Goldman Sachs Placement Papers 2025 - Latest Questions & Solutions

Goldman Sachs 2025 Placement Papers

This page contains actual Goldman Sachs placement papers from 2025 with latest HackerRank assessment questions.

2025 Updates

Goldman Sachs 2025 added system design questions for senior roles. More emphasis on live coding during interviews. Finance domain questions increased.

Goldman Sachs 2025 Assessment Pattern

Round	Duration	Content
HackerRank OA	90 min	2 coding + 10 aptitude MCQs
Technical 1	60 min	DSA + live coding
Technical 2	60 min	System Design basics + CS
HR	45 min	Behavioral + finance knowledge

Aptitude MCQs (2025)

Question 1: Expected Value

A game: Win $100 with 30% probability, lose $50 with 70% probability. Expected value?

Answer: 0.3 × 100 + 0.7 × (-50) = 30 - 35 = -$5

Question 2: Time Value of Money

Present value of $10,000 received after 2 years at 10% discount rate?

Answer: 10000 / (1.1)² = $8,264.46

Question 3: Combinations

Select 3 people from 10 for a committee. How many ways?

Answer: C(10,3) = 10!/(3!×7!) = 120

Question 4: Bonds

A bond with face value $1000 trades at $950. Is it trading at premium or discount?

Answer: Discount (price < face value)

Coding Problems (2025)

Problem 1: Maximum Subarray Sum

Problem

Difficulty: Medium
Time: 25 minutes

Find maximum sum of contiguous subarray (Kadane’s algorithm).

Input: [-2,1,-3,4,-1,2,1,-5,4]
Output: 6

Java

public int maxSubArray(int[] nums) {
    int maxSum = nums[0];
    int currentSum = nums[0];
    for (int i = 1; i < nums.length; i++) {
        currentSum = Math.max(nums[i], currentSum + nums[i]);
        maxSum = Math.max(maxSum, currentSum);
    }
    return maxSum;
}

Python

def max_subarray(nums):
    max_sum = current_sum = nums[0]
    for num in nums[1:]:
        current_sum = max(num, current_sum + num)
        max_sum = max(max_sum, current_sum)
    return max_sum

Problem 2: Stock Buy Sell with Cooldown

Problem

Difficulty: Medium
Time: 35 minutes

Maximum profit with unlimited transactions but must wait 1 day after selling before buying again.

Input: [1,2,3,0,2]
Output: 3

Java

public int maxProfit(int[] prices) {
    if (prices.length <= 1) return 0;

    int hold = Integer.MIN_VALUE;
    int sold = 0;
    int rest = 0;

    for (int price : prices) {
        int prevSold = sold;
        sold = hold + price;
        hold = Math.max(hold, rest - price);
        rest = Math.max(rest, prevSold);
    }
    return Math.max(sold, rest);
}

Python

def max_profit(prices):
    if len(prices) <= 1:
        return 0

    hold = float('-inf')
    sold = 0
    rest = 0

    for price in prices:
        prev_sold = sold
        sold = hold + price
        hold = max(hold, rest - price)
        rest = max(rest, prev_sold)

    return max(sold, rest)

Problem 3: Coin Change

Problem

Difficulty: Medium
Time: 30 minutes

Minimum coins needed to make amount.

Input: coins = [1,2,5], amount = 11
Output: 3 (5+5+1)

Java

public int coinChange(int[] coins, int amount) {
    int[] dp = new int[amount + 1];
    Arrays.fill(dp, amount + 1);
    dp[0] = 0;

    for (int i = 1; i <= amount; i++) {
        for (int coin : coins) {
            if (coin <= i) {
                dp[i] = Math.min(dp[i], dp[i - coin] + 1);
            }
        }
    }
    return dp[amount] > amount ? -1 : dp[amount];
}

Python

def coin_change(coins, amount):
    dp = [float('inf')] * (amount + 1)
    dp[0] = 0

    for i in range(1, amount + 1):
        for coin in coins:
            if coin <= i:
                dp[i] = min(dp[i], dp[i - coin] + 1)

    return dp[amount] if dp[amount] != float('inf') else -1

Problem 4: Word Break

Problem

Difficulty: Medium
Time: 30 minutes

Can string be segmented into dictionary words?

Input: s = “leetcode”, wordDict = [“leet”,“code”]
Output: true

Java

public boolean wordBreak(String s, List<String> wordDict) {
    Set<String> wordSet = new HashSet<>(wordDict);
    boolean[] dp = new boolean[s.length() + 1];
    dp[0] = true;

    for (int i = 1; i <= s.length(); i++) {
        for (int j = 0; j < i; j++) {
            if (dp[j] && wordSet.contains(s.substring(j, i))) {
                dp[i] = true;
                break;
            }
        }
    }
    return dp[s.length()];
}

Python

def word_break(s, word_dict):
    word_set = set(word_dict)
    dp = [False] * (len(s) + 1)
    dp[0] = True

    for i in range(1, len(s) + 1):
        for j in range(i):
            if dp[j] and s[j:i] in word_set:
                dp[i] = True
                break

    return dp[len(s)]

Key Insights 2025

1. Finance Emphasis: More questions on time value, expected value, bonds
2. Live Coding: Interview rounds include live coding with discussion
3. System Design: Added for experienced candidates
4. DP Focus: Dynamic programming problems are common
5. Clean Code: Emphasis on production-quality code

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Last updated: February 2026