1049

1049. Last Stone Weight II (Medium)

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

Example 3:

Input: stones = [1,2]
Output: 1

 

Constraints:

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

Related Topics:
Dynamic Programming

Solution 1.

// OJ: https://leetcode.com/problems/last-stone-weight-ii/
// Author: github.com/lzl124631x
// Time: O(N + SUM(A))
// Space: O(SUM(A))
// Ref: https://leetcode.com/problems/last-stone-weight-ii/discuss/294888/JavaC%2B%2BPython-Easy-Knapsacks-DP
class Solution {
public:
    int lastStoneWeightII(vector<int>& A) {
        bitset<1501> dp = {1};
        int sum = 0;
        for (int a : A) {
            sum += a;
            for (int i = min(1500, sum); i >= a; --i) {
                dp[i] = dp[i] + dp[i - a];
            }
        }
        for (int i = sum / 2; i >= 0; --i) {
            if (dp[i]) return sum - i - i;
        }
        return 0;
    }
};