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Last Stone Weight II⚓︎

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Description⚓︎

You are given an array of integers stones where stones[i] is the weight of the i-th 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

Solution⚓︎

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class Solution {
public:
    int lastStoneWeightII(vector<int>& stones) {
        int sum = 0;
        for (int x : stones) sum += x;
        vector<int> dp(sum + 1, 0);
        for (int i = 0; i < stones.size(); i++) {
            for (int j = sum / 2; j >= stones[i]; j--)
                dp[j] = max(dp[j], dp[j - stones[i]] + stones[i]);
        }
        return sum - 2 * dp[sum / 2];
    }
};