Last Stone Weight

Problem Statement:

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.


Solution:
class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        while len(stones)>1:
            stones.sort()
            a,b=map(int,stones[-2:])
            print(a,b)
            if a<=b:
                if a==b:
                    stones.remove(a)
                    stones.remove(b)
                else:
                    stones.remove(a)
                    stones[stones.index(b)]=b-a
        if stones:
            return(stones[0])
        else:
            return 0