1049. Last Stone Weight II

https://leetcode.com/problems/last-stone-weight-ii/

Problem

We have a collection of rocks, each rock has a positive integer weight.

Each turn, we choose any two rocks 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 totally destroyed;

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

At the end, there is at most 1 stone left. Return the smallest possible weight of this stone (the weight is 0 if there are no stones left.)

Example 1:

Input: [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.

Note:

1. 1 <= stones.length <= 30

2. 1 <= stones[i] <= 100

Solution

class Solution {
public:
    int lastStoneWeightII(vector<int>& stones) {
        int sum = 0;
        for (const auto n: stones) sum += n;
        const int S = sum / 2 + 1;
        bool dp[S];
        memset(dp, 0, S);
        dp[0] = true;
        for (int i = 0; i < stones.size(); ++i) {
            for (int s = S - 1; s >= stones[i]; s--) {
                dp[s] = dp[s] || dp[s - stones[i]];
            }
        }
        for (int s = S - 1; s > 0; --s) {
            if (dp[s]) return sum - 2 * s;
        }
        return -1;
    }
};

• #knapsack

• #dp

• #math