Last Stone Weight (LeetCode Day 12 Question) | Programming Tutorials

Given 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 is less than equal to 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.)


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.

This problem is the day 12 challenge of LeetCode 30 day challenge.

[Shishiraithal]

Also we can reverse the priority queue ordering If we add elements with - negetive sign