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:
x == y
, both stones are totally destroyed;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 <= stones.length <= 30
1 <= stones[i] <= 100
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);
}