Capgemini 2024 Papers
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Capgemini placement papers 2025 with latest pseudocode questions, hands-on coding problems, and solutions from 2025 recruitment.
This page contains actual Capgemini placement papers from 2025 with latest game-based assessment questions.
| 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 |
FUNCTION sum(n) IF n == 0 THEN RETURN 0 END IF RETURN n + sum(n - 1)END FUNCTIONPRINT sum(5)Answer: 15
Explanation: 5+4+3+2+1+0 = 15
SET arr = [1, 3, 5, 7, 9]SET target = 7SET low = 0, high = 4, count = 0WHILE 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 IFEND WHILEAnswer: 2 (mid=2 gives 5, then mid=3 gives 7)
SET matrix = [[1,2],[3,4]]SET sum = 0FOR i FROM 0 TO 1 FOR j FROM 0 TO 1 SET sum = sum + matrix[i][j] END FOREND FORPRINT sumAnswer: 10
Explanation: 1+2+3+4 = 10
SET str = "CAPGEMINI"SET count = 0FOR i FROM 0 TO LENGTH(str) - 1 IF str[i] == 'I' THEN SET count = count + 1 END IFEND FORPRINT countAnswer: 2
Explanation: Two ‘I’s in CAPGEMINI
CREATE STACK sPUSH(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.
FUNCTION gcd(a, b) IF b == 0 THEN RETURN a END IF RETURN gcd(b, a MOD b)END FUNCTIONPRINT gcd(48, 18)Answer: 6
Explanation: gcd(48,18)→gcd(18,12)→gcd(12,6)→gcd(6,0)→6
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 IFEND FORAnswer: [3, 5, 4, 8]
Explanation: 5>3 swap→[3,5,8,4], 5<8 no swap, 8>4 swap→[3,5,4,8]
FUNCTION power(x, n) IF n == 0 THEN RETURN 1 END IF RETURN x * power(x, n - 1)END FUNCTIONPRINT power(2, 4)Answer: 16
Explanation: 2^4 = 16
Rotate array right by k positions.
Input: [1,2,3,4,5], k=2
Output: [4,5,1,2,3]
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; }}def rotate(nums, k): n = len(nums) k = k % n nums[:] = nums[-k:] + nums[:-k]Find indices of two numbers that add up to target.
Input: [2,7,11,15], target=9
Output: [0,1]
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[]{};}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 []Remove duplicates from sorted array in-place.
Input: [1,1,2,2,3]
Output: 3 (array becomes [1,2,3,…])
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;}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 + 1Capgemini 2024 Papers
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Last updated: February 2026