Last Stone Weight

Asked byVisaNvidiaAmazonSalesforce

Problem

You have a collection of stones with given weights. Repeatedly smash together the two heaviest stones: if they're equal, both are destroyed; otherwise the lighter one is destroyed and the heavier one's weight is reduced by the lighter's weight. Return the weight of the last remaining stone, or 0 if none remain.

Examples

Example 1

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

Output:1

Constraints

1 <= stones.length <= 30

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How to approach it: the Heap / Priority Queue pattern

A heap keeps the minimum (or maximum) element accessible in O(1), with O(log n) insert and remove. It's the tool whenever you repeatedly need "the smallest/largest remaining item" without needing everything fully sorted.

Look for this pattern when

You need the top-k largest/smallest elements, not a full sort.
You're merging multiple sorted sequences (always take the smallest available head).
You need a running min/max/median as data streams in.

Read the full Heap / Priority Queue guide →

Video walkthroughs

Original problem on LeetCode ↗