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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();
    }
};

Java

• Java
• C++
• Python

• 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:
      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]
  

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