1046. Last Stone Weight

Question

LeetCode Problem

You are given an array of integers stones where stones[i] is the weight of the i th 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

int lastStoneWeight(vector<int> &stones) {
	// this problem doesn't mention the stones are sorted.
    // we can use priority queue and use push/pop is more easy.
    // Time Complexity: O(NlogN).
    // Space Complexity: O(N).
    priority_queue<int> pq(stones.begin(), stones.end());
    while (pq.size() > 1) {
      int y = pq.top();
      pq.pop();
      int x = pq.top();
      pq.pop();
      if (y > x)
        pq.push(y - x);
    }

    return pq.empty() ? 0 : pq.top();
}