Solved! Leetcode 1046. Last Stone Weight

source: https://leetcode.com/problems/last-stone-weight/description/

Last Stone Weight

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

Let’s say n is the number of stones in an array
Time Complexity
O(nlogn)

Space Complexity
O(n)

1/5 - (6 votes)

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