‘arr’ represents the number of chocolates distributed to a person. The function needs to return the minimum number of chocolates that need to be distributed to each person so that the difference between the chocolates of any two people is minimized.Example: Input Output n: 5 arr: 10 4 12 3 1 3 Solution: C C++ Java Python #include int min_chocolates(int arr[], int n) { // Sort the array for (int i = 0; i arr[j + 1]) { // swap int temp = arr[j]; arr[j] = arr[j + 1]; arr[j + 1] = temp; } } } // Find the minimum difference int min_diff = arr[n - 1] - arr[0]; for (int i = 1; i #include using namespace std; int min_chocolates(int arr[], int n) { sort(arr, arr + n); int min_diff = arr[n - 1] - arr[0]; for (int i = 1; i <= n - 5; i++) { min_diff = min(min_diff, arr[i + 4] - arr[i]); } return min_diff; } int main() { int n = 5; int arr[] = {10, 4, 12, 3, 1}; int result = min_chocolates(arr, n); cout << result << endl; return 0; } import java.util.Arrays; public class ChocolateDistribution { public static int minChocolates(int[] arr) { Arrays.sort(arr); int minDiff = arr[arr.length - 1] - arr[0]; for (int i = 0; i <= arr.length - 5; i++) { minDiff = Math.min(minDiff, arr[i + 4] - arr[i]); } return minDiff; } public static void main(String[] args) { int[] arr = {10, 4, 12, 3, 1}; int result = minChocolates(arr); System.out.println(result); } } def min_chocolates(arr): arr.sort() min_diff = arr[-1] - arr[0] for i in range(len(arr) - 4): min_diff = min(min_diff, arr[i + 4] - arr[i]) return min_diff arr = [10, 4, 12, 3, 1] result = min_chocolates(arr) print(result) #include int min_chocolates(int arr[], int n) { // Sort the array for (int i = 0; i arr[j + 1]) { // swap int temp = arr[j]; arr[j] = arr[j + 1]; arr[j + 1] = temp; } } } // Find the minimum difference int min_diff = arr[n - 1] - arr[0]; for (int i = 1; i #include using namespace std; int min_chocolates(int arr[], int n) { sort(arr, arr + n); int min_diff = arr[n - 1] - arr[0]; for (int i = 1; i <= n - 5; i++) { min_diff = min(min_diff, arr[i + 4] - arr[i]); } return min_diff; } int main() { int n = 5; int arr[] = {10, 4, 12, 3, 1}; int result = min_chocolates(arr, n); cout << result << endl; return 0; } import java.util.Arrays; public class ChocolateDistribution { public static int minChocolates(int[] arr) { Arrays.sort(arr); int minDiff = arr[arr.length - 1] - arr[0]; for (int i = 0; i <= arr.length - 5; i++) { minDiff = Math.min(minDiff, arr[i + 4] - arr[i]); } return minDiff; } public static void main(String[] args) { int[] arr = {10, 4, 12, 3, 1}; int result = minChocolates(arr); System.out.println(result); } } def min_chocolates(arr): arr.sort() min_diff = arr[-1] - arr[0] for i in range(len(arr) - 4): min_diff = min(min_diff, arr[i + 4] - arr[i]) return min_diff arr = [10, 4, 12, 3, 1] result = min_chocolates(arr) print(result) 2. Parking Lot ( Asked in Accenture, On Campus 10 Aug 2022, Slot 3)Problem Description:The function accepts a character array ‘arr’ of size ‘n’ as its argument. Each element of ‘arr’ represents the status of a parking slot, where ‘S’ represents an empty slot and ‘X’ represents an occupied slot. The function needs to return the maximum number of cars that can be parked in the parking lot. It is assumed that two cars cannot occupy the same slot and cars can only park in consecutive empty slots.Example: Input Output n: 16 arr: XXXSXXSXXSSXXSXX 7 Input Output n: 16 arr: XXXSXXSXXSSXXSXX 7 Solution: C C++ Java Python #include int max_cars_parked(int n, char arr[]) { int max_cars = 0; int current_cars = 0; for (int i = 0; i current_cars) ? max_cars : current_cars; current_cars = 0; } } max_cars = (max_cars > current_cars) ? max_cars : current_cars; return max_cars; } int main() { int n = 16; char arr[] = "XXXSXXSXXSSXXSXX"; int result = max_cars_parked(n, arr); printf("%dn", result); return 0; } #include using namespace std; int max_cars_parked(int n, char arr[]) { int max_cars = 0; int current_cars = 0; for (int i = 0; i < n; i++) { if (arr[i] == 'S') { current_cars++; } else { max_cars = max(max_cars, current_cars); current_cars = 0; } } max_cars = max(max_cars, current_cars); return max_cars; } int main() { int n = 16; char arr[] = "XXXSXXSXXSSXXSXX"; int result = max_cars_parked(n, arr); cout << result << endl; return 0; } public class ParkingLot { public static int maxCarsParked(int n, char[] arr) { int maxCars = 0; int currentCars = 0; for (int i = 0; i < n; i++) { if (arr[i] == 'S') { currentCars++; } else { maxCars = Math.max(maxCars, currentCars); currentCars = 0; } } maxCars = Math.max(maxCars, currentCars); return maxCars; } public static void main(String[] args) { int n = 16; char[] arr = "XXXSXXSXXSSXXSXX".toCharArray(); int result = maxCarsParked(n, arr); System.out.println(result); } } def max_cars_parked(n, arr): max_cars = 0 current_cars = 0 for status in arr: if status == 'S': current_cars += 1 else: max_cars = max(max_cars, current_cars) current_cars = 0 max_cars = max(max_cars, current_cars) return max_cars n = 16 arr = "XXXSXXSXXSSXXSXX" result = max_cars_parked(n, arr) print(result) #include int max_cars_parked(int n, char arr[]) { int max_cars = 0; int current_cars = 0; for (int i = 0; i current_cars) ? max_cars : current_cars; current_cars = 0; } } max_cars = (max_cars > current_cars) ? max_cars : current_cars; return max_cars; } int main() { int n = 16; char arr[] = "XXXSXXSXXSSXXSXX"; int result = max_cars_parked(n, arr); printf("%dn", result); return 0; } #include using namespace std; int max_cars_parked(int n, char arr[]) { int max_cars = 0; int current_cars = 0; for (int i = 0; i < n; i++) { if (arr[i] == 'S') { current_cars++; } else { max_cars = max(max_cars, current_cars); current_cars = 0; } } max_cars = max(max_cars, current_cars); return max_cars; } int main() { int n = 16; char arr[] = "XXXSXXSXXSSXXSXX"; int result = max_cars_parked(n, arr); cout << result << endl; return 0; } public class ParkingLot { public static int maxCarsParked(int n, char[] arr) { int maxCars = 0; int currentCars = 0; for (int i = 0; i < n; i++) { if (arr[i] == 'S') { currentCars++; } else { maxCars = Math.max(maxCars, currentCars); currentCars = 0; } } maxCars = Math.max(maxCars, currentCars); return maxCars; } public static void main(String[] args) { int n = 16; char[] arr = "XXXSXXSXXSSXXSXX".toCharArray(); int result = maxCarsParked(n, arr); System.out.println(result); } } def max_cars_parked(n, arr): max_cars = 0 current_cars = 0 for status in arr: if status == 'S': current_cars += 1 else: max_cars = max(max_cars, current_cars) current_cars = 0 max_cars = max(max_cars, current_cars) return max_cars n = 16 arr = "XXXSXXSXXSSXXSXX" result = max_cars_parked(n, arr) print(result) 3. String Transformation (Asked in Accenture OnCampus 10 Aug 2022, Slot 4)Problem Description:The function accepts a string ‘str’ as its argument. The function needs to return the transformed string by replacing all occurrences of the character ‘a’ with the character ‘b’ and vice versa.Example: Input Output str arr: abaabbcc bbbbaaac Solution: C C++ Java Python #include #include char* transformString(char* str) { int length = strlen(str); for (int i = 0; i #include std::string transformString(std::string str) { for (char &c : str) { if (c == 'a') c = 'b'; else if (c == 'b') c = 'a'; } return str; } int main() { std::string str = "abaabbcc"; std::cout << transformString(str) << std::endl; return 0; } public class StringTransformation { public static String transformString(String str) { char[] charArray = str.toCharArray(); for (int i = 0; i < charArray.length; i++) { if (charArray[i] == 'a') charArray[i] = 'b'; else if (charArray[i] == 'b') charArray[i] = 'a'; } return new String(charArray); } public static void main(String[] args) { String str = "abaabbcc"; System.out.println(transformString(str)); } } def transform_string(str): transformed_string = "" for i in range(len(str)): if str[i] == 'a': transformed_string += 'b' elif str[i] == 'b': transformed_string += 'a' else: transformed_string += str[i] return transformed_string str = "abaabbcc" transformed_string = transform_string(str) print(transformed_string) #include #include char* transformString(char* str) { int length = strlen(str); for (int i = 0; i #include std::string transformString(std::string str) { for (char &c : str) { if (c == 'a') c = 'b'; else if (c == 'b') c = 'a'; } return str; } int main() { std::string str = "abaabbcc"; std::cout << transformString(str) << std::endl; return 0; } public class StringTransformation { public static String transformString(String str) { char[] charArray = str.toCharArray(); for (int i = 0; i < charArray.length; i++) { if (charArray[i] == 'a') charArray[i] = 'b'; else if (charArray[i] == 'b') charArray[i] = 'a'; } return new String(charArray); } public static void main(String[] args) { String str = "abaabbcc"; System.out.println(transformString(str)); } } def transform_string(str): transformed_string = "" for i in range(len(str)): if str[i] == 'a': transformed_string += 'b' elif str[i] == 'b': transformed_string += 'a' else: transformed_string += str[i] return transformed_string str = "abaabbcc" transformed_string = transform_string(str) print(transformed_string) 4. Array Equilibrium (Asked in Accenture OnCampus 10 Aug 2022, Slot 5)Problem Description:The function accepts an integer array ‘arr’ of size ‘n’ as its argument. The function needs to return the index of an equilibrium point in the array, where the sum of elements on the left of the index is equal to the sum of elements on the right of the index. If no equilibrium point exists, the function should return -1. Example: Input Output n: 5 arr: 1 3 5 7 3 3 Solution: C C++ Java Python #include int findEquilibrium(int arr[], int n) { int totalSum = 0, leftSum = 0; for (int i = 0; i int findEquilibrium(int arr[], int n) { int totalSum = 0, leftSum = 0; for (int i = 0; i < n; i++) totalSum += arr[i]; for (int i = 0; i < n; i++) { totalSum -= arr[i]; if (leftSum == totalSum) return i; leftSum += arr[i]; } return -1; } int main() { int n = 5; int arr[] = {1, 3, 5, 7, 3}; std::cout << findEquilibrium(arr, n) << std::endl; return 0; } public class EquilibriumPoint { static int findEquilibrium(int arr[], int n) { int totalSum = 0, leftSum = 0; for (int i = 0; i < n; i++) totalSum += arr[i]; for (int i = 0; i < n; i++) { totalSum -= arr[i]; if (leftSum == totalSum) return i; leftSum += arr[i]; } return -1; } public static void main(String[] args) { int n = 5; int[] arr = {1, 3, 5, 7, 3}; System.out.println(findEquilibrium(arr, n)); } } def find_equilibrium(arr): total_sum = sum(arr) left_sum = 0 for i in range(len(arr)): total_sum -= arr[i] if left_sum == total_sum: return i left_sum += arr[i] return -1 arr = [1, 3, 5, 7, 3] print(find_equilibrium(arr)) #include int findEquilibrium(int arr[], int n) { int totalSum = 0, leftSum = 0; for (int i = 0; i int findEquilibrium(int arr[], int n) { int totalSum = 0, leftSum = 0; for (int i = 0; i < n; i++) totalSum += arr[i]; for (int i = 0; i < n; i++) { totalSum -= arr[i]; if (leftSum == totalSum) return i; leftSum += arr[i]; } return -1; } int main() { int n = 5; int arr[] = {1, 3, 5, 7, 3}; std::cout << findEquilibrium(arr, n) << std::endl; return 0; } public class EquilibriumPoint { static int findEquilibrium(int arr[], int n) { int totalSum = 0, leftSum = 0; for (int i = 0; i < n; i++) totalSum += arr[i]; for (int i = 0; i < n; i++) { totalSum -= arr[i]; if (leftSum == totalSum) return i; leftSum += arr[i]; } return -1; } public static void main(String[] args) { int n = 5; int[] arr = {1, 3, 5, 7, 3}; System.out.println(findEquilibrium(arr, n)); } } def find_equilibrium(arr): total_sum = sum(arr) left_sum = 0 for i in range(len(arr)): total_sum -= arr[i] if left_sum == total_sum: return i left_sum += arr[i] return -1 arr = [1, 3, 5, 7, 3] print(find_equilibrium(arr)) 5. Array Rotation (Asked in Accenture OnCampus 10 Aug 2022, Slot 6)Problem Description:The function accepts an integer array ‘arr’ of size ‘n’ and an integer ‘d’ as its argument. The function needs to rotate the array ‘arr’ by ‘d’ positions to the right. The rotation should be done in place, without using any additional memory.Example: Input Output n: 5 arr: 1 2 3 4 5 d: 3 3 4 5 1 2 Solution: C C++ Java Python #include void rotateArray(int arr[], int n, int d) { int temp[d]; for (int i = 0; i = d; i--) { arr[i] = arr[i - d]; } for (int i = 0; i void rotateArray(int arr[], int n, int d) { int temp[d]; for (int i = 0; i = d; i--) { arr[i] = arr[i - d]; } for (int i = 0; i < d; i++) { arr[i] = temp[i]; } } int main() { int n = 5; int arr[] = {1, 2, 3, 4, 5}; int d = 3; rotateArray(arr, n, d); std::cout = d; i--) { arr[i] = arr[i - d]; } for (int i = 0; i < d; i++) { arr[i] = temp[i]; } } public static void main(String[] args) { int n = 5; int[] arr = {1, 2, 3, 4, 5}; int d = 3; rotateArray(arr, n, d); System.out.print("Output: "); for (int i = 0; i void rotateArray(int arr[], int n, int d) { int temp[d]; for (int i = 0; i = d; i--) { arr[i] = arr[i - d]; } for (int i = 0; i void rotateArray(int arr[], int n, int d) { int temp[d]; for (int i = 0; i = d; i--) { arr[i] = arr[i - d]; } for (int i = 0; i < d; i++) { arr[i] = temp[i]; } } int main() { int n = 5; int arr[] = {1, 2, 3, 4, 5}; int d = 3; rotateArray(arr, n, d); std::cout = d; i--) { arr[i] = arr[i - d]; } for (int i = 0; i < d; i++) { arr[i] = temp[i]; } } public static void main(String[] args) { int n = 5; int[] arr = {1, 2, 3, 4, 5}; int d = 3; rotateArray(arr, n, d); System.out.print("Output: "); for (int i = 0; i < n; i++) { System.out.print(arr[i] + " "); } } } def rotate_array(arr, n, d): temp = arr[-d:] for i in range(n - 1, d - 1, -1): arr[i] = arr[i - d] arr[:d] = temp n = 5 arr = [1, 2, 3, 4, 5] d = 3 rotate_array(arr, n, d) print("Output:", *arr) 6. Substring Search (Asked in Accenture On Campus 10 Aug 2022, Slot 7)Problem Description:The function accepts two strings ‘str1’ and ‘str2’ as its argument. The function needs to return the index of the first occurrence of substring ‘str2’ in string ‘str1’ or -1 if the substring is not found.Example: Input Output str1: “Hello, World!” str2: “World” 7 Solution: C C++ Java Python #include #include int substringSearch(char str1[], char str2[]) { char *ptr = strstr(str1, str2); if (ptr != NULL) { return ptr - str1; } else { return -1; } } int main() { char str1[] = "Hello, World!"; char str2[] = "World"; printf("%dn", substringSearch(str1, str2)); return 0; } #include #include int substringSearch(const char* str1, const char* str2) { const char* ptr = std::strstr(str1, str2); if (ptr != nullptr) { return ptr - str1; } else { return -1; } } int main() { const char* str1 = "Hello, World!"; const char* str2 = "World"; std::cout << substringSearch(str1, str2) << std::endl; return 0; } public class SubstringSearch { public static int substringSearch(String str1, String str2) { int index = str1.indexOf(str2); return index != -1 ? index : -1; } public static void main(String[] args) { String str1 = "Hello, World!"; String str2 = "World"; System.out.println(substringSearch(str1, str2)); } } def substring_search(str1, str2): index = str1.find(str2) return index if index != -1 else -1 str1 = "Hello, World!" str2 = "World" print(substring_search(str1, str2)) #include #include int substringSearch(char str1[], char str2[]) { char *ptr = strstr(str1, str2); if (ptr != NULL) { return ptr - str1; } else { return -1; } } int main() { char str1[] = "Hello, World!"; char str2[] = "World"; printf("%dn", substringSearch(str1, str2)); return 0; } #include #include int substringSearch(const char* str1, const char* str2) { const char* ptr = std::strstr(str1, str2); if (ptr != nullptr) { return ptr - str1; } else { return -1; } } int main() { const char* str1 = "Hello, World!"; const char* str2 = "World"; std::cout << substringSearch(str1, str2) << std::endl; return 0; } public class SubstringSearch { public static int substringSearch(String str1, String str2) { int index = str1.indexOf(str2); return index != -1 ? index : -1; } public static void main(String[] args) { String str1 = "Hello, World!"; String str2 = "World"; System.out.println(substringSearch(str1, str2)); } } def substring_search(str1, str2): index = str1.find(str2) return index if index != -1 else -1 str1 = "Hello, World!" str2 = "World" print(substring_search(str1, str2)) 7. Palindrome Check (Asked in Accenture OnCampus 10 Aug 2022, Slot 8)Problem Description:The function accepts a string ‘str’ as its argument. The function needs to determine whether the string is a palindrome or not. A palindrome is a word or phrase that reads the same backward as forward. Example: Input Output str: “madam” 1 Solution: C C++ Java Python #include #include int isPalindrome(char *str) { int length = strlen(str); for (int i = 0; i #include using namespace std; bool isPalindrome(string str) { int length = str.length(); for (int i = 0; i < length / 2; i++) { if (str[i] != str[length - i - 1]) { return false; // Not a palindrome } } return true; // Palindrome } int main() { string str = "madam"; if (isPalindrome(str)) { cout << "1" << endl; } else { cout << "0" << endl; } return 0; } public class PalindromeCheck { static boolean isPalindrome(String str) { int length = str.length(); for (int i = 0; i < length / 2; i++) { if (str.charAt(i) != str.charAt(length - i - 1)) { return false; // Not a palindrome } } return true; // Palindrome } public static void main(String[] args) { String str = "madam"; if (isPalindrome(str)) { System.out.println("1"); } else { System.out.println("0"); } } } def isPalindrome(s): return s == s[::-1] str_input = "madam" if isPalindrome(str_input): print("1") else: print("0") #include #include int isPalindrome(char *str) { int length = strlen(str); for (int i = 0; i #include using namespace std; bool isPalindrome(string str) { int length = str.length(); for (int i = 0; i < length / 2; i++) { if (str[i] != str[length - i - 1]) { return false; // Not a palindrome } } return true; // Palindrome } int main() { string str = "madam"; if (isPalindrome(str)) { cout << "1" << endl; } else { cout << "0" << endl; } return 0; } public class PalindromeCheck { static boolean isPalindrome(String str) { int length = str.length(); for (int i = 0; i < length / 2; i++) { if (str.charAt(i) != str.charAt(length - i - 1)) { return false; // Not a palindrome } } return true; // Palindrome } public static void main(String[] args) { String str = "madam"; if (isPalindrome(str)) { System.out.println("1"); } else { System.out.println("0"); } } } def isPalindrome(s): return s == s[::-1] str_input = "madam" if isPalindrome(str_input): print("1") else: print("0") 8. Reverse Words (Asked in Accenture OnCampus 10 Aug 2022, Slot 9)Problem Description:The function accepts a string ‘str’ as its argument. The function needs to reverse the order of the words in the string.Example: Input Output str: “Hello, World!” !dlroW ,olleH Solution: C C++ Java Python #include #include void reverseWords(char* str) { char* word = strtok(str, " "); if(word != NULL) { reverseWords(word + strlen(word) + 1); printf("%s ", word); } } int main() { char str[] = "Hello, World!"; reverseWords(str); return 0; } #include #include #include void reverseWords(std::string& str) { std::istringstream iss(str); std::string word; if (iss >> word) { reverseWords(iss.str()); std::cout = 0; i--) { System.out.print(words[i] + " "); } } public static void main(String[] args) { String str = "Hello, World!"; reverseWords(str); } } def reverse_words(s): words = s.split() reversed_words = " ".join(reversed(words)) print(reversed_words) str_input = "Hello, World!" reverse_words(str_input) #include #include void reverseWords(char* str) { char* word = strtok(str, " "); if(word != NULL) { reverseWords(word + strlen(word) + 1); printf("%s ", word); } } int main() { char str[] = "Hello, World!"; reverseWords(str); return 0; } #include #include #include void reverseWords(std::string& str) { std::istringstream iss(str); std::string word; if (iss >> word) { reverseWords(iss.str()); std::cout = 0; i--) { System.out.print(words[i] + " "); } } public static void main(String[] args) { String str = "Hello, World!"; reverseWords(str); } } def reverse_words(s): words = s.split() reversed_words = " ".join(reversed(words)) print(reversed_words) str_input = "Hello, World!" reverse_words(str_input) 9. Find Two Numbers with Sum NProblem Description:Given an array of integers and an integer sum, find a pair of numbers (a, b) in the array where a + b = sum.Example: Input Output An array of integers An integer sum An array of two integers representing the pair (a, b) or -1 if no such pair exists Explanation:Given an array of integers, such as [5, 2, 4, 1, 3], and an integer sum, such as 9, the algorithm should determine that the pair (a, b) = (2, 7) or (4, 5) satisfies the condition a + b = sum. If no such pair exists, the algorithm should return -1. Solution: C C++ Java Python #include void findPair(int arr[], int size, int sum) { for (int i = 0; i < size - 1; i++) { for (int j = i + 1; j void findPair(int arr[], int size, int sum) { for (int i = 0; i < size - 1; i++) { for (int j = i + 1; j < size; j++) { if (arr[i] + arr[j] == sum) { std::cout << "Pair found: " << arr[i] << ", " << arr[j] << std::endl; return; } } } std::cout << "Pair not found" << std::endl; } int main() { int arr[] = {1, 4, 7, 8, 3, 9}; int sum = 10; int size = sizeof(arr) / sizeof(arr[0]); findPair(arr, size, sum); return 0; } public class FindPair { static void findPair(int arr[], int sum) { for (int i = 0; i < arr.length - 1; i++) { for (int j = i + 1; j < arr.length; j++) { if (arr[i] + arr[j] == sum) { System.out.println("Pair found: " + arr[i] + ", " + arr[j]); return; } } } System.out.println("Pair not found"); } public static void main(String[] args) { int arr[] = {1, 4, 7, 8, 3, 9}; int sum = 10; findPair(arr, sum); } } def find_pair(arr, sum): for i in range(len(arr) - 1): for j in range(i + 1, len(arr)): if arr[i] + arr[j] == sum: print(f"Pair found: {arr[i]}, {arr[j]}") return print("Pair not found") arr = [1, 4, 7, 8, 3, 9] sum_val = 10 find_pair(arr, sum_val) #include void findPair(int arr[], int size, int sum) { for (int i = 0; i < size - 1; i++) { for (int j = i + 1; j void findPair(int arr[], int size, int sum) { for (int i = 0; i < size - 1; i++) { for (int j = i + 1; j < size; j++) { if (arr[i] + arr[j] == sum) { std::cout << "Pair found: " << arr[i] << ", " << arr[j] << std::endl; return; } } } std::cout << "Pair not found" << std::endl; } int main() { int arr[] = {1, 4, 7, 8, 3, 9}; int sum = 10; int size = sizeof(arr) / sizeof(arr[0]); findPair(arr, size, sum); return 0; } public class FindPair { static void findPair(int arr[], int sum) { for (int i = 0; i < arr.length - 1; i++) { for (int j = i + 1; j < arr.length; j++) { if (arr[i] + arr[j] == sum) { System.out.println("Pair found: " + arr[i] + ", " + arr[j]); return; } } } System.out.println("Pair not found"); } public static void main(String[] args) { int arr[] = {1, 4, 7, 8, 3, 9}; int sum = 10; findPair(arr, sum); } } def find_pair(arr, sum): for i in range(len(arr) - 1): for j in range(i + 1, len(arr)): if arr[i] + arr[j] == sum: print(f"Pair found: {arr[i]}, {arr[j]}") return print("Pair not found") arr = [1, 4, 7, 8, 3, 9] sum_val = 10 find_pair(arr, sum_val) 10. Maximum Subarray SumProblem Description:Given an array of integers, find the maximum subarray sum. A subarray is a contiguous subsequence of the array.Eample: Input Output An array of integers An integer representing the maximum subarray sum Explanation:Given an array of integers, such as [-2, 1, -3, 4, -1, 2, 1, -5, 4], the algorithm should determine that the maximum subarray sum is 6 ([4, -1, 2, 1]). Solution: C C++ Java Python #include int maxSubArraySum(int arr[], int size) { int max_so_far = arr[0]; int max_ending_here = arr[0]; for (int i = 1; i arr[i]) ? max_ending_here + arr[i] : arr[i]; max_so_far = (max_so_far > max_ending_here) ? max_so_far : max_ending_here; } return max_so_far; } int main() { int arr[] = {-2, -3, 4, -1, -2, 1, 5, -3}; int size = sizeof(arr) / sizeof(arr[0]); printf("Maximum Subarray Sum is %dn", maxSubArraySum(arr, size)); return 0; } #include using namespace std; int maxSubArraySum(int arr[], int size) { int max_so_far = arr[0]; int max_ending_here = arr[0]; for (int i = 1; i arr[i]) ? max_ending_here + arr[i] : arr[i]; max_so_far = (max_so_far > max_ending_here) ? max_so_far : max_ending_here; } return max_so_far; } int main() { int arr[] = {-2, -3, 4, -1, -2, 1, 5, -3}; int size = sizeof(arr) / sizeof(arr[0]); cout << "Maximum Subarray Sum is " << maxSubArraySum(arr, size) << endl; return 0; } public class MaxSubArraySum { public static int maxSubArraySum(int[] arr) { int max_so_far = arr[0]; int max_ending_here = arr[0]; for (int i = 1; i < arr.length; i++) { max_ending_here = Math.max(max_ending_here + arr[i], arr[i]); max_so_far = Math.max(max_so_far, max_ending_here); } return max_so_far; } public static void main(String[] args) { int[] arr = {-2, -3, 4, -1, -2, 1, 5, -3}; System.out.println("Maximum Subarray Sum is " + maxSubArraySum(arr)); } } def max_sub_array_sum(arr): max_so_far = max_ending_here = arr[0] for num in arr[1:]: max_ending_here = max(num, max_ending_here + num) max_so_far = max(max_so_far, max_ending_here) return max_so_far arr = [-2, -3, 4, -1, -2, 1, 5, -3] print("Maximum Subarray Sum is", max_sub_array_sum(arr)) #include int maxSubArraySum(int arr[], int size) { int max_so_far = arr[0]; int max_ending_here = arr[0]; for (int i = 1; i arr[i]) ? max_ending_here + arr[i] : arr[i]; max_so_far = (max_so_far > max_ending_here) ? max_so_far : max_ending_here; } return max_so_far; } int main() { int arr[] = {-2, -3, 4, -1, -2, 1, 5, -3}; int size = sizeof(arr) / sizeof(arr[0]); printf("Maximum Subarray Sum is %dn", maxSubArraySum(arr, size)); return 0; } #include using namespace std; int maxSubArraySum(int arr[], int size) { int max_so_far = arr[0]; int max_ending_here = arr[0]; for (int i = 1; i arr[i]) ? max_ending_here + arr[i] : arr[i]; max_so_far = (max_so_far > max_ending_here) ? max_so_far : max_ending_here; } return max_so_far; } int main() { int arr[] = {-2, -3, 4, -1, -2, 1, 5, -3}; int size = sizeof(arr) / sizeof(arr[0]); cout << "Maximum Subarray Sum is " << maxSubArraySum(arr, size) << endl; return 0; } public class MaxSubArraySum { public static int maxSubArraySum(int[] arr) { int max_so_far = arr[0]; int max_ending_here = arr[0]; for (int i = 1; i < arr.length; i++) { max_ending_here = Math.max(max_ending_here + arr[i], arr[i]); max_so_far = Math.max(max_so_far, max_ending_here); } return max_so_far; } public static void main(String[] args) { int[] arr = {-2, -3, 4, -1, -2, 1, 5, -3}; System.out.println("Maximum Subarray Sum is " + maxSubArraySum(arr)); } } def max_sub_array_sum(arr): max_so_far = max_ending_here = arr[0] for num in arr[1:]: max_ending_here = max(num, max_ending_here + num) max_so_far = max(max_so_far, max_ending_here) return max_so_far arr = [-2, -3, 4, -1, -2, 1, 5, -3] print("Maximum Subarray Sum is", max_sub_array_sum(arr)) 11. Character ReplacementProblem Description:Given a string str, a character ch1, and a character ch2, replace all occurrences of ch1 in str with ch2 and vice versa. Input Output A string str A character ch1 A character ch2 The modified string str where all occurrences of ch1 are replaced with ch2 and vice versa Example: Input Output str = “apples”, ch1 = ‘a’, ch2 = ‘p’ str = “paales” Solution: C C++ Java Python #include #include void replaceCharacters(char str[], char ch1, char ch2) { int len = strlen(str); for (int i = 0; i #include void replaceCharacters(std::string &str, char ch1, char ch2) { for (char &c : str) { if (c == ch1) c = ch2; else if (c == ch2) c = ch1; } } int main() { std::string str = "apples"; char ch1 = 'a'; char ch2 = 'p'; replaceCharacters(str, ch1, ch2); std::cout << "Modified string: " << str << std::endl; return 0; } public class CharacterReplacement { public static String replaceCharacters(String str, char ch1, char ch2) { char[] charArray = str.toCharArray(); for (int i = 0; i < charArray.length; ++i) { if (charArray[i] == ch1) charArray[i] = ch2; else if (charArray[i] == ch2) charArray[i] = ch1; } return new String(charArray); } public static void main(String[] args) { String str = "apples"; char ch1 = 'a'; char ch2 = 'p'; String modifiedStr = replaceCharacters(str, ch1, ch2); System.out.println("Modified string: " + modifiedStr); } } def replace_characters(input_str, ch1, ch2): return input_str.replace(ch1, 'temp').replace(ch2, ch1).replace('temp', ch2) str_value = "apples" ch1_value = 'a' ch2_value = 'p' modified_str = replace_characters(str_value, ch1_value, ch2_value) print("Modified string:", modified_str) #include #include void replaceCharacters(char str[], char ch1, char ch2) { int len = strlen(str); for (int i = 0; i #include void replaceCharacters(std::string &str, char ch1, char ch2) { for (char &c : str) { if (c == ch1) c = ch2; else if (c == ch2) c = ch1; } } int main() { std::string str = "apples"; char ch1 = 'a'; char ch2 = 'p'; replaceCharacters(str, ch1, ch2); std::cout << "Modified string: " << str << std::endl; return 0; } public class CharacterReplacement { public static String replaceCharacters(String str, char ch1, char ch2) { char[] charArray = str.toCharArray(); for (int i = 0; i < charArray.length; ++i) { if (charArray[i] == ch1) charArray[i] = ch2; else if (charArray[i] == ch2) charArray[i] = ch1; } return new String(charArray); } public static void main(String[] args) { String str = "apples"; char ch1 = 'a'; char ch2 = 'p'; String modifiedStr = replaceCharacters(str, ch1, ch2); System.out.println("Modified string: " + modifiedStr); } } def replace_characters(input_str, ch1, ch2): return input_str.replace(ch1, 'temp').replace(ch2, ch1).replace('temp', ch2) str_value = "apples" ch1_value = 'a' ch2_value = 'p' modified_str = replace_characters(str_value, ch1_value, ch2_value) print("Modified string:", modified_str) 12. Find the Minimum Value and Its Index in the ArrayProblem Description:Given an integer array, find the minimum value and its index in the array. Input Output An integer array The minimum value and its index, separated by a newline character Example: Input Output [5, 2, 4, 1, 3] 1 3 Solution: C C++ Java Python #include int main() { int arr[] = {5, 2, 4, 1, 3}; int n = sizeof(arr) / sizeof(arr[0]); int min = arr[0], index = 0; for (int i = 1; i using namespace std; int main() { int arr[] = {5, 2, 4, 1, 3}; int n = sizeof(arr) / sizeof(arr[0]); int min = arr[0], index = 0; for (int i = 1; i < n; ++i) { if (arr[i] < min) { min = arr[i]; index = i; } } cout << min << " " << index << endl; return 0; } public class MinValueAndIndex { public static void main(String[] args) { int[] arr = {5, 2, 4, 1, 3}; int min = arr[0], index = 0; for (int i = 1; i < arr.length; ++i) { if (arr[i] < min) { min = arr[i]; index = i; } } System.out.println(min + " " + index); } } def find_min_and_index(arr): min_value = arr[0] min_index = 0 for i in range(1, len(arr)): if arr[i] int main() { int arr[] = {5, 2, 4, 1, 3}; int n = sizeof(arr) / sizeof(arr[0]); int min = arr[0], index = 0; for (int i = 1; i using namespace std; int main() { int arr[] = {5, 2, 4, 1, 3}; int n = sizeof(arr) / sizeof(arr[0]); int min = arr[0], index = 0; for (int i = 1; i < n; ++i) { if (arr[i] < min) { min = arr[i]; index = i; } } cout << min << " " << index << endl; return 0; } public class MinValueAndIndex { public static void main(String[] args) { int[] arr = {5, 2, 4, 1, 3}; int min = arr[0], index = 0; for (int i = 1; i < arr.length; ++i) { if (arr[i] < min) { min = arr[i]; index = i; } } System.out.println(min + " " + index); } } def find_min_and_index(arr): min_value = arr[0] min_index = 0 for i in range(1, len(arr)): if arr[i] < min_value: min_value = arr[i] min_index = i return f"{min_value} {min_index}” 13. Find the Average of All Positive Numbers in an ArrayProblem Description:Given an array of integers, find the average of all positive numbers in the array. Input Output An integer array The average of all positive numbers, or -1 if there are no positive numbers Example: Input Output [5, 2, -4, 1, 3] 3 Solution: C C++ Java Python #include float averagePositive(int arr[], int size) { int sum = 0, count = 0; for (int i = 0; i 0) { sum += arr[i]; count++; } } return (count > 0) ? (float)sum / count : -1; } int main() { int arr[] = {5, 2, -4, 1, 3}; int size = sizeof(arr) / sizeof(arr[0]); float result = averagePositive(arr, size); printf("Output: %.2fn", result); return 0; } #include float averagePositive(int arr[], int size) { int sum = 0, count = 0; for (int i = 0; i 0) { sum += arr[i]; count++; } } return (count > 0) ? (float)sum / count : -1; } int main() { int arr[] = {5, 2, -4, 1, 3}; int size = sizeof(arr) / sizeof(arr[0]); float result = averagePositive(arr, size); std::cout 0) { sum += num; count++; } } return (count > 0) ? (float) sum / count : -1; } public static void main(String[] args) { int[] arr = {5, 2, -4, 1, 3}; float result = averagePositive(arr); System.out.println("Output: " + result); } } def average_positive(arr): total = count = 0 for num in arr: if num > 0: total += num count += 1 return total / count if count > 0 else -1 arr = [5, 2, -4, 1, 3] result = average_positive(arr) print("Output:", result) #include float averagePositive(int arr[], int size) { int sum = 0, count = 0; for (int i = 0; i 0) { sum += arr[i]; count++; } } return (count > 0) ? (float)sum / count : -1; } int main() { int arr[] = {5, 2, -4, 1, 3}; int size = sizeof(arr) / sizeof(arr[0]); float result = averagePositive(arr, size); printf("Output: %.2fn", result); return 0; } #include float averagePositive(int arr[], int size) { int sum = 0, count = 0; for (int i = 0; i 0) { sum += arr[i]; count++; } } return (count > 0) ? (float)sum / count : -1; } int main() { int arr[] = {5, 2, -4, 1, 3}; int size = sizeof(arr) / sizeof(arr[0]); float result = averagePositive(arr, size); std::cout 0) { sum += num; count++; } } return (count > 0) ? (float) sum / count : -1; } public static void main(String[] args) { int[] arr = {5, 2, -4, 1, 3}; float result = averagePositive(arr); System.out.println("Output: " + result); } } def average_positive(arr): total = count = 0 for num in arr: if num > 0: total += num count += 1 return total / count if count > 0 else -1 arr = [5, 2, -4, 1, 3] result = average_positive(arr) print("Output:", result) 14. Count the Occurrences of a Given Element in an ArrayProblem Description:Given an integer array and an integer element, count the number of occurrences of the element in the array. Input Output An integer array An integer element The number of occurrences of the element Example: Input Output [5, 2, 4, 1, 2], 2 2 Solution: C C++ Java Python #include int count_occurrences(int arr[], int n, int element) { int count = 0; for (int i = 0; i using namespace std; int count_occurrences(int arr[], int n, int element) { int count = 0; for (int i = 0; i < n; i++) { if (arr[i] == element) { count++; } } return count; } int main() { int arr[] = {5, 2, 4, 1, 2}; int n = sizeof(arr) / sizeof(arr[0]); int element = 2; int result = count_occurrences(arr, n, element); cout << result << endl; return 0; } public class OccurrenceCounter { public static int countOccurrences(int[] arr, int element) { int count = 0; for (int num : arr) { if (num == element) { count++; } } return count; } public static void main(String[] args) { int[] arr = {5, 2, 4, 1, 2}; int element = 2; int result = countOccurrences(arr, element); System.out.println(result); } } def count_occurrences(arr, element): count = 0 for num in arr: if num == element: count += 1 return count arr = [5, 2, 4, 1, 2] element = 2 result = count_occurrences(arr, element) print(result) #include int count_occurrences(int arr[], int n, int element) { int count = 0; for (int i = 0; i using namespace std; int count_occurrences(int arr[], int n, int element) { int count = 0; for (int i = 0; i < n; i++) { if (arr[i] == element) { count++; } } return count; } int main() { int arr[] = {5, 2, 4, 1, 2}; int n = sizeof(arr) / sizeof(arr[0]); int element = 2; int result = count_occurrences(arr, n, element); cout << result << endl; return 0; } public class OccurrenceCounter { public static int countOccurrences(int[] arr, int element) { int count = 0; for (int num : arr) { if (num == element) { count++; } } return count; } public static void main(String[] args) { int[] arr = {5, 2, 4, 1, 2}; int element = 2; int result = countOccurrences(arr, element); System.out.println(result); } } def count_occurrences(arr, element): count = 0 for num in arr: if num == element: count += 1 return count arr = [5, 2, 4, 1, 2] element = 2 result = count_occurrences(arr, element) print(result) Also Read: 20 IBM Coding Questions with Answers 15. Check if an Array Contains a Given ElementProblem Description:Given an integer array and an integer element, check if the array contains the element. Input Output An integer array An integer element True if the element is found, False otherwise Example: Input Output [5, 2, 4, 1, 3], 2 True Solution: C C++ Java Python #include int containsElement(int arr[], int size, int element) { for (int i = 0; i using namespace std; bool containsElement(int arr[], int size, int element) { for (int i = 0; i < size; i++) { if (arr[i] == element) { return true; } } return false; } int main() { int arr[] = {5, 2, 4, 1, 3}; int size = sizeof(arr) / sizeof(arr[0]); int element = 2; bool result = containsElement(arr, size, element); cout << (result ? "True" : "False") << endl; return 0; } public class ArrayContainsElement { public static boolean containsElement(int[] arr, int element) { for (int num : arr) { if (num == element) { return true; } } return false; } public static void main(String[] args) { int[] arr = {5, 2, 4, 1, 3}; int element = 2; boolean result = containsElement(arr, element); System.out.println(result ? "True" : "False"); } } def contains_element(arr, element): return element in arr arr = [5, 2, 4, 1, 3] element = 2 result = contains_element(arr, element) print("True" if result else "False") #include int containsElement(int arr[], int size, int element) { for (int i = 0; i using namespace std; bool containsElement(int arr[], int size, int element) { for (int i = 0; i < size; i++) { if (arr[i] == element) { return true; } } return false; } int main() { int arr[] = {5, 2, 4, 1, 3}; int size = sizeof(arr) / sizeof(arr[0]); int element = 2; bool result = containsElement(arr, size, element); cout << (result ? "True" : "False") << endl; return 0; } public class ArrayContainsElement { public static boolean containsElement(int[] arr, int element) { for (int num : arr) { if (num == element) { return true; } } return false; } public static void main(String[] args) { int[] arr = {5, 2, 4, 1, 3}; int element = 2; boolean result = containsElement(arr, element); System.out.println(result ? "True" : "False"); } } def contains_element(arr, element): return element in arr arr = [5, 2, 4, 1, 3] element = 2 result = contains_element(arr, element) print("True" if result else "False") 16. Problem Statement: Calculate Prime SumImplement the function:Int CalculatePrimeSum(int m, int n);Calculate and return the sum of prime numbers between 'm' and 'n' (inclusive).Note: 0 < m <= n Example: Input Output m : 10 n : 50 158 Solution: C C++ Java Python #include #include bool is_prime(int num) { if (num <= 1) { return false; } for (int i = 2; i * i <= num; i++) { if (num % i == 0) { return false; } } return true; } int calculate_prime_sum(int m, int n) { int sum = 0; for (int i = m; i #include using namespace std; bool is_prime(int num) { if (num <= 1) { return false; } for (int i = 2; i * i <= num; i++) { if (num % i == 0) { return false; } } return true; } int calculate_prime_sum(int m, int n) { int sum = 0; for (int i = m; i <= n; i++) { if (is_prime(i)) { sum += i; } } return sum; } int main() { int m = 10, n = 50; int result = calculate_prime_sum(m, n); cout << result << endl; return 0; } public class CalculatePrimeSum { static boolean isPrime(int num) { if (num <= 1) { return false; } for (int i = 2; i * i <= num; i++) { if (num % i == 0) { return false; } } return true; } static int calculatePrimeSum(int m, int n) { int sum = 0; for (int i = m; i <= n; i++) { if (isPrime(i)) { sum += i; } } return sum; } public static void main(String[] args) { int m = 10, n = 50; int result = calculatePrimeSum(m, n); System.out.println(result); } } def is_prime(num): if num #include bool is_prime(int num) { if (num <= 1) { return false; } for (int i = 2; i * i <= num; i++) { if (num % i == 0) { return false; } } return true; } int calculate_prime_sum(int m, int n) { int sum = 0; for (int i = m; i #include using namespace std; bool is_prime(int num) { if (num <= 1) { return false; } for (int i = 2; i * i <= num; i++) { if (num % i == 0) { return false; } } return true; } int calculate_prime_sum(int m, int n) { int sum = 0; for (int i = m; i <= n; i++) { if (is_prime(i)) { sum += i; } } return sum; } int main() { int m = 10, n = 50; int result = calculate_prime_sum(m, n); cout << result << endl; return 0; } public class CalculatePrimeSum { static boolean isPrime(int num) { if (num <= 1) { return false; } for (int i = 2; i * i <= num; i++) { if (num % i == 0) { return false; } } return true; } static int calculatePrimeSum(int m, int n) { int sum = 0; for (int i = m; i <= n; i++) { if (isPrime(i)) { sum += i; } } return sum; } public static void main(String[] args) { int m = 10, n = 50; int result = calculatePrimeSum(m, n); System.out.println(result); } } def is_prime(num): if num <= 1: return False for i in range(2, int(num**0.5) + 1): if num % i == 0: return False return True def calculate_prime_sum(m, n): prime_sum = sum(i for i in range(m, n + 1) if is_prime(i)) return prime_sum m, n = 10, 50 result = calculate_prime_sum(m, n) print(result) 17. Problem Statement: Digit Sum DifferenceImplement the function:Int DigitSumDifference(int m, int n);Calculate and return the absolute difference between the sum of digits of numbers divisible by 4 and the sum of digits of numbers divisible by 7, in the range from 'm' to 'n' (inclusive).Note: 0 < m <= nExample: Input Output m : 50 n : 120 2 Solution: C C++ Java Python #include int digit_sum_difference(int m, int n) { int sum_divisible_by_4 = 0; int sum_divisible_by_7 = 0; for (int i = m; i 0) { sum_divisible_by_4 += num % 10; num /= 10; } } else if (i % 7 == 0) { int num = i; while (num > 0) { sum_divisible_by_7 += num % 10; num /= 10; } } } return abs(sum_divisible_by_4 - sum_divisible_by_7); } int main() { int m = 50; int n = 120; int result = digit_sum_difference(m, n); printf("%dn", result); return 0; } #include using namespace std; int digit_sum_difference(int m, int n) { int sum_divisible_by_4 = 0; int sum_divisible_by_7 = 0; for (int i = m; i 0) { sum_divisible_by_4 += num % 10; num /= 10; } } else if (i % 7 == 0) { int num = i; while (num > 0) { sum_divisible_by_7 += num % 10; num /= 10; } } } return abs(sum_divisible_by_4 - sum_divisible_by_7); } int main() { int m = 50; int n = 120; int result = digit_sum_difference(m, n); cout 0) { sumDivisibleBy4 += num % 10; num /= 10; } } else if (i % 7 == 0) { int num = i; while (num > 0) { sumDivisibleBy7 += num % 10; num /= 10; } } } return Math.abs(sumDivisibleBy4 - sumDivisibleBy7); } public static void main(String[] args) { int m = 50; int n = 120; int result = digitSumDifference(m, n); System.out.println(result); } } def digit_sum_difference(m, n): sum_divisible_by_4 = 0 sum_divisible_by_7 = 0 for i in range(m, n + 1): if i % 4 == 0: num = i while num > 0: sum_divisible_by_4 += num % 10 num //= 10 elif i % 7 == 0: num = i while num > 0: sum_divisible_by_7 += num % 10 num //= 10 return abs(sum_divisible_by_4 - sum_divisible_by_7) m = 50 n = 120 result = digit_sum_difference(m, n) print(result) #include int digit_sum_difference(int m, int n) { int sum_divisible_by_4 = 0; int sum_divisible_by_7 = 0; for (int i = m; i 0) { sum_divisible_by_4 += num % 10; num /= 10; } } else if (i % 7 == 0) { int num = i; while (num > 0) { sum_divisible_by_7 += num % 10; num /= 10; } } } return abs(sum_divisible_by_4 - sum_divisible_by_7); } int main() { int m = 50; int n = 120; int result = digit_sum_difference(m, n); printf("%dn", result); return 0; } #include using namespace std; int digit_sum_difference(int m, int n) { int sum_divisible_by_4 = 0; int sum_divisible_by_7 = 0; for (int i = m; i 0) { sum_divisible_by_4 += num % 10; num /= 10; } } else if (i % 7 == 0) { int num = i; while (num > 0) { sum_divisible_by_7 += num % 10; num /= 10; } } } return abs(sum_divisible_by_4 - sum_divisible_by_7); } int main() { int m = 50; int n = 120; int result = digit_sum_difference(m, n); cout 0) { sumDivisibleBy4 += num % 10; num /= 10; } } else if (i % 7 == 0) { int num = i; while (num > 0) { sumDivisibleBy7 += num % 10; num /= 10; } } } return Math.abs(sumDivisibleBy4 - sumDivisibleBy7); } public static void main(String[] args) { int m = 50; int n = 120; int result = digitSumDifference(m, n); System.out.println(result); } } def digit_sum_difference(m, n): sum_divisible_by_4 = 0 sum_divisible_by_7 = 0 for i in range(m, n + 1): if i % 4 == 0: num = i while num > 0: sum_divisible_by_4 += num % 10 num //= 10 elif i % 7 == 0: num = i while num > 0: sum_divisible_by_7 += num % 10 num //= 10 return abs(sum_divisible_by_4 - sum_divisible_by_7) m = 50 n = 120 result = digit_sum_difference(m, n) print(result) 18. Problem Statement: Fibonacci SumImplement the function:Int FibonacciSum(int m, int n);Calculate and return the sum of Fibonacci numbers in the range from 'm' to 'n' (inclusive).Note: 0 < m <= nExample: Input Output m : 5 n : 20 52 Solution: C C++ Java Python #include int fibonacciSum(int m, int n) { int a = 0, b = 1, temp, sum = 0; while (b = m) { sum += b; } temp = a + b; a = b; b = temp; } return sum; } int main() { int m = 5, n = 20; int result = fibonacciSum(m, n); printf("%dn", result); return 0; } #include using namespace std; int fibonacciSum(int m, int n) { int a = 0, b = 1, temp, sum = 0; while (b = m) { sum += b; } temp = a + b; a = b; b = temp; } return sum; } int main() { int m = 5, n = 20; int result = fibonacciSum(m, n); cout = m) { sum += b; } temp = a + b; a = b; b = temp; } return sum; } public static void main(String[] args) { int m = 5, n = 20; int result = fibonacciSum(m, n); System.out.println(result); } } def fibonacci_sum(m, n): a, b, total = 0, 1, 0 while b = m: total += b a, b = b, a + b return total m, n = 5, 20 result = fibonacci_sum(m, n) print(result) #include int fibonacciSum(int m, int n) { int a = 0, b = 1, temp, sum = 0; while (b = m) { sum += b; } temp = a + b; a = b; b = temp; } return sum; } int main() { int m = 5, n = 20; int result = fibonacciSum(m, n); printf("%dn", result); return 0; } #include using namespace std; int fibonacciSum(int m, int n) { int a = 0, b = 1, temp, sum = 0; while (b = m) { sum += b; } temp = a + b; a = b; b = temp; } return sum; } int main() { int m = 5, n = 20; int result = fibonacciSum(m, n); cout = m) { sum += b; } temp = a + b; a = b; b = temp; } return sum; } public static void main(String[] args) { int m = 5, n = 20; int result = fibonacciSum(m, n); System.out.println(result); } } def fibonacci_sum(m, n): a, b, total = 0, 1, 0 while b = m: total += b a, b = b, a + b return total m, n = 5, 20 result = fibonacci_sum(m, n) print(result) 19. Problem Statement: Reverse and AddImplement the function:Int ReverseAndAdd(int m, int n);Calculate and return the sum of numbers obtained by reversing the digits of each number in the range from 'm' to 'n' (inclusive).Note: 0 < m <= nExample: Input Output m : 21 n : 35 288 Solution: C C++ Java Python #include int reverseAndAdd(int m, int n) { int sum = 0; for (int i = m; i 0) { reversedNum = reversedNum * 10 + temp % 10; temp /= 10; } sum += reversedNum + i; } return sum; } int main() { int m = 21; int n = 35; int result = reverseAndAdd(m, n); printf("%dn", result); return 0; } #include using namespace std; int reverseAndAdd(int m, int n) { int sum = 0; for (int i = m; i 0) { reversedNum = reversedNum * 10 + temp % 10; temp /= 10; } sum += reversedNum + i; } return sum; } int main() { int m = 21; int n = 35; int result = reverseAndAdd(m, n); cout 0) { reversedNum = reversedNum * 10 + temp % 10; temp /= 10; } sum += reversedNum + i; } return sum; } public static void main(String[] args) { int m = 21; int n = 35; int result = reverseAndAdd(m, n); System.out.println(result); } } def reverse_and_add(m, n): total_sum = 0 for i in range(m, n + 1): reversed_num = int(str(i)[::-1]) total_sum += reversed_num + i return total_sum m = 21 n = 35 result = reverse_and_add(m, n) print(result) #include int reverseAndAdd(int m, int n) { int sum = 0; for (int i = m; i 0) { reversedNum = reversedNum * 10 + temp % 10; temp /= 10; } sum += reversedNum + i; } return sum; } int main() { int m = 21; int n = 35; int result = reverseAndAdd(m, n); printf("%dn", result); return 0; } #include using namespace std; int reverseAndAdd(int m, int n) { int sum = 0; for (int i = m; i 0) { reversedNum = reversedNum * 10 + temp % 10; temp /= 10; } sum += reversedNum + i; } return sum; } int main() { int m = 21; int n = 35; int result = reverseAndAdd(m, n); cout 0) { reversedNum = reversedNum * 10 + temp % 10; temp /= 10; } sum += reversedNum + i; } return sum; } public static void main(String[] args) { int m = 21; int n = 35; int result = reverseAndAdd(m, n); System.out.println(result); } } def reverse_and_add(m, n): total_sum = 0 for i in range(m, n + 1): reversed_num = int(str(i)[::-1]) total_sum += reversed_num + i return total_sum m = 21 n = 35 result = reverse_and_add(m, n) print(result) 20. Problem Statement: Square Root DifferenceImplement the function:Int SquareRootDifference(int m, int n);Calculate and return the difference between the sum of square roots of even numbers and the sum of square roots of odd numbers in the range from 'm' to 'n' (inclusive).Note: 0 < m <= nExample: Input Output m : 1 n : 10 2.29416 Solution: C C++ Java Python #include #include double squareRootDifference(int m, int n) { double evenSum = 0, oddSum = 0; for (int i = m; i #include using namespace std; double squareRootDifference(int m, int n) { double evenSum = 0, oddSum = 0; for (int i = m; i <= n; ++i) { double squareRoot = sqrt(i); (i % 2 == 0) ? evenSum += squareRoot : oddSum += squareRoot; } return evenSum - oddSum; } int main() { int m = 1, n = 10; double result = squareRootDifference(m, n); cout << fixed << result << endl; return 0; } public class SquareRootDifference { public static double squareRootDifference(int m, int n) { double evenSum = 0, oddSum = 0; for (int i = m; i <= n; ++i) { double squareRoot = Math.sqrt(i); if (i % 2 == 0) { evenSum += squareRoot; } else { oddSum += squareRoot; } } return evenSum - oddSum; } public static void main(String[] args) { int m = 1, n = 10; double result = squareRootDifference(m, n); System.out.printf("%.5fn", result); } } import math def square_root_difference(m, n): even_sum = sum(math.sqrt(i) for i in range(m, n + 1) if i % 2 == 0) odd_sum = sum(math.sqrt(i) for i in range(m, n + 1) if i % 2 != 0) return even_sum - odd_sum m, n = 1, 10 result = square_root_difference(m, n) print(f"{result:.5f}") #include #include double squareRootDifference(int m, int n) { double evenSum = 0, oddSum = 0; for (int i = m; i #include using namespace std; double squareRootDifference(int m, int n) { double evenSum = 0, oddSum = 0; for (int i = m; i <= n; ++i) { double squareRoot = sqrt(i); (i % 2 == 0) ? evenSum += squareRoot : oddSum += squareRoot; } return evenSum - oddSum; } int main() { int m = 1, n = 10; double result = squareRootDifference(m, n); cout << fixed << result << endl; return 0; } public class SquareRootDifference { public static double squareRootDifference(int m, int n) { double evenSum = 0, oddSum = 0; for (int i = m; i <= n; ++i) { double squareRoot = Math.sqrt(i); if (i % 2 == 0) { evenSum += squareRoot; } else { oddSum += squareRoot; } } return evenSum - oddSum; } public static void main(String[] args) { int m = 1, n = 10; double result = squareRootDifference(m, n); System.out.printf("%.5fn", result); } } import math def square_root_difference(m, n): even_sum = sum(math.sqrt(i) for i in range(m, n + 1) if i % 2 == 0) odd_sum = sum(math.sqrt(i) for i in range(m, n + 1) if i % 2 != 0) return even_sum - odd_sum m, n = 1, 10 result = square_root_difference(m, n) print(f"{result:.5f}") Accenture Coding Questions with solutions (Faq) Does Accenture ask coding questions?Accenture does ask coding questions during the interview process for the specific roles. They ask the questions to assess the candidate’s problem-solving skills, coding proficiency, and ability to write clean and efficient code. Are Accenture coding questions repeated?Accenture does not repeat the coding questions from previous interviews as they keep updating the questions to ensure the problems are relevant with today’s trends and challenging for the candidates. Although similar questions could’ve been included for the Accenture coding round. How many coding questions are asked in an accenture interview?There are 2 questions asked in the Accenture coding round. All Accenture Coding Questions should be solved and the candidate is required to use algorithms, data structures, and design patterns. Conclusion We hope you liked this post and grab free Accenture Coding Questions pdf download and receive valuable information from this. Let us know what you think is the hardest part of an Accenture interview process. 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20 Accenture Coding Questions With Proven Solutions (2025)

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