Last Stone Weight

April 12, 2020

Introduction

We have a collection of stones, each stone has a positive integer weight.

Each turn, we choose the two heaviest 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 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 weight of this stone (or 0 if there are no stones left.)

Example 1:

Input: [2,7,4,1,8,1]

Output: 1 Explanation:

We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,
we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,
we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of last stone.

Note:

1 <= stones.length <= 30
1 <= stones[i] <= 1000

Solution

The simple solution with sorting gives pretty good performance.

70 / 70 test cases passed.
Status: Accepted
Runtime: 0 ms
Memory Usage: 2 MB

Therefore, it is ok to use it:

func lastStoneWeight(stones []int) int {
    n := len(stones)
    switch n {
        case 0:
            return 0
        case 1:
            return stones[0]
        case 2:
            return abs(stones[0], stones[1])
        default:
            sort.Ints(stones)
            stones[n-2] = abs(stones[n-1], stones[n-2])
            return lastStoneWeight(stones[:n-1])
    }
}

func abs(a, b int) int {
    x := a - b
    if x < 0 {
        x = -x
    }
    return x
}