Deloitte Coding Questions and Answers for Freshers (2026)

24 min read

Deloitte coding questions are the part of the hiring process most candidates underestimate. Deloitte is not a product company, so applicants often assume the technical bar is low — then open the assessment and find two live coding problems sharing a 90-minute clock with 65 multiple-choice questions.

This page collects ten practice problems that match the format candidate accounts consistently describe, each with a working solution in C++, Java and Python, plus a plain explanation of what the NLA, PYQ papers and the “coding challenge” actually refer to.

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 you’re 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.

In this article, we’ve listed the most asked Deloitte coding questions for freshers that you can practice for better preparation. Let’s dive in!

Deloitte Coding Questions for Freshers: Exam Overview

On a phone, scroll the table sideways to see every column. Figures above come from published NLA syllabi and candidate accounts, not from Deloitte. Confirm them on your own invitation email and instruction screen.

Top 10 Deloitte Coding Questions and Answers in 2026

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

Deloitte Coding Questions #1 — Express a Number as the Sum of Two Primes

Problem Statement:

How can a number be expressed as a sum of two prime numbers? Write a program.

Answer:

Sample Input: 34

Solutions in C++, Java and Python below.

#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)

Output:

’34 can be expressed as the sum of 3 and 31.’

Deloitte Coding Questions #2 — Check Whether a Number Is Perfect

Problem Statement:

How do you check whether a number is a perfect number or not? Write a program.

Answer:

Sample Input: 28

Solutions in C++, Java and Python below.

#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)

Output:

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

Deloitte Coding Questions #3 — Drop Binary Digits from Each Number

Problem Statement:

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.

Answer:

Sample Input:

6

10 20 30 40 50 60

3

Solutions in C++, Java and Python below.

#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)

Output:

‘1 2 3 5 6 7’

Deloitte Coding Questions #4 — Sort Words and Find the Vowel-Starting Ones

Problem Statement:

Write a program that takes a string containing multiple words as input. Sort these words in ascending alphabetical order, then print all the words that start with a vowel along with their positions in the sorted list.

Answer:

Sample Input: apple banana cherry orange mango

Solutions in C++, Java and Python below.

#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

Also Read: 20 IBM Coding Questions with Answers

Deloitte Coding Questions #5 — Choosing the Class Monitor

Problem Statement:

After JEE Mains, students have been admitted to an engineering college, forming a class of ‘n’ 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 student’s rank is lower than that of the previous student met, the HOD cuts the previous student’s name and writes the new student’s 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 ‘N’.

The second line contains ‘N’ space-separated ranks the HOD receives each time.

Output Format: 

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

Answer: 

Sample Input: 

7

5 8 6 3 9 4 2

Solutions in C++, Java and Python below.

#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)

Output: 3

Deloitte Coding Questions #6 — Detect a Rearranged-Letter Copy

Problem Statement:

Rahul is known for copying in exams from his adjacent students, but he’s 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 Rahul’s words.

Output Format:

0 if Rahul did not copy.

1 if Rahul copied.

Constraints:

1 <= Length of string <= 10^6

Answer: 

Sample Input:

HELLO

EHLLO

Solutions in C++, Java and Python below.

#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

Deloitte Coding Questions #7 — Print a Trapezium Pattern

Problem Statement:

Anirudh is attending an astronomy lecture, but he’s 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?

Answer: 

Sample Input:

N = 5

Solutions in C++, Java and Python below.

#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()

Output: 

**..**

*….*

……

*….*

**..**

Deloitte Coding Questions #8 — Maximum of the Minimums in Every Window

Problem Statement:

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

Answer: 

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.

Solutions in C++, Java and Python below.

#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)

Output: 4

Deloitte Coding Questions #9 — Interleave Two Strings into a Password

Problem Statement:

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 ‘a’ and from string ‘b’. 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.

Example:

If a = ‘hackerrank’ and b = ‘mountain’, the resulting password is ‘hmaocuknetrariannk’.

Function Description:

Complete the function newPassword which takes two strings ‘a’ and ‘b’ as input and returns the new password as a string.

Parameters:

Str a: String ‘a’

Str b: String ‘b’

Returns:

Str: New password using two strings

Answer: 

Sample Input:

a = “open”

b = “source”

Solutions in C++, Java and Python below.

#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))

Output: ‘ospuoerce’

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

Deloitte Coding Questions #10 — Lexicographically Smallest Longest Common Subsequence

Problem Statement:

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 winner’s 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.

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: 

Sample Input:

GEEK

EKEG

Solutions in C++, Java and Python below.

#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))

Output: ‘EG’

Previous-year papers are pattern evidence, not a question bank — Deloitte rotates its pool.

Deloitte NLA, PYQ Papers and the Coding Challenge Explained

Three terms come up constantly in searches for Deloitte coding questions, and they are not interchangeable. Knowing which one your invitation refers to changes what you should revise.

What is the Deloitte NLA?

NLA stands for National Level Assessment — Deloitte’s large-scale hiring assessment, most commonly associated with fresher and campus recruitment. Deloitte does not publish the NLA syllabus or the specific Deloitte coding questions used in any round. The closest first-party description of how its graduate assessments are structured is Deloitte’s own early careers assessment page, which sets out an application stage, an immersive online assessment and a job simulation — a different geography from the Indian NLA, but the same multi-stage shape. Candidate reports describe it as a multi-section test rather than a pure coding round, typically combining aptitude and technical sections with a coding component. If your invitation says NLA, expect more than algorithms alone.

Deloitte PYQ and previous-year papers

PYQ simply means “previous year questions”. These circulate widely for the NLA in particular, and they are genuinely useful — but as pattern evidence, not as a question bank. Deloitte rotates its pool, so the value is in recognising which topics recur (strings, number theory, arrays, basic DP) rather than memorising specific problems. Every question on this page is drawn from that recurring pattern.

One thing worth being clear about: collections sold or shared as “Deloitte previous year coding questions” are reconstructions written from memory by earlier candidates, not official papers. Expect the wording to drift from whatever was actually on screen. They are good for judging the difficulty band and the topic mix, and bad for memorising answers — if you find yourself recognising solutions rather than deriving them, the previous-year set has stopped helping you.

The Deloitte coding challenge vs the coding round

These names are often used loosely for the same thing. “Coding round” usually refers to the programming section inside a larger assessment; “coding challenge” sometimes describes a standalone timed contest used for specific hiring drives. Neither term has a fixed published definition, so treat your invitation email as the authority rather than any guide.

Deloitte Python coding questions

Python appears in every candidate account that names an accepted language list, and every problem on this page carries a Python solution alongside C++ and Java. Two practical notes if Python is your choice.

First, the Deloitte coding questions in the assessment are usually full-program format, so you read input with input() and print the result yourself rather than returning from a function. Second, none of these problems has a constraint tight enough for a clean Python solution to time out — so pick the language you write fastest in, not the one you think reads as more impressive.

Is the Deloitte coding round different for experienced hires?

Reports suggest experienced-hire processes lean more on a technical interview discussing your own projects and less on a standardised assessment, while fresher and campus routes centre on the NLA. The underlying programming fundamentals are the same; the weighting differs.

Deloitte Coding Questions and Answers – FAQs

Does Deloitte actually ask coding questions?

Yes. A coding component appears both in the online assessment and again in the technical interview. Candidate write-ups of the NLA describe two live coding questions inside the online test, and interviewers frequently ask a further one or two basic problems on the spot, usually in whichever language your resume claims.

How many Deloitte coding questions are in the online test?

Two, in almost every account. Preparation sites that break the NLA down list 67 questions in total — 65 MCQs across aptitude, logical reasoning, English and technical skills, plus 2 coding questions — inside a 90-minute window. One candidate account on GeeksforGeeks describes the coding portion as a separate 30-minute block; other sources describe the 90 minutes as a single pooled budget with no per-section timer. Treat the timing as unconfirmed and plan to finish both problems inside 30 minutes regardless.

Which programming languages can I use?

Deloitte does not publish an official language list. Prep sites and candidate reports consistently name C, C++, C#, Java and Python, which is why the Deloitte coding questions on this page carry solutions in C++, Java and Python. If your invitation email names a platform, check its editor for the language dropdown before test day rather than assuming.

Are Deloitte coding questions repeated from previous years’ papers?

Not verbatim, and treating PYQ collections as a leaked question bank is the most common preparation mistake. What repeats is the pattern — string manipulation, array traversal, number theory, simple matrix work — not the exact prompts. Practising previous-year sets is useful for calibrating difficulty and speed; memorising their answers is not.

How hard are the coding questions?

Moderate. They sit closer to LeetCode Easy and lower-Medium than to competitive programming. The difficulty is rarely the algorithm; it is finishing two problems and 65 MCQs inside one time budget without leaving a section blank.

What is the difference between the NLA and a regular Deloitte coding test?

The NLA is the large-scale national assessment used for fresher and campus intake. Experienced-hire and lateral processes usually skip it and go straight to a technical screen or a role-specific assessment. If your invitation does not say NLA, do not prepare for the NLA sectional structure.

Do I need to handle input and output myself?

Usually yes. Most reported Deloitte coding rounds use a full-program format where you read from standard input and print to standard output, rather than filling in a function body. Practise writing the complete program, including the input parsing, because losing marks to an I/O mismatch on a correct algorithm is a genuinely common failure.

Is there negative marking?

Reported syllabi for the NLA say no negative marking on the MCQ sections, which means leaving a question unanswered is strictly worse than guessing. Deloitte itself does not publish this, so confirm it on your own instruction screen before the timer starts.

Deloitte Coding Questions and Answers PDF

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Wrapping Up: How to Actually Use These Deloitte Coding Questions

The ten problems above cover the patterns that show up again and again in reported Deloitte assessments: string manipulation, array and matrix traversal, basic number theory, and small sliding-window logic. None of them is hard in isolation. The pressure comes from solving two of them while 65 MCQs are eating the same clock.

So practise for the constraint, not just the algorithm. Give yourself 12 minutes per problem, write the full program including input parsing, and run it against your own edge cases before you look at the solution. If you can clear these Deloitte coding questions cold in under 25 minutes total, the coding section will not be what decides your result — the technical MCQ block will, and that is where the rest of your revision time belongs.

One honest caveat: Deloitte does not publish its question bank, so nothing here is a leaked paper. These are practice problems built to match the format and difficulty that candidate accounts consistently describe.

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