Last Stone Weight II

May 20, 2019

Introduction

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 <= stones.length <= 30
1 <= stones[i] <= 100

Solution

Golang:

func lastStoneWeightII(stones []int) int {
    
    n := len(stones)
    if n == 0 {
        return 0
    }

    sum := 0
    for _, v := range stones {
        sum += v
    }

    target := sum / 2
    dp := make([]bool, target + 1)
    dp[0] = true
    
    for _, v := range stones {
        for j := target; j >= v; j-- {
            dp[j] = dp[j] || dp[j - v]
        }
    }

    for i := target; i > 0; i-- {
        if dp[i] {
            return sum - 2 * i
        }
    }

    return 0
}

Explanation

This problem can be transformed in to Partition a set into two subsets such that the difference of subset sums is minimum and solved by using dynamic programming.