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Deloitte Coding Questions : Deloitte is one of the largest UK-based multinational finance and professional services providers in the world. Most of the graduates with an accounting degree had a dream of working for Deloitte. Itโs a top name in a number of employees and market size as per the Gartner report.

Deloitte is a perfect place to boost your professional career to new heights. However, the hiring process of Deloitte is complex and depends on various factors including skills, education, and extensive technical coding knowledge experience.

So, if youโre planning for the Deloitte interview then you must prepare for a technical coding round because it plays an important part. I suggest you prepare with the most asked Deloitte coding questions for analyst trainee to get a better idea about the exam and increase your chance of getting selected.

Explore Deloitte Coding Questions for freshers

Discover comprehensive information about the Deloitte hiring process, encompassing Coding Questions featured in NLA (National Level Assessment) and Technical Interviews, and other details on this Deloitte Coding Questions and Answers blog page

In this article, weโve listed the most asked **Deloitte coding questions**ย for freshers that you can practice for better preparation. Letโs dive in!

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Here, weโve listed some commonly asked Deloitte coding questions and answers including deloitte coding questions in python , java and c++ Languages.

**Q.1 ****How can a number be expressed as a sum of two prime numbers? Write a Program.**

**Answer:**

**Sample Input: 34**

#include <iostream> using namespace std; bool isPrime(int n) { if (n <= 1) return false; for (int i = 2; i * i <= n; i++) { if (n % i == 0) return false; } return true; } void findPrimeSum(int num) { bool found = false; for (int i = 2; i <= num / 2; i++) { if (isPrime(i) && isPrime(num - i)) { cout << num << " can be expressed as the sum of " << i << " and " << num-i << "." << endl; found = true; break; // Uncomment this to find all pairs } } if (!found) cout << num << " cannot be expressed as the sum of two prime numbers." << endl; } int main() { int num = 34; findPrimeSum(num); return 0; }

public class PrimeSum { public static boolean isPrime(int n) { if (n <= 1) return false; for (int i = 2; i <= Math.sqrt(n); i++) { if (n % i == 0) return false; } return true; } public static void findPrimeSum(int num) { boolean found = false; for (int i = 2; i <= num / 2; i++) { if (isPrime(i) && isPrime(num - i)) { System.out.println(num + " can be expressed as the sum of " + i + " and " + (num-i) + "."); found = true; break; // Uncomment this to find all pairs } } if (!found) System.out.println(num + " cannot be expressed as the sum of two prime numbers."); } public static void main(String[] args) { int num = 34; findPrimeSum(num); } }

def is_prime(n): if n <= 1: return False for i in range(2, int(n**0.5) + 1): if n % i == 0: return False return True def find_prime_sum(n): for i in range(2, n//2 + 1): if is_prime(i) and is_prime(n - i): return f"{n} can be expressed as the sum of {i} and {n - i}." return f"{n} cannot be expressed as the sum of two prime numbers." # Sample input n = 34 find_prime_sum(n)

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**Output:**

โ34 can be expressed as the sum of 3 and 31.โ

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**Answer:**

**Sample Input: 28**

#include <iostream> using namespace std; bool isPerfect(int n) { int sum = 1; for (int i = 2; i * i <= n; i++) { if (n % i == 0) { sum += i; if (i != n / i) sum += n / i; } } return sum == n && n != 1; } int main() { int n = 28; // Sample input if (isPerfect(n)) cout << n << " is a perfect number." << endl; else cout << n << " is not a perfect number." << endl; return 0; }

public class PerfectNumber { static boolean isPerfect(int n) { int sum = 1; for (int i = 2; i * i <= n; i++) { if (n % i == 0) { sum += i; if (i != n / i) sum += n / i; } } return sum == n && n != 1; } public static void main(String[] args) { int n = 28; // Sample input if (isPerfect(n)) System.out.println(n + " is a perfect number."); else System.out.println(n + " is not a perfect number."); } }

def is_perfect(n): sum = 1 for i in range(2, int(n**0.5) + 1): if n % i == 0: sum += i if i != n // i: sum += n // i return sum == n and n != 1 # Sample input n = 28 if is_perfect(n): result = f"{n} is a perfect number." else: result = f"{n} is not a perfect number." print(result)

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**Output:**

28 is a perfect number because its divisors are 1, 2, 4, 7, and 14, and the sum of these divisors is 28.

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**There is a list of decimal numbers, as well as an infection that has a specific amount of spikes. Each spike represents the amount of binary digits that a virus will eat from the right-hand part of a decimal number. Write a program that will determine the final status of each number following the virus that has consumed the specified amount of binary digits from each one of them.**

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**Answer:**

**Sample Input:**

**6**

**10 20 30 40 50 60**

**3**

#include <iostream> #include <vector> using namespace std; int main() { int N, n; cin >> N; vector<int> v(N); for (int i = 0; i < N; i++) cin >> v[i]; cin >> n; for (auto i : v) cout << (i >> n) << " "; return 0; }

import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int N = sc.nextInt(); int[] arr = new int[N]; for (int i = 0; i < N; i++) { arr[i] = sc.nextInt(); } int n = sc.nextInt(); for (int i : arr) { System.out.print((i >> n) + " "); } } }

N = int(input()) a = list(map(int, input().split())) n = int(input()) result = " ".join(str(i >> n) for i in a) print(result)

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**Output:**

โ1 2 3 5 6 7โ

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**Answer:**

**Sample Input: apple banana cherry orange mango**

#include <iostream> #include <vector> #include <algorithm> using namespace std; bool isVowel(char c) { return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' || c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U'; } int main() { string input; getline(cin, input); vector<string> words; string word = ""; for (char c : input) { if (c == ' ') { words.push_back(word); word = ""; } else { word += c; } } words.push_back(word); sort(words.begin(), words.end()); for (int i = 0; i < words.size(); i++) { if (isVowel(words[i][0])) { cout << i + 1 << ": " << words[i] << endl; } } return 0; }

import java.util.Arrays; public class Main { public static void main(String[] args) { String input = "apple banana cherry orange mango"; String[] words = input.split(" "); Arrays.sort(words); for (int i = 0; i < words.length; i++) { if (isVowel(words[i].charAt(0))) { System.out.println((i + 1) + ": " + words[i]); } } } public static boolean isVowel(char c) { return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' || c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U'; } }

def is_vowel(c): return c in 'aeiouAEIOU' input_str = "apple banana cherry orange mango" words = input_str.split() words.sort() for i, word in enumerate(words, 1): if is_vowel(word[0]): print(f"{i}: {word}")

**Output:**

1: apple

2: orange

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**Also Read: ****20 IBM Coding Questions with Answers**

After JEE Mains, students have been admitted to an engineering college, forming a class of โnโ students. The Head of the Department (HOD) is tasked with selecting a class monitor. However, the HOD meets the students one by one and has a peculiar way of choosing. Each time the HOD meets a student, if the studentโs rank is lower than that of the previous student met, the HOD cuts the previous studentโs name and writes the new studentโs name along with their rank in a register. Given the ranks the HOD receives each time they meet a student, predict how many names are cut from the list.

**Constraints:**

Number of visits <= 10^9

Ranks <= 10000

**Input Format:**

The first line contains the number of visits โNโ.

The second line contains โNโ space-separated ranks the HOD receives each time.

**Output Format:ย **

Single N space-separated integers denoting the final situation with the array v.

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**Answer:ย **

**Sample Input:ย **

7

5 8 6 3 9 4 2

#include <iostream> using namespace std; int main() { int N; cin >> N; int prevRank, currRank, cuts = 0; cin >> prevRank; for (int i = 1; i < N; i++) { cin >> currRank; if (currRank < prevRank) { cuts++; } prevRank = currRank; } cout << cuts << endl; return 0; }

import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int N = sc.nextInt(); int prevRank = sc.nextInt(); int cuts = 0; for (int i = 1; i < N; i++) { int currRank = sc.nextInt(); if (currRank < prevRank) { cuts++; } prevRank = currRank; } System.out.println(cuts); } }

N = int(input()) ranks = list(map(int, input().split())) cuts = 0 prev_rank = ranks[0] for rank in ranks[1:]: if rank < prev_rank: cuts += 1 prev_rank = rank print(cuts)

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**Output**: 3

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Rahul is known for copying in exams from his adjacent students, but heโs smart about it. Instead of directly copying the words, he changes the positions of letters while keeping the letters constant. As the examiner, you need to determine if Rahul has copied a certain word from the adjacent student who is giving the same exam. You should provide Rahul with the appropriate markings based on your findings.

**Note**: Uppercase and lowercase letters are considered the same.

**Input Format:**

The first line contains the word of the adjacent student.

The second line contains Rahulโs words.

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**Output Format:**

0 if Rahul did not copy.

1 if Rahul copied.

Constraints:

1 <= Length of string <= 10^6

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**Answer:ย **

**Sample Input:**

HELLO

EHLLO

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#include <iostream> #include <algorithm> using namespace std; int main() { string adjacentWord, rahulWord; cin >> adjacentWord >> rahulWord; sort(adjacentWord.begin(), adjacentWord.end()); sort(rahulWord.begin(), rahulWord.end()); if (adjacentWord == rahulWord) { cout << 1 << endl; } else { cout << 0 << endl; } return 0; }

import java.util.Arrays; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); String adjacentWord = sc.next(); String rahulWord = sc.next(); char[] adjacentArray = adjacentWord.toCharArray(); char[] rahulArray = rahulWord.toCharArray(); Arrays.sort(adjacentArray); Arrays.sort(rahulArray); if (Arrays.equals(adjacentArray, rahulArray)) { System.out.println(1); } else { System.out.println(0); } } }

adjacent_word = input() rahul_word = input() if sorted(adjacent_word.lower()) == sorted(rahul_word.lower()): print(1) else: print(0)

**Output: **1

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Anirudh is attending an astronomy lecture, but heโs struggling with understanding the material. His very strict professor asks students to write a program to print a trapezium pattern using stars (*) and dots (.) as shown below. Since Anirudh is not proficient in astronomy, can you help him?

ย

**Answer:ย **

**Sample Input:**

N = 5

#include <iostream> using namespace std; int main() { int N; cin >> N; for (int i = 0; i < N; i++) { for (int j = 0; j < N * 2; j++) { if (j < N - i || j >= N + i + 1) { cout << "."; } else { cout << "*"; } } cout << endl; } return 0; }

import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int N = sc.nextInt(); for (int i = 0; i < N; i++) { for (int j = 0; j < N * 2; j++) { if (j < N - i || j >= N + i + 1) { System.out.print("."); } else { System.out.print("*"); } } System.out.println(); } } }

N = int(input()) for i in range(N): for j in range(N * 2): if j < N - i or j >= N + i + 1: print(".", end="") else: print("*", end="") print()

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**Output**:ย

**..**

*โฆ.*

โฆโฆ

*โฆ.*

**..**

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You are given an array, and you need to choose a contiguous subarray of length โkโ. Then, find the minimum value within that subarray and return the maximum of those minimum values.

**Answer:ย **

**Sample Input:**

- Length of the subarray: 2
- Size of the array: 6
- Array elements: [3, 1, 4, 6, 2, 5]

**Explanation:**

The subarrays of size 2 are: [3, 1], [1, 4], [4, 6], [6, 2], and [2, 5].

The minimum values within these subarrays are: 1, 1, 4, 2, and 2 respectively.

The maximum of these minimum values is 4.

#include <iostream> #include <vector> #include <algorithm> using namespace std; int main() { int k, n; cin >> k >> n; vector<int> arr(n); for (int i = 0; i < n; ++i) { cin >> arr[i]; } int max_min = INT_MIN; for (int i = 0; i <= n - k; ++i) { int min_val = *min_element(arr.begin() + i, arr.begin() + i + k); max_min = max(max_min, min_val); } cout << max_min << endl; return 0; }

import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int k = sc.nextInt(); int n = sc.nextInt(); int[] arr = new int[n]; for (int i = 0; i < n; ++i) { arr[i] = sc.nextInt(); } int maxMin = Integer.MIN_VALUE; for (int i = 0; i <= n - k; ++i) { int minVal = Integer.MAX_VALUE; for (int j = i; j < i + k; ++j) { minVal = Math.min(minVal, arr[j]); } maxMin = Math.max(maxMin, minVal); } System.out.println(maxMin); } }

k, n = map(int, input().split()) arr = list(map(int, input().split())) max_min = float('-inf') for i in range(n - k + 1): min_val = min(arr[i:i + k]) max_min = max(max_min, min_val) print(max_min)

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**Output: **4

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A password manager wants to create new passwords using two strings given by the user, then combines them to create a harder-to-guess combination. Given two strings, the task is to interleave the characters of the strings to create a new string. Begin with an empty string and alternately append a character from string โaโ and from string โbโ. If one of the strings is exhausted before the other, append the remaining letters from the other string all at once. The resulting string is the new password.

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**Example:**

If a = โhackerrankโ and b = โmountainโ, the resulting password is โhmaocuknetrariannkโ.

**Function Description:**

Complete the function newPassword which takes two strings โaโ and โbโ as input and returns the new password as a string.

**Parameters:**

Str a: String โaโ

Str b: String โbโ

**Returns:**

Str: New password using two strings

Answer:ย

**Sample Input:**

a = โopenโ

b = โsourceโ

#include <iostream> #include <string> using namespace std; string newPassword(string a, string b) { string result = ""; int i = 0, j = 0; while (i < a.length() && j < b.length()) { result += a[i++]; result += b[j++]; } while (i < a.length()) { result += a[i++]; } while (j < b.length()) { result += b[j++]; } return result; } int main() { string a, b; cin >> a >> b; cout << newPassword(a, b) << endl; return 0; }

import java.util.Scanner; public class Main { public static String newPassword(String a, String b) { StringBuilder result = new StringBuilder(); int i = 0, j = 0; while (i < a.length() && j < b.length()) { result.append(a.charAt(i++)); result.append(b.charAt(j++)); } while (i < a.length()) { result.append(a.charAt(i++)); } while (j < b.length()) { result.append(b.charAt(j++)); } return result.toString(); } public static void main(String[] args) { Scanner sc = new Scanner(System.in); String a = sc.next(); String b = sc.next(); System.out.println(newPassword(a, b)); } }

def newPassword(a, b): result = "" i, j = 0, 0 while i < len(a) and j < len(b): result += a[i] + b[j] i += 1 j += 1 while i < len(a): result += a[i] i += 1 while j < len(b): result += b[j] j += 1 return result a = input() b = input() print(newPassword(a, b))

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**Output**: โospuoerceโ

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**Learn More: ****Top Infosys Coding Questions with Answers For Specialist Programmer**

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Anish, known for his laziness, is tasked with writing the name of the winner in a game where two people participate. However, instead of writing the winnerโs name directly, he simply writes the longest common subsequence (LCS) of the two names. This allows him to minimize the number of changes or avoid using the backspace key while editing the name. Given two names, your task is to predict what Anish will write on his computer before the start of the game. If there are multiple longest subsequences possible, choose the one with the smallest lexicographic value.

ย

**Input Format:**

Two lines containing two strings representing the names (all letters are in capital case).

**Output Format:**

A single line containing the lexicographically smallest possible longest common subsequence.

Answer:ย

**Sample Input:**

GEEK

EKEG

#include <iostream> #include <vector> #include <algorithm> using namespace std; int main() { string s1, s2; cin >> s1 >> s2; vector<vector<int>> dp(s1.size() + 1, vector<int>(s2.size() + 1, 0)); for (int i = 1; i <= s1.size(); ++i) { for (int j = 1; j <= s2.size(); ++j) { if (s1[i - 1] == s2[j - 1]) { dp[i][j] = dp[i - 1][j - 1] + 1; } else { dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]); } } } string lcs = ""; int i = s1.size(), j = s2.size(); while (i > 0 && j > 0) { if (s1[i - 1] == s2[j - 1]) { lcs = s1[i - 1] + lcs; i--; j--; } else if (dp[i - 1][j] > dp[i][j - 1]) { i--; } else { j--; } } cout << lcs << endl; return 0; }

import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); String s1 = sc.next(); String s2 = sc.next(); int[][] dp = new int[s1.length() + 1][s2.length() + 1]; for (int i = 1; i <= s1.length(); ++i) { for (int j = 1; j <= s2.length(); ++j) { if (s1.charAt(i - 1) == s2.charAt(j - 1)) { dp[i][j] = dp[i - 1][j - 1] + 1; } else { dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]); } } } StringBuilder lcs = new StringBuilder(); int i = s1.length(), j = s2.length(); while (i > 0 && j > 0) { if (s1.charAt(i - 1) == s2.charAt(j - 1)) { lcs.insert(0, s1.charAt(i - 1)); i--; j--; } else if (dp[i - 1][j] > dp[i][j - 1]) { i--; } else { j--; } } System.out.println(lcs.toString()); } }

def lcs(s1, s2): m, n = len(s1), len(s2) dp = [[0] * (n + 1) for _ in range(m + 1)] for i in range(1, m + 1): for j in range(1, n + 1): if s1[i - 1] == s2[j - 1]: dp[i][j] = dp[i - 1][j - 1] + 1 else: dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) lcs = "" i, j = m, n while i > 0 and j > 0: if s1[i - 1] == s2[j - 1]: lcs = s1[i - 1] + lcs i -= 1 j -= 1 elif dp[i - 1][j] > dp[i][j - 1]: i -= 1 else: j -= 1 return lcs s1 = input() s2 = input() print(lcs(s1, s2))

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**Output**: โEGโ

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Yes, Deloitte asks coding questions in their technical interview. These coding questions are complex and require good knowledge of the technical background to answer coding questions efficiently.

Some experts say that you should practice the previous year's coding questions to understand the concept and complexity of questions and efficiently prepare for the Deloitte interview.

Deloitte follows a standard format for their interview, and these are slightly difficult and require extensive knowledge and practice to crack the interview.

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These are some of the most commonly asked **Deloitte coding questions for Freshers, **and if you practice, then it will increase your chances of getting selected and make your exam preparation faster. We hope this article helps you in coding exam preparation, and now you can start practicing previous Deloitte coding questions. Drop your comment, and let me know if you find these Deloitte coding questions helpful for your preparation.

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