Formatted question description: https://leetcode.ca/all/1046.html

1046. Last Stone Weight (Easy)

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

Related Topics:
Heap, Greedy

Solution 1. Max-root Heap

// OJ: https://leetcode.com/problems/last-stone-weight/
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    int lastStoneWeight(vector<int>& A) {
        priority_queue<int> pq(begin(A), end(A));
        while (pq.size() > 1) {
            int a = pq.top();
            pq.pop();
            int b = pq.top();
            pq.pop();
            pq.push(a - b);
        }
        return pq.top();
    }
};

class Solution {
    public int lastStoneWeight(int[] stones) {
        PriorityQueue<Integer> priorityQueue = new PriorityQueue<Integer>(new Comparator<Integer>() {
            public int compare(Integer num1, Integer num2) {
                return num1 == null || num2 == null ? 0 : num2 - num1;
            }
        });
        for (int stone : stones)
            priorityQueue.offer(stone);
        while (priorityQueue.size() > 1) {
            int stone1 = priorityQueue.poll();
            int stone2 = priorityQueue.poll();
            if (stone1 > stone2)
                priorityQueue.offer(stone1 - stone2);
        }
        return priorityQueue.size() == 0 ? 0 : priorityQueue.poll();
    }
}

############

class Solution {
    public int lastStoneWeight(int[] stones) {
        PriorityQueue<Integer> q = new PriorityQueue<>((a, b) -> b - a);
        for (int v : stones) {
            q.offer(v);
        }
        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();
    }
}

// OJ: https://leetcode.com/problems/last-stone-weight/
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    int lastStoneWeight(vector<int>& A) {
        priority_queue<int> pq(begin(A), end(A));
        while (pq.size() > 1) {
            int a = pq.top(); pq.pop();
            int b = pq.top(); pq.pop();
            int r = abs(a - b);
            if (r) pq.push(r);
        }
        return pq.size() ? pq.top() : 0;
    }
};

class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        h = [-v for v in stones]
        heapify(h)
        while len(h) > 1:
            x = heappop(h)
            y = heappop(h)
            if x != y:
                heappush(h, x - y)
        return 0 if len(h) == 0 else -h[0]

############

class Solution(object):
    def lastStoneWeight(self, stones):
        """
        :type stones: List[int]
        :rtype: int
        """
        stones = map(lambda x : -x, stones)
        heapq.heapify(stones)
        while len(stones) > 1:
            x = heapq.heappop(stones)
            if stones:
                y = heapq.heappop(stones)
                if x != y:
                    heapq.heappush(stones, -abs(x - y))
        return 0 if not stones else -stones[0]

func lastStoneWeight(stones []int) int {
	q := &hp{stones}
	heap.Init(q)
	for q.Len() > 1 {
		x, y := q.pop(), q.pop()
		if x != y {
			q.push(x - y)
		}
	}
	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 interface{}) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() interface{} {
	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) }

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

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

1046 - Last Stone Weight

1046. Last Stone Weight (Easy)

Solution 1. Max-root Heap

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1046. Last Stone Weight (Easy)

Solution 1. Max-root Heap

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