Capgemini Coding Questions

Practice with 25+ Capgemini placement paper coding questions covering the Capgemini game-based assessment and hands-on coding test. These questions are representative of what you’ll encounter in Capgemini’s online assessment.

What’s Included:

• 25+ Coding Problems: Easy and Medium level problems with solutions
• Multiple Language Solutions: C, C++, Java, and Python solutions
• Time Complexity Analysis: Every solution includes complexity analysis
• Capgemini-Specific Patterns: Focus on pseudocode, basic programming, arrays, strings

Capgemini Coding Pattern 2025

Capgemini assessment has 2 sections: Game-Based Assessment (Pseudocode - 15 MCQs in 25 min) and Hands-on Coding (2 problems in 45 min). Master pseudocode reading and basic programming in one language.

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Capgemini coding section overview

Capgemini Assessment Coding Section Breakdown:

Section	Questions	Time	Difficulty	Focus Areas
Game-Based (Pseudocode)	15 MCQs	25 min	Easy-Medium	Pseudocode, logic, tracing
Hands-on Coding	2	45 min	Easy-Medium	Arrays, strings, basic algorithms

Languages Allowed: C, C++, Java, Python

Pseudocode practice questions

Pseudocode MCQs

Q1: Loop Output Prediction

Question: What is the output of the following pseudocode?

SET x = 5
SET y = 0
WHILE x > 0 DO
    SET y = y + x
    SET x = x - 1
ENDWHILE
PRINT y

Options:

• A) 10

• B) 15

• C) 20

• D) 5

  Answer: B) 15

  Explanation:

• x=5: y = 0+5 = 5, x = 4

• x=4: y = 5+4 = 9, x = 3

• x=3: y = 9+3 = 12, x = 2

• x=2: y = 12+2 = 14, x = 1

• x=1: y = 14+1 = 15, x = 0

• Loop ends, y = 15

< >
Q2: Conditional Statement

Question: What is the output?

SET a = 10
SET b = 20
IF a > b THEN
    PRINT a
ELSE
    IF a == 10 THEN
        PRINT b + a
    ELSE
        PRINT b - a
    ENDIF
ENDIF

Options:

• A) 10

• B) 20

• C) 30

• D) -10

  Answer: C) 30

  Explanation: a (10) is not > b (20), so else executes. a == 10 is true, so b + a = 30.

< >
Q3: Array Traversal

Question: What is the output?

SET arr = [1, 2, 3, 4, 5]
SET sum = 0
FOR i = 0 TO 4 DO
    IF arr[i] MOD 2 == 0 THEN
        SET sum = sum + arr[i]
    ENDIF
ENDFOR
PRINT sum

Options:

• A) 6

• B) 9

• C) 15

• D) 5

  Answer: A) 6

  Explanation: Even numbers in array are 2 and 4. Sum = 2 + 4 = 6.

< >
Q4: Nested Loop

Question: What is the output?

SET count = 0
FOR i = 1 TO 3 DO
    FOR j = 1 TO i DO
        SET count = count + 1
    ENDFOR
ENDFOR
PRINT count

Options:

• A) 3

• B) 6

• C) 9

• D) 4

  Answer: B) 6

  Explanation:

• i=1: j runs 1 time (count=1)

• i=2: j runs 2 times (count=3)

• i=3: j runs 3 times (count=6)

< >
Q5: String Operation

Question: What is the output?

SET str = "CAPGEMINI"
SET count = 0
FOR i = 0 TO LENGTH(str) - 1 DO
    IF str[i] == 'A' OR str[i] == 'E' OR str[i] == 'I' THEN
        SET count = count + 1
    ENDIF
ENDFOR
PRINT count

Options:

• A) 2

• B) 3

• C) 4

• D) 5

  Answer: C) 4

  Explanation: Vowels in “CAPGEMINI”: A, E, I, I = 4 vowels.

Hands-on coding problems

Array Problems

Sum of Array Elements

Problem: Find the sum of all elements in an array.

Example:

Input: [1, 2, 3, 4, 5]
Output: 15

Solution (Java):

public int arraySum(int[] arr) {
    int sum = 0;
    for (int num : arr) {
        sum += num;
    }
    return sum;
}

Solution (Python):

def array_sum(arr):
    return sum(arr)

Solution (C):

int arraySum(int arr[], int n) {
    int sum = 0;
    for (int i = 0; i < n; i++) {
        sum += arr[i];
    }
    return sum;
}

Time Complexity: O(n) | Space Complexity: O(1)

Find Second Largest

Problem: Find second largest element in array.

Example:

Input: [12, 35, 1, 10, 34]
Output: 34

Solution (Java):

public int secondLargest(int[] arr) {
    int first = Integer.MIN_VALUE;
    int second = Integer.MIN_VALUE;

    for (int num : arr) {
        if (num > first) {
            second = first;
            first = num;
        } else if (num > second && num != first) {
            second = num;
        }
    }
    return second;
}

Solution (Python):

def second_largest(arr):
    first = second = float('-inf')
    for num in arr:
        if num > first:
            second = first
            first = num
        elif num > second and num != first:
            second = num
    return second

Time Complexity: O(n) | Space Complexity: O(1)

Reverse Array

Problem: Reverse an array in-place.

Example:

Input: [1, 2, 3, 4, 5]
Output: [5, 4, 3, 2, 1]

Solution (Java):

public void reverseArray(int[] arr) {
    int left = 0, right = arr.length - 1;
    while (left < right) {
        int temp = arr[left];
        arr[left] = arr[right];
        arr[right] = temp;
        left++;
        right--;
    }
}

Solution (Python):

def reverse_array(arr):
    left, right = 0, len(arr) - 1
    while left < right:
        arr[left], arr[right] = arr[right], arr[left]
        left += 1
        right -= 1

Time Complexity: O(n) | Space Complexity: O(1)

Find Duplicates

Problem: Find duplicate elements in array.

Solution (Java):

public List<Integer> findDuplicates(int[] arr) {
    Set<Integer> seen = new HashSet<>();
    List<Integer> duplicates = new ArrayList<>();

    for (int num : arr) {
        if (seen.contains(num)) {
            if (!duplicates.contains(num)) {
                duplicates.add(num);
            }
        } else {
            seen.add(num);
        }
    }
    return duplicates;
}

Solution (Python):

def find_duplicates(arr):
    seen = set()
    duplicates = set()
    for num in arr:
        if num in seen:
            duplicates.add(num)
        seen.add(num)
    return list(duplicates)

Time Complexity: O(n) | Space Complexity: O(n)

Rotate Array

Problem: Rotate array to the left by k positions.

Example:

Input: arr = [1, 2, 3, 4, 5], k = 2
Output: [3, 4, 5, 1, 2]

Solution (Java):

public void rotateLeft(int[] arr, int k) {
    int n = arr.length;
    k = k % n;
    reverse(arr, 0, k - 1);
    reverse(arr, k, n - 1);
    reverse(arr, 0, n - 1);
}

private void reverse(int[] arr, int start, int end) {
    while (start < end) {
        int temp = arr[start];
        arr[start] = arr[end];
        arr[end] = temp;
        start++;
        end--;
    }
}

Solution (Python):

def rotate_left(arr, k):
    n = len(arr)
    k = k % n
    return arr[k:] + arr[:k]

Time Complexity: O(n) | Space Complexity: O(1)

String Problems

Reverse String

Problem: Reverse a given string.

Example:

Input: "hello"
Output: "olleh"

Solution (Java):

public String reverseString(String s) {
    char[] chars = s.toCharArray();
    int left = 0, right = chars.length - 1;
    while (left < right) {
        char temp = chars[left];
        chars[left] = chars[right];
        chars[right] = temp;
        left++;
        right--;
    }
    return new String(chars);
}

Solution (Python):

def reverse_string(s):
    return s[::-1]

Time Complexity: O(n) | Space Complexity: O(n)

Palindrome Check

Problem: Check if a string is palindrome.

Example:

Input: "madam"
Output: true

Solution (Java):

public boolean isPalindrome(String s) {
    s = s.toLowerCase();
    int left = 0, right = s.length() - 1;
    while (left < right) {
        if (s.charAt(left) != s.charAt(right)) {
            return false;
        }
        left++;
        right--;
    }
    return true;
}

Solution (Python):

def is_palindrome(s):
    s = s.lower()
    return s == s[::-1]

Time Complexity: O(n) | Space Complexity: O(1)

Count Vowels and Consonants

Problem: Count vowels and consonants in a string.

Solution (Java):

public int[] countVowelsConsonants(String s) {
    int vowels = 0, consonants = 0;
    s = s.toLowerCase();
    for (char c : s.toCharArray()) {
        if (Character.isLetter(c)) {
            if ("aeiou".indexOf(c) != -1) {
                vowels++;
            } else {
                consonants++;
            }
        }
    }
    return new int[]{vowels, consonants};
}

Solution (Python):

def count_vowels_consonants(s):
    vowels = set('aeiouAEIOU')
    v = c = 0
    for char in s:
        if char.isalpha():
            if char in vowels:
                v += 1
            else:
                c += 1
    return v, c

Time Complexity: O(n) | Space Complexity: O(1)

Anagram Check

Problem: Check if two strings are anagrams.

Example:

Input: s1 = "listen", s2 = "silent"
Output: true

Solution (Java):

public boolean isAnagram(String s1, String s2) {
    if (s1.length() != s2.length()) return false;
    int[] count = new int[26];
    for (int i = 0; i < s1.length(); i++) {
        count[s1.charAt(i) - 'a']++;
        count[s2.charAt(i) - 'a']--;
    }
    for (int c : count) {
        if (c != 0) return false;
    }
    return true;
}

Solution (Python):

def is_anagram(s1, s2):
    return sorted(s1.lower()) == sorted(s2.lower())

Time Complexity: O(n) | Space Complexity: O(1)

Remove Duplicates from String

Problem: Remove duplicate characters from string.

Example:

Input: "programming"
Output: "progamin"

Solution (Java):

public String removeDuplicates(String s) {
    StringBuilder result = new StringBuilder();
    Set<Character> seen = new LinkedHashSet<>();
    for (char c : s.toCharArray()) {
        if (!seen.contains(c)) {
            seen.add(c);
            result.append(c);
        }
    }
    return result.toString();
}

Solution (Python):

def remove_duplicates(s):
    seen = set()
    result = []
    for char in s:
        if char not in seen:
            seen.add(char)
            result.append(char)
    return ''.join(result)

Time Complexity: O(n) | Space Complexity: O(n)

Math Problems

Prime Number Check

Problem: Determine if a number is prime.

Solution (Java):

public boolean isPrime(int n) {
    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    for (int i = 5; i * i <= n; i += 6) {
        if (n % i == 0 || n % (i + 2) == 0) return false;
    }
    return true;
}

Solution (Python):

def is_prime(n):
    if n <= 1:
        return False
    if n <= 3:
        return True
    if n % 2 == 0 or n % 3 == 0:
        return False
    i = 5
    while i * i <= n:
        if n % i == 0 or n % (i + 2) == 0:
            return False
        i += 6
    return True

Time Complexity: O(√n) | Space Complexity: O(1)

Factorial

Problem: Calculate factorial.

Solution (Java):

public long factorial(int n) {
    if (n <= 1) return 1;
    long result = 1;
    for (int i = 2; i <= n; i++) {
        result *= i;
    }
    return result;
}

Solution (Python):

def factorial(n):
    result = 1
    for i in range(2, n + 1):
        result *= i
    return result

Time Complexity: O(n) | Space Complexity: O(1)

Fibonacci Series

Problem: Generate Fibonacci series up to n terms.

Solution (Java):

public int[] fibonacci(int n) {
    if (n <= 0) return new int[]{};
    int[] fib = new int[n];
    fib[0] = 0;
    if (n > 1) fib[1] = 1;
    for (int i = 2; i < n; i++) {
        fib[i] = fib[i-1] + fib[i-2];
    }
    return fib;
}

Solution (Python):

def fibonacci(n):
    if n <= 0:
        return []
    fib = [0, 1]
    for i in range(2, n):
        fib.append(fib[-1] + fib[-2])
    return fib[:n]

Time Complexity: O(n) | Space Complexity: O(n)

GCD of Two Numbers

Problem: Find GCD using Euclidean algorithm.

Solution (Java):

public int gcd(int a, int b) {
    while (b != 0) {
        int temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}

Solution (Python):

def gcd(a, b):
    while b:
        a, b = b, a % b
    return a

Time Complexity: O(log(min(a,b))) | Space Complexity: O(1)

Armstrong Number

Problem: Check if number is Armstrong number.

Example:

Input: 153
Output: true (1³ + 5³ + 3³ = 153)

Solution (Java):

public boolean isArmstrong(int n) {
    int original = n;
    int digits = String.valueOf(n).length();
    int sum = 0;
    while (n > 0) {
        int digit = n % 10;
        sum += Math.pow(digit, digits);
        n /= 10;
    }
    return sum == original;
}

Solution (Python):

def is_armstrong(n):
    digits = len(str(n))
    return sum(int(d)**digits for d in str(n)) == n

Time Complexity: O(d) | Space Complexity: O(1)

Practice tips for Capgemini assessment

Master pseudocode

• Practice tracing code
• Understand loops and conditionals
• Know array indexing
• Read carefully for edge cases

Master one language

• Choose C, C++, Java, or Python
• Know standard library functions
• Practice writing code quickly
• Handle I/O properly

Time management

• Pseudocode: ~1.5 min/question
• Coding: ~20 min/problem
• Read problem carefully
• Test with given examples

Common mistakes to avoid

• Off-by-one errors
• Not handling empty inputs
• Wrong output format
• Not considering edge cases

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