## Algorithm

Problem Name: 1049. Last Stone Weight II

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`

## Code Examples

### #1 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public int lastStoneWeightII(int[] stones) {
int sum = 0;
for (int stone : stones) {
sum += stone;
}
int[] dp = new int[sum / 2 + 1];
for (int i = 1; i  < = stones.length; i++) {
for (int j = sum / 2; j >= stones[i - 1]; j--) {
dp[j] = Math.max(dp[j], dp[j - stones[i - 1]] + stones[i - 1]);
}
}
return sum - 2 * dp[sum / 2];
}
}
``````
Copy The Code &

Input

cmd
stones = [2,7,4,1,8,1]

Output

cmd
1

### #2 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const lastStoneWeightII = function(stones) {
let sum=stones.reduce((a,b)=>a+b)
let dp=Array(sum+1).fill(0)
dp[0]=1
for(let i=0;i < stones.length;i++){
let cur=stones[i]
for(let j=dp.length-1;j>=0;j--){
if(j-stones[i]<0)break
if(dp[j-stones[i]]){
dp[j]=1
}
}
}

let minLen=Infinity
for(let i=0;i < dp.length;i++){
if(dp[i]>{
if(i*2-sum>=0)minLen=Math.min(minLen,i*2-sum)
}
}
return minLen
};
``````
Copy The Code &

Input

cmd
stones = [2,7,4,1,8,1]

Output

cmd
1

### #3 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def lastStoneWeightII(self, A: List[int]) -> int:
dp = {0}
sumA = sum(A)
for a in A:
dp |= {a + i for i in dp}
return min(abs(sumA - i - i) for i in dp)
``````
Copy The Code &

Input

cmd
stones = [31,26,33,21,40]

Output

cmd
5