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();
    }
};

Java

  • 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();
        }
    }
    
  • // 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(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]
    

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