1046. Last Stone Weight

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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.

Note:
  1. 1 <= stones.length <= 30
  2. 1 <= stones[i] <= 1000
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Intuition
We need a data structure that keeps weights ordered in descending
We can pick up last 2 biggest elements
If same discard
if different, add bigger - smaller back into the ordered data structure

Use Max Heap
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Time - O(n log n)
Space - O(n)
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