004_Last_Stone_Weight.py

# Problem: https://leetcode.com/problems/last-stone-weight/
"""
We have a collection of rocks, each rock has a positive integer weight.

Each turn, we choose the two heaviest rocks 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."""

import heapq
class Solution:
    def lastStoneWeight(self, stones):
        # we will use heappop which pops smallest elmenent
        # Since we need largest, we are adding minus(-) and pick it using heappop
        pq = [-x for x in stones]
        heapq.heapify(pq)
        for i in range(len(stones)-1):
            x, y = -heapq.heappop(pq), -heapq.heappop(pq)
            heapq.heappush(pq, -abs(x-y))
        return -pq[0]