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 }