LeetCode 1049: Last Stone Weight II Solution

Master LeetCode problem 1049 (Last Stone Weight II), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1049. Last Stone Weight II

Medium

Array Dynamic Programming

Problem Explanation

Explanation

To solve this problem, we can reframe it as a subset sum problem. We need to divide the stones into two groups such that their weights are as close as possible. This is equivalent to finding a subset with the sum closest to half of the total sum of all stones. We can use dynamic programming to solve this problem efficiently.

Calculate the total sum of all stones.
Initialize a dp array of size half of the total sum + 1 to represent if a subset sum can be achieved using stones.
For each stone, update the dp array with the possible subset sums.
The answer will be the difference between the total sum and twice the largest subset sum that can be achieved.

Time complexity: O(n * sum/2), where n is the number of stones and sum is the total sum of all stones. Space complexity: O(sum/2) for the dp array.

Solution Code

class Solution {
    public int lastStoneWeightII(int[] stones) {
        int totalSum = 0;
        for (int stone : stones) {
            totalSum += stone;
        }

        int target = totalSum / 2;
        boolean[] dp = new boolean[target + 1];
        dp[0] = true;

        for (int stone : stones) {
            for (int i = target; i >= stone; i--) {
                dp[i] = dp[i] || dp[i - stone];
            }
        }

        for (int i = target; i >= 0; i--) {
            if (dp[i]) {
                return totalSum - 2 * i;
            }
        }

        return totalSum;
    }
}

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