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Infosys Coding Questions : When I asked aspiring engineering freshers what their dream IT Company in India was? I found that Infosys is the most common name that every engineer wants to join. It is one of the most valuable companies in the world with a market cap of $85.42 billion. And now it has become a goal of every aspiring engineer to get a job. Because it can provide career stability, competitive salary, and much more.
However, Infosys has multiple interview rounds in different domains such as written, aptitude, logical, technical and programming skills for the selection process.
One of the most important parts of the entire selection process is the Infosys coding questions round. And if someone can have good hands-on experience with most-asked coding questions, then the chances of getting selected are higher.
Thatโs why weโve compiled a list of the top 10 Infosys coding questions with answers that you can practice for smart preparation.
No. of Questions For The Coding Round | 3 Questions |
Type Of Test | Online Assessment Test |
Time Duration | 90 minutes |
Negative Marking | No Negative Marking |
Hereโs the list of some commonly asked Infosys coding questions with answers:
Answer:
Sample Input
A[ ] = {21, 22, 23, 24}
B[ ] = {25, 26, 27, 28}
Sample Output
A[ ] = {25, 26, 27, 28}
B[ ] = {21, 22, 23, 24}
#include <iostream> using namespace std; void swapArrays(int* a, int* b, int size) { for (int i = 0; i < size; i++) { int temp = a[i]; a[i] = b[i]; b[i] = temp; } } int main() { int a[] = {21, 22, 23, 24}; int b[] = {25, 26, 27, 28}; int n = sizeof(a)/sizeof(a[0]); swapArrays(a, b, n); cout << "a[] = "; for (int i = 0; i < n; i++) cout << a[i] << ", "; cout << "\nb[] = "; for (int i = 0; i < n; i++) cout << b[i] << ", "; return 0; }
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Output
A[ ] = {25, 26, 27, 28}
B[ ] = {21, 22, 23, 24}
infosys coding questions for specialist programmer
Answer:ย
Sample Input
{5, 67, 2, 88, 9, 76, 52, 4}
#include <iostream> #include <vector> #include <algorithm> using namespace std; // Custom comparator to decide which concatenation forms a larger number struct Comparator { bool operator()(int a, int b) { string ab = to_string(a) + to_string(b); string ba = to_string(b) + to_string(a); return ab > ba; } }; int main() { vector<int> nums = {5, 67, 2, 88, 9, 76, 52, 4}; sort(nums.begin(), nums.end(), Comparator()); for (int num : nums) { cout << num; } cout << endl; return 0; }
import java.util.*; public class LargestNumberSimple { public static void main(String[] args) { // Input array of integers String[] numbers = {"5", "67", "2", "88", "9", "76", "52", "4"}; // Sort the array using a custom Comparator Arrays.sort(numbers, new Comparator<String>() { public int compare(String a, String b) { // Compare two concatenations to decide which should come first return (b + a).compareTo(a + b); } }); // Check for a case where the largest number is 0 if (numbers[0].equals("0")) { System.out.println("0"); return; } // Concatenate the sorted numbers String largestNumber = ""; for (String num : numbers) { largestNumber += num; } // Print the largest number System.out.println(largestNumber); } }
def largestNumber(arr): # Convert integers to strings to enable custom sorting arr = sorted(map(str, arr), key=lambda x: x*10, reverse=True) # Join and return the sorted array, handling leading zeros result = ''.join(arr) return result if result[0] != '0' else '0' # New input array arr = [5, 67, 2, 88, 9, 76, 52, 4] print(largestNumber(arr))
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Output
988766752452
infosys coding questions for specialist programmer
Answer:
Sample Input
[3, 1, 56, 34, 12, 9, 98, 23, 4]
#include <iostream> #include <vector> using namespace std; int main() { vector<int> arr = {3, 1, 56, 34, 12, 9, 98, 23, 4}; int minVal = arr[0]; int maxVal = arr[0]; for(int i = 1; i < arr.size(); i++) { if(arr[i] > maxVal) maxVal = arr[i]; if(arr[i] < minVal) minVal = arr[i]; } cout << "Smallest Number: " << minVal << endl; cout << "Largest Number: " << maxVal << endl; return 0; }
public class Main { public static void main(String[] args) { int[] arr = {3, 1, 56, 34, 12, 9, 98, 23, 4}; int minVal = arr[0]; int maxVal = arr[0]; for(int i = 1; i < arr.length; i++) { if(arr[i] > maxVal) maxVal = arr[i]; if(arr[i] < minVal) minVal = arr[i]; } System.out.println("Smallest Number: " + minVal); System.out.println("Largest Number: " + maxVal); } }
arr = [3, 1, 56, 34, 12, 9, 98, 23, 4] minVal = min(arr) maxVal = max(arr) print(f"Smallest Number: {minVal}") print(f"Largest Number: {maxVal}")
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Output
Smallest Number: 1
Largest Number: 98
infosys coding questions for specialist programmer
Answer:ย
Sample Input
s=โdcdโ
#include <iostream> #include <algorithm> using namespace std; void swap(char* a, char* b) { if (*a == *b) return; *a ^= *b; *b ^= *a; *a ^= *b; } void reverseString(string& s, int start, int end) { while (start < end) { swap(s[start], s[end]); start++; end--; } } bool nextPermutation(string& s) { int n = s.length(); int i = n - 2; while (i >= 0 && s[i] >= s[i + 1]) i--; if (i < 0) return false; int j = n - 1; while (s[j] <= s[i]) j--; swap(s[i], s[j]); reverseString(s, i + 1, n - 1); return true; } int main() { string s = "dcd"; // Sample Input bool val = nextPermutation(s); if (!val) cout << "No next permutation possible" << endl; else cout << "Next permutation: " << s << endl; return 0; }
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Output
Next permutation: ddc
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infosys coding questions for specialist programmer
Answer:
Sample Input:
1 2 3
4 5 6
7 8 9
#include <iostream> #include <vector> using namespace std; void rotateMatrix(vector<vector<int>>& matrix) { int n = matrix.size(); // Transpose the matrix for (int i = 0; i < n; ++i) { for (int j = i + 1; j < n; ++j) { swap(matrix[i][j], matrix[j][i]); } } // Reverse each row for (int i = 0; i < n; ++i) { int left = 0, right = n - 1; while (left < right) { swap(matrix[i][left], matrix[i][right]); left++; right--; } } } int main() { // Input matrix vector<vector<int>> matrix = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}; // Print original matrix cout << "Original Matrix:" << endl; for (const auto& row : matrix) { for (int val : row) { cout << val << " "; } cout << endl; } // Rotate matrix rotateMatrix(matrix); // Print rotated matrix cout << "Rotated Matrix:" << endl; for (const auto& row : matrix) { for (int val : row) { cout << val << " "; } cout << endl; } return 0; }
import java.util.Arrays; public class RotateMatrix { public static void rotateMatrix(int[][] matrix) { int n = matrix.length; // Transpose the matrix for (int i = 0; i < n; ++i) { for (int j = i + 1; j < n; ++j) { int temp = matrix[i][j]; matrix[i][j] = matrix[j][i]; matrix[j][i] = temp; } } // Reverse each row for (int i = 0; i < n; ++i) { int left = 0, right = n - 1; while (left < right) { int temp = matrix[i][left]; matrix[i][left] = matrix[i][right]; matrix[i][right] = temp; left++; right--; } } } public static void main(String[] args) { // Input matrix int[][] matrix = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}; // Print original matrix System.out.println("Original Matrix:"); for (int[] row : matrix) { System.out.println(Arrays.toString(row)); } // Rotate matrix rotateMatrix(matrix); // Print rotated matrix System.out.println("Rotated Matrix:"); for (int[] row : matrix) { System.out.println(Arrays.toString(row)); } } }
def rotate_matrix(matrix): n = len(matrix) # Transpose the matrix for i in range(n): for j in range(i + 1, n): matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j] # Reverse each row for i in range(n): matrix[i] = matrix[i][::-1] # Input matrix matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] # Print original matrix print("Original Matrix:") for row in matrix: print(row) # Rotate matrix rotate_matrix(matrix) # Print rotated matrix print("Rotated Matrix:") for row in matrix: print(row)
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Output:
Rotated Matrix:
7 4 1ย
8 5 2ย
9 6 3ย
ย
Also Read: 20 IBM Coding Questions with Answers (Assessment by Expert)
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infosys coding questions for specialist programmer
Answer:
Sample Input: โA quick movement of the enemy will jeopardize six gunboatsโ
#include <iostream> #include <string> #include <unordered_set> #include <algorithm> std::string findMissingCharacters(const std::string& s) { std::unordered_set<char> alphabets; for (char c = 'a'; c <= 'z'; ++c) { alphabets.insert(c); } for (char c : s) { alphabets.erase(tolower(c)); } std::string missing; for (char c : alphabets) { missing.push_back(c); } std::sort(missing.begin(), missing.end()); return missing; } int main() { std::string str = "A quick movement of the enemy will jeopardize six gunboats"; std::cout << "Missing characters: " << findMissingCharacters(str) << std::endl; return 0; }
import java.util.HashSet; import java.util.Set; public class PangramChecker { public static String findMissingCharacters(String s) { Set<Character> alphabets = new HashSet<>(); for (char c = 'a'; c <= 'z'; c++) { alphabets.add(c); } for (char c : s.toLowerCase().toCharArray()) { alphabets.remove(c); } StringBuilder missing = new StringBuilder(); for (char c : alphabets) { missing.append(c); } return missing.toString(); } public static void main(String[] args) { String str = "A quick movement of the enemy will jeopardize six gunboats"; System.out.println("Missing characters: " + findMissingCharacters(str)); } }
def find_missing_characters_for_pangram(s): alphabets = set('abcdefghijklmnopqrstuvwxyz') for char in s.lower(): alphabets.discard(char) return ''.join(sorted(alphabets)) # Example usage string = "A quick movement of the enemy will jeopardize six gunboats" missing_characters = find_missing_characters_for_pangram(string) print("Missing characters:", missing_characters)
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Output
โMissing characters: flrโ
infosys coding questions for specialist programmer
Answer:ย
Sample Input: โHello, World!โ
#include <iostream> #include <unordered_set> int countUniqueCharacters(const std::string& str) { std::unordered_set<char> uniqueChars; for (char c : str) { uniqueChars.insert(c); } return uniqueChars.size(); } int main() { std::string input = "Hello, World!"; std::cout << "Number of unique characters: " << countUniqueCharacters(input) << std::endl; return 0; }
import java.util.HashSet; import java.util.Set; public class UniqueCharacterCounter { public static int countUniqueCharacters(String str) { Set<Character> uniqueChars = new HashSet<>(); for (char c : str.toCharArray()) { uniqueChars.add(c); } return uniqueChars.size(); } public static void main(String[] args) { String input = "Hello, World!"; System.out.println("Number of unique characters: " + countUniqueCharacters(input)); } }
def count_unique_characters(s): return len(set(s)) # Example usage input_string = "Hello, World!" unique_character_count = count_unique_characters(input_string) print("Number of unique characters:", unique_character_count)
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Output
Number of unique characters is 10.
infosys coding questions for specialist programmer
Answer:
Sample Input:ย
a)
1 2
3 4
b)
4 3
2 1
#include <iostream> #include <vector> std::vector<std::vector<int>> subtractMatrices(const std::vector<std::vector<int>>& A, const std::vector<std::vector<int>>& B) { std::vector<std::vector<int>> result(A.size(), std::vector<int>(A[0].size())); for (size_t i = 0; i < A.size(); ++i) { for (size_t j = 0; j < A[0].size(); ++j) { result[i][j] = A[i][j] - B[i][j]; } } return result; } int main() { std::vector<std::vector<int>> A = {{1, 2}, {3, 4}}; std::vector<std::vector<int>> B = {{4, 3}, {2, 1}}; std::vector<std::vector<int>> result = subtractMatrices(A, B); for (const auto& row : result) { for (int val : row) { std::cout << val << " "; } std::cout << std::endl; } return 0; }
public class MatrixSubtraction { public static int[][] subtractMatrices(int[][] A, int[][] B) { int[][] result = new int[A.length][A[0].length]; for (int i = 0; i < A.length; i++) { for (int j = 0; j < A[0].length; j++) { result[i][j] = A[i][j] - B[i][j]; } } return result; } public static void main(String[] args) { int[][] A = {{1, 2}, {3, 4}}; int[][] B = {{4, 3}, {2, 1}}; int[][] result = subtractMatrices(A, B); for (int[] row : result) { for (int val : row) { System.out.print(val + " "); } System.out.println(); } } }
def subtract_matrices(A, B): return [[A[i][j] - B[i][j] for j in range(len(A[0]))] for i in range(len(A))] # Example usage A = [[1, 2], [3, 4]] B = [[4, 3], [2, 1]] result = subtract_matrices(A, B) for row in result: print(" ".join(map(str, row)))
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Output
-3 -1
ย 1ย 3
infosys coding questions for specialist programmer
Answer:ย
A:
1 2
3 4
B:
5 6
7 8
#include <iostream> #include <vector> std::vector<std::vector<int>> multiplyMatrices(const std::vector<std::vector<int>>& A, const std::vector<std::vector<int>>& B) { size_t rows = A.size(); size_t cols = B[0].size(); size_t common = B.size(); std::vector<std::vector<int>> result(rows, std::vector<int>(cols, 0)); for (size_t i = 0; i < rows; ++i) { for (size_t j = 0; j < cols; ++j) { for (size_t k = 0; k < common; ++k) { result[i][j] += A[i][k] * B[k][j]; } } } return result; } int main() { std::vector<std::vector<int>> A = {{1, 2}, {3, 4}}; std::vector<std::vector<int>> B = {{5, 6}, {7, 8}}; std::vector<std::vector<int>> result = multiplyMatrices(A, B); for (const auto& row : result) { for (int val : row) { std::cout << val << " "; } std::cout << std::endl; } return 0; }
public class MatrixMultiplication { public static int[][] multiplyMatrices(int[][] A, int[][] B) { int rows = A.length; int cols = B[0].length; int common = B.length; int[][] result = new int[rows][cols]; for (int i = 0; i < rows; i++) { for (int j = 0; j < cols; j++) { for (int k = 0; k < common; k++) { result[i][j] += A[i][k] * B[k][j]; } } } return result; } public static void main(String[] args) { int[][] A = {{1, 2}, {3, 4}}; int[][] B = {{5, 6}, {7, 8}}; int[][] result = multiplyMatrices(A, B); for (int[] row : result) { for (int val : row) { System.out.print(val + " "); } System.out.println(); } } }
def multiply_matrices(A, B): rows_A, cols_A = len(A), len(A[0]) rows_B, cols_B = len(B), len(B[0]) result = [[0 for _ in range(cols_B)] for _ in range(rows_A)] for i in range(rows_A): for j in range(cols_B): for k in range(cols_A): result[i][j] += A[i][k] * B[k][j] return result # Example usage A = [[1, 2], [3, 4]] B = [[5, 6], [7, 8]] result = multiply_matrices(A, B) for row in result: print(" ".join(map(str, row)))
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Output
The output of multiplying the given matrices A and B is:
19 22
43 50
infosys coding questions for specialist programmer
Answer:
Sample Input: 29
#include <iostream> #include <string> #include <algorithm> std::string decimalToBinary(int n) { std::string binary = ""; while (n > 0) { binary += std::to_string(n % 2); n /= 2; } std::reverse(binary.begin(), binary.end()); return binary; } int main() { int decimal = 29; std::cout << "Binary of " << decimal << " is " << decimalToBinary(decimal) << std::endl; return 0; }
public class DecimalToBinary { public static String decimalToBinary(int n) { StringBuilder binary = new StringBuilder(); while (n > 0) { binary.insert(0, n % 2); n /= 2; } return binary.toString(); } public static void main(String[] args) { int decimal = 29; System.out.println("Binary of " + decimal + " is " + decimalToBinary(decimal)); } }
def decimal_to_binary(n): binary = "" while n > 0: binary = str(n % 2) + binary n //= 2 return binary # Example usage decimal = 29 binary = decimal_to_binary(decimal) print(f"Binary of {decimal} is {binary}")
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Output
The binary representation of the decimal number 29 is โ11101โ.
Learn more aboutย โ 20 Accenture Coding Questions With Proven Solutions
Hope this article gives you a brief idea about what kind of coding questions asked in Infosys, how to answer them and what level of preparation you need to crack the Infosys interview. Now, itโs your turn to start preparing and focus on practicing previous Infosys coding questions like these given to succeed in the interview. Drop your comments to let us know if you find these Infosys coding questions helpful for your exam preparation.
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