[LeetCode]Last Stone Weight II@Golang

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

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.

Solution

func lastStoneWeightII(stones []int) int {
    hassum := [3001]bool{}
    sum := 0
    hassum[0] = true
    
    for _,v := range stones{
        sum += v
        for i:=sum;i>=v;i-- {
            hassum[i] = hassum[i] || hassum[i-v]
        }
    }
    
    less := sum/2
    for less>=0 && !hassum[less] {
        less --
    }
    return sum-less-less
}