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1049. Last Stone Weight II

Description

You are given an array of integers stones where stones[i] is the weight of the ith stone.

We are playing a game with the stones. On each turn, we choose any two stones and smash them together. Suppose the stones have weights x and y with x <= y. The result of this smash is:

  • If x == y, both stones are destroyed, and
  • If x != y, the stone of weight x is destroyed, and the stone of weight y has new weight y - x.

At the end of the game, there is at most one stone left.

Return the smallest possible weight of the left stone. If there are no stones left, return 0.

 

Example 1:

Input: stones = [2,7,4,1,8,1]
Output: 1
Explanation:
We can combine 2 and 4 to get 2, so the array converts to [2,7,1,8,1] then,
we can combine 7 and 8 to get 1, so the array converts to [2,1,1,1] then,
we can combine 2 and 1 to get 1, so the array converts to [1,1,1] then,
we can combine 1 and 1 to get 0, so the array converts to [1], then that's the optimal value.

Example 2:

Input: stones = [31,26,33,21,40]
Output: 5

 

Constraints:

  • 1 <= stones.length <= 30
  • 1 <= stones[i] <= 100

Solutions

Solution 1

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class Solution:
    def lastStoneWeightII(self, stones: List[int]) -> int:
        s = sum(stones)
        m, n = len(stones), s >> 1
        dp = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(1, m + 1):
            for j in range(n + 1):
                dp[i][j] = dp[i - 1][j]
                if stones[i - 1] <= j:
                    dp[i][j] = max(
                        dp[i][j], dp[i - 1][j - stones[i - 1]] + stones[i - 1]
                    )
        return s - 2 * dp[-1][-1]

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class Solution {
    public int lastStoneWeightII(int[] stones) {
        int s = 0;
        for (int v : stones) {
            s += v;
        }
        int m = stones.length;
        int n = s >> 1;
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 1; i <= m; ++i) {
            for (int j = 0; j <= n; ++j) {
                dp[i][j] = dp[i - 1][j];
                if (stones[i - 1] <= j) {
                    dp[i][j] = Math.max(dp[i][j], dp[i - 1][j - stones[i - 1]] + stones[i - 1]);
                }
            }
        }
        return s - dp[m][n] * 2;
    }
}

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class Solution {
public:
    int lastStoneWeightII(vector<int>& stones) {
        int s = accumulate(stones.begin(), stones.end(), 0);
        int m = stones.size(), n = s >> 1;
        vector<vector<int>> dp(m + 1, vector<int>(n + 1));
        for (int i = 1; i <= m; ++i) {
            for (int j = 0; j <= n; ++j) {
                dp[i][j] = dp[i - 1][j];
                if (stones[i - 1] <= j) dp[i][j] = max(dp[i][j], dp[i - 1][j - stones[i - 1]] + stones[i - 1]);
            }
        }
        return s - dp[m][n] * 2;
    }
};

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func lastStoneWeightII(stones []int) int {
    s := 0
    for _, v := range stones {
        s += v
    }
    m, n := len(stones), s>>1
    dp := make([][]int, m+1)
    for i := range dp {
        dp[i] = make([]int, n+1)
    }
    for i := 1; i <= m;