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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. 1 <= stones.length <= 30
  2. 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;
    }
    
    

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