Capgemini Placement Papers 2025

This page contains actual Capgemini placement papers from 2025 with latest game-based assessment questions.

2025 Updates

Capgemini 2025 introduced adaptive difficulty - harder questions if you answer correctly. More focus on recursion and data structures in pseudocode.

Capgemini 2025 assessment pattern

Round	Questions	Time	Notes
Game-Based	15 MCQs	25 min	Adaptive difficulty
Hands-on Coding	2 problems	45 min	Partial marking
Technical	-	30-45 min	CS fundamentals
HR	-	20-30 min	Behavioral

Pseudocode questions (2025 latest)

Question 1: recursive sum

What is the output?

FUNCTION sum(n)
    IF n == 0 THEN
        RETURN 0
    END IF
    RETURN n + sum(n - 1)
END FUNCTION
PRINT sum(5)

Answer: 15

Explanation: 5+4+3+2+1+0 = 15

Question 2: binary search

How many comparisons for arr=[1,3,5,7,9] searching for 7?

SET arr = [1, 3, 5, 7, 9]
SET target = 7
SET low = 0, high = 4, count = 0
WHILE low <= high
    SET mid = (low + high) / 2
    SET count = count + 1
    IF arr[mid] == target THEN
        PRINT count
        EXIT
    ELSE IF arr[mid] < target THEN
        SET low = mid + 1
    ELSE
        SET high = mid - 1
    END IF
END WHILE

Answer: 2 (mid=2 gives 5, then mid=3 gives 7)

Question 3: 2d array

What is the output?

SET matrix = [[1,2],[3,4]]
SET sum = 0
FOR i FROM 0 TO 1
    FOR j FROM 0 TO 1
        SET sum = sum + matrix[i][j]
    END FOR
END FOR
PRINT sum

Answer: 10

Explanation: 1+2+3+4 = 10

Question 4: string operations

What is the output?

SET str = "CAPGEMINI"
SET count = 0
FOR i FROM 0 TO LENGTH(str) - 1
    IF str[i] == 'I' THEN
        SET count = count + 1
    END IF
END FOR
PRINT count

Answer: 2

Explanation: Two ’I’s in CAPGEMINI

Question 5: stack operations

What is at top after operations?

CREATE STACK s
PUSH(s, 10)
PUSH(s, 20)
POP(s)
PUSH(s, 30)
PRINT TOP(s)

Answer: 30

Explanation: Push 10, push 20, pop (removes 20), push 30. Top is 30.

Question 6: gcd

What is the output for gcd(48, 18)?

FUNCTION gcd(a, b)
    IF b == 0 THEN
        RETURN a
    END IF
    RETURN gcd(b, a MOD b)
END FUNCTION
PRINT gcd(48, 18)

Answer: 6

Explanation: gcd(48,18)→gcd(18,12)→gcd(12,6)→gcd(6,0)→6

Question 7: bubble sort pass

Array after first pass?

SET arr = [5, 3, 8, 4]
FOR i FROM 0 TO 2
    IF arr[i] > arr[i+1] THEN
        SWAP arr[i] AND arr[i+1]
    END IF
END FOR

Answer: [3, 5, 4, 8]

Explanation: 5>3 swap→[3,5,8,4], 5<8 no swap, 8>4 swap→[3,5,4,8]

Question 8: power function

What is the output?

FUNCTION power(x, n)
    IF n == 0 THEN
        RETURN 1
    END IF
    RETURN x * power(x, n - 1)
END FUNCTION
PRINT power(2, 4)

Answer: 16

Explanation: 2^4 = 16

Hands-on coding (2025)

Problem 1: rotate array

Problem

Rotate array right by k positions.

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

Java

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

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

Python

def rotate(nums, k):
    n = len(nums)
    k = k % n
    nums[:] = nums[-k:] + nums[:-k]

Problem 2: two sum

Problem

Find indices of two numbers that add up to target.

Input: [2,7,11,15], target=9
Output: [0,1]

Java

public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement)) {
            return new int[]{map.get(complement), i};
        }
        map.put(nums[i], i);
    }
    return new int[]{};
}

Python

def two_sum(nums, target):
    seen = {}
    for i, num in enumerate(nums):
        complement = target - num
        if complement in seen:
            return [seen[complement], i]
        seen[num] = i
    return []

Problem 3: remove duplicates

Problem

Remove duplicates from sorted array in-place.

Input: [1,1,2,2,3]
Output: 3 (array becomes [1,2,3,…])

Java

public int removeDuplicates(int[] nums) {
    if (nums.length == 0) return 0;
    int i = 0;
    for (int j = 1; j < nums.length; j++) {
        if (nums[j] != nums[i]) {
            nums[++i] = nums[j];
        }
    }
    return i + 1;
}

Python

def remove_duplicates(nums):
    if not nums: return 0
    i = 0
    for j in range(1, len(nums)):
        if nums[j] != nums[i]:
            i += 1
            nums[i] = nums[j]
    return i + 1

Key insights 2025

1. Adaptive Questions: Questions get harder based on performance
2. Recursion Heavy: More recursive pseudocode questions
3. Data Structure Pseudocode: Stack, queue operations in pseudocode
4. Partial Marking: Coding has partial marks now
5. 35% Success Rate: Pseudocode round filters most candidates

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