What is the optimal time complexity of solving Last Stone Weight?

The optimized solution achieves a time complexity of O(n) by using sorting, mathematical shortcuts, or hash table lookups.

What is the auxiliary space complexity for Last Stone Weight?

The space complexity is O(1) since we only use a minimal/constant amount of extra memory for tracking variables (or auxiliary space if dynamic structures are allocated).

What are the typical edge cases we must handle in Last Stone Weight?

Key edge cases include empty collections, negative or large bounds causing integer overflow, arrays with only one element, or inputs where target criteria are not met.

How can we optimize the brute force approach for Last Stone Weight?

Brute force methods often inspect all possible combinations. We optimize this by sorting the input list to enable binary search, or tracking processed items in a set to avoid redundant nested loop lookups.

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Last Stone Weight

Learn how to solve the 'Last Stone Weight' problem. This detailed resource details brute force and optimized approaches.

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Problem Statement

Easy

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,

- 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 no stones are left, return 0.

Write a function lastStoneWeight(stones: List[int]) -> int.

Constraints

•1 <= len(stones) <= 30
•1 <= stones[i] <= 1000

Examples

Example 1

Input

stones = [2,7,4,1,8,1]

Output

Explanation

Smash 7 and 8 to get 1, array becomes [2,4,1,1,1]. Smash 2 and 4 to get 2, array becomes [2,1,1,1]. Smash 2 and 1 to get 1, array becomes [1,1,1]. Smash 1 and 1 to get 0, array becomes [1]. The last remaining stone is 1.

Example 2

Input

stones = [1]

Output

Explanation

Only one stone, so weight is 1.

Need a Hint?

Analyze the input constraints. Try sorting first (O(n log n)) or using a hash map/set to track seen elements in O(n) time.

Edge Cases to Watch

Empty list or null input variables
Single item lists/arrays
Extremely large input bounds causing integer or stack overflow

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