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.
implSolution{#[allow(dead_code)]pubfnlast_stone_weight_ii(stones:Vec<i32>)->i32{letn=stones.len();letmutsum=0;forein&stones{sum+=*e;}letm=(sum/2)asusize;letmutdp:Vec<Vec<i32>>=vec![vec![0;m+1];n+1];// Begin the actual dp processforiin1..=n{forjin1..=m{dp[i][j]=ifstones[i-1]>(jasi32){dp[i-1][j]}else{std::cmp::max(dp[i-1][j],dp[i-1][j-(stones[i-1]asusize)]+stones[i-1],)};}}sum-2*dp[n][m]}}