1049. Last Stone Weight II

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

1049. Last Stone Weight II
struct Solution;

impl Solution {
    fn last_stone_weight_ii(stones: Vec<i32>) -> i32 {
        let sum = stones.iter().sum::<i32>() as usize;
        let mut dp = vec![false; sum + 1];
        dp[0] = true;
        let n = stones.len();
        for i in 0..n {
            for j in (1..=sum).rev() {
                if j >= stones[i] as usize && dp[j - stones[i] as usize] {
                    dp[j] = true;
                }
            }
        }
        let mut res = sum;
        for i in 0..=sum / 2 {
            if dp[i] {
                res = res.min(sum - 2 * i);
            }
        }
        res as i32
    }
}

#[test]
fn test() {
    let stones = vec![2, 7, 4, 1, 8, 1];
    let res = 1;
    assert_eq!(Solution::last_stone_weight_ii(stones), res);
}