# LC 1049. Last Stone Weight II

### [Problem link](https://leetcode.com/problems/last-stone-weight-ii/)

###### tags: `leedcode` `python` `c++` `medium` `DP`

You are given an array of integers <code>stones</code> where <code>stones[i]</code> is the weight of the <code>ith</code> 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 <code>x</code> and <code>y</code> with <code>x <= y</code>. The result of this smash is:

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

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 <code>0</code>.

**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:**

- <code>1 <= stones.length <= 30</code>
- <code>1 <= stones[i] <= 100</code>

## Solution 1 - DP
#### Python

class Solution:
    def lastStoneWeightII(self, stones: List[int]) -> int:
        target = sum(stones) // 2

        dp = [0] * (target + 1)

        for i in range(len(stones)):
            for j in range(target, stones[i] - 1, -1):
                dp[j] = max(dp[j], dp[j - stones[i]] + stones[i])

        return sum(stones) - dp[-1] * 2

#### C++

class Solution {
public:
    int lastStoneWeightII(vector<int>& stones) {
        int sum = accumulate(stones.begin(), stones.end(), 0);
        int target = sum / 2;

        vector<int> dp(target + 1, 0);
        for (int i = 0; i < stones.size(); i++) {
            for (int j = target; j >= stones[i]; j--) {
                dp[j] = max(dp[j], dp[j - stones[i]] + stones[i]);
            }
        }
        return sum - dp[target] - dp[target];
    }
};

>### Complexity
>m = int(sum(stones) / 2)
>n = stones.length
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(mn) | O(m) |

## Note
[reference](https://github.com/youngyangyang04/leetcode-master/blob/master/problems/1049.%E6%9C%80%E5%90%8E%E4%B8%80%E5%9D%97%E7%9F%B3%E5%A4%B4%E7%9A%84%E9%87%8D%E9%87%8FII.md)