What is the time complexity of Last Stone Weight?

The optimal solution has a time complexity of O(nlogn).

What data structures are used to solve Last Stone Weight?

This problem uses Array, Heap (Priority Queue). Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Last Stone Weight?

Leveraging Python's `heapq` as a Max-Heap: Since `heapq` is a min-heap, transforming all stone weights to their negative counterparts allows `heapq.heappop` to consistently retrieve the largest original stone weight. This is a crucial adaptation for problems requiring maximum elements from a heap in Python, avoiding the need for a custom max-heap implementation. The 'Heaviest Two' Implies a Priority Queue: The problem's core operation of repeatedly selecting the two heaviest stones strongly points towards using a priority queue (heap). It efficiently provides the largest elements (in `O(log N)` time per extraction) without needing to sort the entire collection repeatedly, which would be far less efficient for a dynamic set.

Which companies ask LeetCode 1046. Last Stone Weight in interviews?

This problem is frequently asked at Flipkart, Nvidia, PayPal, Rippling.

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Last Stone Weight - LeetCode 1046 Solution

Q: How to solve Last Stone Weight?

The core challenge is consistently identifying the 'heaviest two stones' from a constantly changing collection. This screams for a priority queue, specifically one that can give us the maximum elements. Python's `heapq` module implements a *min-heap*. To efficiently get the two largest positive weights, the solution employs a clever trick: it stores the *negative* of all stone weights in the heap. When you extract the smallest (most negative) elements from this modified min-heap, you are effectively getting the largest (most positive) original stone weights. The algorithm initializes the heap by converting all stone weights to their negatives and then heapifying the list. It then enters a loop that continues as long as there are at least two stones available to smash. In each iteration, it extracts the two smallest (most negative) elements, negates them back to their true positive values (`y` and `x`), and simulates the smash. If `x` and `y` are different, the resulting `y - x` stone is created, and its *negative* weight is pushed back into the heap. If `x` and `y` are the same, nothing is pushed back, correctly simulating both stones being destroyed. Finally, once the loop finishes, if any stone remains (meaning the heap is not empty), its original positive weight is returned; otherwise, 0 is returned.

Last Stone Weight - Complete Solution Guide

Last Stone Weight is LeetCode problem 1046, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an array of integers stones where stones[i] is the weight of the i th 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. Retu

Detailed Explanation

Imagine you're playing a peculiar game where you have a pile of stones, each with a specific weight. Your goal is to keep smashing them together until you're left with at most one stone. The rules for smashing are quite specific: in each turn, you *must* pick the two heaviest stones currently available. Let's call their weights 'x' and 'y', making sure 'x' is always less than or equal to 'y'. What happens after the smash? If 'x' and 'y' are identical, both stones vanish into dust. If they are different, the lighter stone ('x') is destroyed, but the heavier stone ('y') survives, albeit with a new, reduced weight of 'y - x'. This new stone is then thrown back into the pile. The game continues this cycle – pick two heaviest, smash, potentially add a new one – until there are either zero or one stone left. Your task is to report the weight of that final stone, or 0 if all stones were destroyed. The interesting aspect here is the dynamic nature: stones are removed, and a new stone might be added with a different weight. This isn't a simple one-time operation but a sequence of choices where each choice (the two heaviest stones) depends on the previous outcomes. You can't just pick any two; it has to be the largest, which means the set of available stones is constantly changing in a way that necessitates efficient access to the maximum values.

Solution Approach

The core challenge is consistently identifying the 'heaviest two stones' from a constantly changing collection. This screams for a priority queue, specifically one that can give us the maximum elements. Python's `heapq` module implements a *min-heap*. To efficiently get the two largest positive weights, the solution employs a clever trick: it stores the *negative* of all stone weights in the heap. When you extract the smallest (most negative) elements from this modified min-heap, you are effectively getting the largest (most positive) original stone weights. The algorithm initializes the heap by converting all stone weights to their negatives and then heapifying the list. It then enters a loop that continues as long as there are at least two stones available to smash. In each iteration, it extracts the two smallest (most negative) elements, negates them back to their true positive values (`y` and `x`), and simulates the smash. If `x` and `y` are different, the resulting `y - x` stone is created, and its *negative* weight is pushed back into the heap. If `x` and `y` are the same, nothing is pushed back, correctly simulating both stones being destroyed. Finally, once the loop finishes, if any stone remains (meaning the heap is not empty), its original positive weight is returned; otherwise, 0 is returned.

Step-by-Step Algorithm

Step 1: Create a priority queue (max-heap).
Step 2: Add all the stone weights to the priority queue.
Step 3: While the priority queue contains more than one element:
Step 4: Remove the two largest stones (x and y, with y >= x).
Step 5: If x != y, add (y - x) back to the priority queue.
Step 6: If the priority queue is empty, return 0; otherwise, return the weight of the remaining stone.

Key Insights

Leveraging Python's `heapq` as a Max-Heap: Since `heapq` is a min-heap, transforming all stone weights to their negative counterparts allows `heapq.heappop` to consistently retrieve the largest original stone weight. This is a crucial adaptation for problems requiring maximum elements from a heap in Python, avoiding the need for a custom max-heap implementation.
The 'Heaviest Two' Implies a Priority Queue: The problem's core operation of repeatedly selecting the two heaviest stones strongly points towards using a priority queue (heap). It efficiently provides the largest elements (in `O(log N)` time per extraction) without needing to sort the entire collection repeatedly, which would be far less efficient for a dynamic set.
Dynamic Stone Population Management: The solution elegantly handles the changing number of stones within the `while len(stones) > 1` loop. Each iteration correctly reduces the stone count by one (two removed, one potentially added) or two (both destroyed). The loop naturally concludes when zero or one stone remains, directly mapping to the problem's termination conditions and final result.

Complexity Analysis

Time Complexity: O(nlogn)

Space Complexity: O(n)

Topics

This problem involves: Array, Heap (Priority Queue).

Companies

Asked at: Flipkart, Nvidia, PayPal, Rippling.

Last Stone Weight | LeetCode 1046

Easy

Array Heap (Priority Queue)

Time: O(nlogn)Space: O(n)

Solution Code

Available in:

import java.util.PriorityQueue;

class Solution {
    public int lastStoneWeight(int[] stones) {
        PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
        for (int stone : stones) {
            pq.offer(stone);
        }

        while (pq.size() > 1) {
            int y = pq.poll();
            int x = pq.poll();
            if (x != y) {
                pq.offer(y - x);
            }
        }

        return pq.isEmpty() ? 0 : pq.poll();
    }
}

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Problem Info

Description

Problem 1046: Last Stone Weight - Detailed explanation and solution approaches.

Topics

Array Heap (Priority Queue)

Companies

FlipkartNvidiaPayPalRippling

Stats

Difficulty:Easy

Languages:4

Companies:4

Problem #:1046

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Create a priority queue (max-heap).
2
Step 2: Add all the stone weights to the priority queue.
3
Step 3: While the priority queue contains more than one element:
4
Step 4: Remove the two largest stones (x and y, with y >= x).
5
Step 5: If x != y, add (y - x) back to the priority queue.
6
Step 6: If the priority queue is empty, return 0; otherwise, return the weight of the remaining stone.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(nlogn)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

Adding n elements to the heap takes O(n log n) time. Each removal and insertion operation on the heap takes O(log n) time, and in the worst case, we perform this operation n-1 times. Therefore, the overall time complexity is dominated by O(n log n). The space complexity is O(n) because we store at most n elements in the heap.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Sorting-based approach

Sort the array in descending order. Iteratively pick the top two elements, smash them, insert the result back, and re-sort. This is less efficient.

Time: O(n^2 log n)Space: O(n)

Trade-offs: This approach is less efficient due to repeated sorting. The heap-based approach is significantly faster for larger inputs.

✨

Key Insights

1Leveraging Python's `heapq` as a Max-Heap: Since `heapq` is a min-heap, transforming all stone weights to their negative counterparts allows `heapq.heappop` to consistently retrieve the largest original stone weight. This is a crucial adaptation for problems requiring maximum elements from a heap in Python, avoiding the need for a custom max-heap implementation.
2The 'Heaviest Two' Implies a Priority Queue: The problem's core operation of repeatedly selecting the two heaviest stones strongly points towards using a priority queue (heap). It efficiently provides the largest elements (in `O(log N)` time per extraction) without needing to sort the entire collection repeatedly, which would be far less efficient for a dynamic set.
3Dynamic Stone Population Management: The solution elegantly handles the changing number of stones within the `while len(stones) > 1` loop. Each iteration correctly reduces the stone count by one (two removed, one potentially added) or two (both destroyed). The loop naturally concludes when zero or one stone remains, directly mapping to the problem's termination conditions and final result.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: Empty input array. The solution correctly handles this by returning 0.
Edge case 2: Input array with only one stone. The solution correctly returns the weight of that single stone.
Edge case 3: All stones have the same weight. The solution correctly handles this, resulting in 0 being returned.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using a priority queue and instead repeatedly searching for the maximum element which leads to O(n^2) time complexity.

✗Common mistake 2: Incorrectly handling the case where the two heaviest stones have equal weight, leading to an incorrect result.

✗Common mistake 3: Forgetting to handle the edge case of an empty input array.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation problems where you need to find the optimal way to combine resources of different sizes (e.g., combining tasks in a scheduling system).

🏦Real-world scenario 2: In certain physical simulations, the collision of objects could be modelled using similar logic where bigger objects impact smaller ones and change momentum.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 25, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Last Stone Weight"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether heap (priority queue) techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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