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1046. Last Stone Weight

Description

You are given an array of integers stones where stones[i] is the weight of the ith stone.

We are playing a game with the stones. On each turn, we choose the heaviest two stones and smash them together. Suppose the heaviest two stones have weights x and y with x <= y. The result of this smash is:

  • If x == y, both stones are destroyed, and
  • If x != y, the stone of weight x is destroyed, and the stone of weight y has new weight y - x.

At the end of the game, there is at most one stone left.

Return the weight of the last remaining stone. If there are no stones left, return 0.

 

Example 1:

Input: stones = [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 the last stone.

Example 2:

Input: stones = [1]
Output: 1

 

Constraints:

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

Solutions

Solution 1

Thinking

Each smash uses the two heaviest stones. Scanning for the maximum every time repeats comparisons; a heap extracts them in logarithmic time.

Negating the weights turns a min-heap into a max-heap. Pop \(y\ge x\) and push \(y-x\) when they differ, until fewer than two stones remain.

An empty heap is \(0\); otherwise the answer is the negated top.

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class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        h = [-x for x in stones]
        heapify(h)
        while len(h) > 1:
            y, x = -heappop(h), -heappop(h)
            if x != y:
                heappush(h, x - y)
        return 0 if not h else -h[0]

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class Solution {
    public int lastStoneWeight(int[] stones) {
        PriorityQueue<Integer> q = new PriorityQueue<>((a, b) -> b - a);
        for (int x : stones) {
            q.offer(x);
        }
        while (q.size() > 1) {
            int y = q.poll();
            int x = q.poll();
            if (x != y) {
                q.offer(y - x);
            }
        }
        return q.isEmpty() ? 0 : q.poll();
    }
}

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class Solution {
public:
    int lastStoneWeight(vector<int>& stones) {
        priority_queue<int> pq;
        for (int x : stones) {
            pq.push(x);
        }
        while (pq.size() > 1) {
            int y = pq.top();
            pq.pop();
            int x = pq.top();
            pq.pop();
            if (x != y) {
                pq.push(y - x);
            }
        }
        return pq.empty() ? 0 : pq.top();
    }
};

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func lastStoneWeight(stones []int) int {
    q := &hp{stones}
    heap.Init(q)
    for q.Len() > 1 {
        y, x := q.pop(), q.pop()
        if x != y {
            q.push(y - x)
        }
    }
    if q.Len() > 0 {
        return q.IntSlice[0]
    }
    return 0
}

type hp struct{ sort.IntSlice }

func (h hp) Less(i, j int) bool { return h.IntSlice[i] > h.IntSlice[j] }
func (h *hp) Push(v any)        { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
    a := h.IntSlice
    v := a[len(a)-1]
    h.IntSlice = a[:len(a)-1]
    return v
}
func (h *hp) push(v int) { heap.Push(h, v) }
func (h *hp) pop() int   { return heap.Pop(h).(int) }

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function lastStoneWeight(stones: number[]): number {
    const pq = new MaxPriorityQueue<number>();
    for (const x of stones) {
        pq.enqueue(x);
    }
    while (pq.size() > 1) {
        const y = pq.dequeue();
        const x = pq.dequeue();
        if (x !== y) {
            pq.enqueue(y - x);
        }
    }
    return pq.isEmpty() ? 0 : pq.dequeue();
}

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/**
 * @param {number[]} stones
 * @return {number}
 */
var lastStoneWeight = function (stones) {
    const pq = new MaxPriorityQueue();
    for (const x of stones) {
        pq.enqueue(x);
    }
    while (pq.size() > 1) {
        const y = pq.dequeue();
        const x = pq.dequeue();
        if (x != y) {
            pq.enqueue(y - x);
        }
    }
    return pq.isEmpty() ? 0 : pq.dequeue();
};

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