What is the time complexity of Minimum Number of Arrows to Burst Balloons?

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

What data structures are used to solve Minimum Number of Arrows to Burst Balloons?

This problem uses Array, Greedy, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Number of Arrows to Burst Balloons?

Insight 1: Sorting the balloons by their end positions allows us to efficiently determine which balloons can be burst by a single arrow. Insight 2: The greedy approach of always shooting an arrow at the end position of the earliest ending balloon guarantees an optimal solution.

Which companies ask LeetCode 452. Minimum Number of Arrows to Burst Balloons in interviews?

This problem is frequently asked at Flipkart, Goldman Sachs, Instacart.

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Minimum Number of Arrows to Burst Balloons - LeetCode 452 Solution

Minimum Number of Arrows to Burst Balloons - Complete Solution Guide

Minimum Number of Arrows to Burst Balloons is LeetCode problem 452, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There are some spherical balloons taped onto a flat wall that represents the XY-plane. The balloons are represented as a 2D integer array points where points[i] = [x start , x end ] denotes a balloon whose horizontal diameter stretches between x start and x end . You do not know the exact y-coordinates of the balloons. Arrows can be shot up directly vertically (in the positive y-direction) from different points along the x-axis. A balloon with x start and x end is burst by an arrow shot at x if

Detailed Explanation

The problem asks us to find the minimum number of arrows needed to burst a set of balloons. Each balloon is represented by a horizontal diameter (x_start, x_end). An arrow shot vertically at any x-coordinate within a balloon's diameter will burst that balloon. The goal is to minimize the number of arrows used to burst all balloons.

Solution Approach

The solution uses a greedy approach. It first sorts the balloons based on their ending x-coordinate. Then, it iterates through the sorted balloons, keeping track of the x-coordinate where the last arrow was shot. For each balloon, it checks if the balloon can be burst by the current arrow. If not, it shoots a new arrow at the ending x-coordinate of the current balloon. This approach works because by shooting at the end of the earliest-ending balloon, we maximize the chance of bursting other balloons as well, as any overlapping balloons must at least overlap with this earliest endpoint.

Step-by-Step Algorithm

Step 1: Sort the `points` array based on the ending x-coordinate (points[i][1]) in ascending order.
Step 2: Initialize the number of arrows to 1, as we need at least one arrow if there are balloons.
Step 3: Initialize the position of the first arrow to be the end coordinate of the first balloon in the sorted list.
Step 4: Iterate through the remaining balloons in the sorted `points` array.
Step 5: For each balloon, compare its starting x-coordinate with the position of the last arrow shot. If the balloon's starting point is greater than the arrow's position, it means the current arrow cannot burst this balloon. Increment the arrow count and update the arrow position to the ending point of the current balloon.
Step 6: Return the total number of arrows required.

Key Insights

Insight 1: Sorting the balloons by their end positions allows us to efficiently determine which balloons can be burst by a single arrow.
Insight 2: The greedy approach of always shooting an arrow at the end position of the earliest ending balloon guarantees an optimal solution.
Insight 3: Overlapping intervals (balloon diameters) indicate balloons that can be burst with a single arrow. Identifying the largest possible overlapping groups is key.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy, Sorting.

Companies

Asked at: Flipkart, Goldman Sachs, Instacart.

Minimum Number of Arrows to Burst Balloons | LeetCode 452

Medium

Array Greedy Sorting

Time: O(n log n)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;
import java.util.Comparator;

class Solution {
    public int findMinArrowShots(int[][] points) {
        if (points == null || points.length == 0) {
            return 0;
        }

        Arrays.sort(points, Comparator.comparingInt(a -> a[1]));

        int arrows = 1;
        int arrowPos = points[0][1];

        for (int[] point : points) {
            int start = point[0];
            int end = point[1];

            if (start > arrowPos) {
                arrows++;
                arrowPos = end;
            }
        }

        return arrows;
    }
}

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Problem Info

Description

Problem 452: Minimum Number of Arrows to Burst Balloons - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting

Companies

FlipkartGoldman SachsInstacart

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:452

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: Sort the `points` array based on the ending x-coordinate (points[i][1]) in ascending order.
2
Step 2: Initialize the number of arrows to 1, as we need at least one arrow if there are balloons.
3
Step 3: Initialize the position of the first arrow to be the end coordinate of the first balloon in the sorted list.
4
Step 4: Iterate through the remaining balloons in the sorted `points` array.
5
Step 5: For each balloon, compare its starting x-coordinate with the position of the last arrow shot. If the balloon's starting point is greater than the arrow's position, it means the current arrow cannot burst this balloon. Increment the arrow count and update the arrow position to the ending point of the current balloon.
6
Step 6: Return the total number of arrows required.

💡

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(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the sorting algorithm used. Most standard sorting algorithms (e.g., merge sort, quicksort) have a time complexity of O(n log n). The space complexity is O(1) because the algorithm modifies the input array in-place and uses only a constant amount of extra space for variables.

📖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 by start time

Sort the balloons by their starting x-coordinate and keep track of the 'current interval'. Extend the current interval if the next balloon overlaps, or start a new interval (and thus need another arrow) if it doesn't.

Time: O(n log n)Space: O(1)

Trade-offs: Sorting by end time can be slightly more intuitive to prove the correctness of the greedy strategy, as we are greedily picking the earliest end point and ensuring the largest possible intersection with future points.

✨

Key Insights

1Insight 1: Sorting the balloons by their end positions allows us to efficiently determine which balloons can be burst by a single arrow.
2Insight 2: The greedy approach of always shooting an arrow at the end position of the earliest ending balloon guarantees an optimal solution.
3Insight 3: Overlapping intervals (balloon diameters) indicate balloons that can be burst with a single arrow. Identifying the largest possible overlapping groups is key.

📚

Key Terms Explained

For Students

Array

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty input `points`. The code correctly handles this by returning 0.
Edge case 2: Balloons with the same end coordinate. The sorting ensures that these are processed together and the greedy approach works correctly.
Edge case 3: Very large coordinate values (close to Integer.MAX_VALUE or Integer.MIN_VALUE in Java/C++). Sorting using `Comparator.comparingInt` in Java can lead to integer overflow, thus sorting by (a[1]-b[1]) is dangerous. Solution using `Comparator.comparingLong` is recommended. But since x_end <= 2^31 - 1, this issue does not occur.
Edge case 4: When starting and ending point of some balloons are same. In this case our algorithm still works correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the balloons before applying the greedy approach. This will likely lead to an incorrect solution.

✗Common mistake 2: Sorting by the starting x-coordinate instead of the ending x-coordinate. This will result in a suboptimal solution.

✗Common mistake 3: Integer overflow when calculating the difference between end points during sorting (in Java/C++). This can be avoided by using a `long` to store the difference, or using Comparator.comparingLong(a -> a[1])

✗Common mistake 4: Forgetting the case when the input list is empty.

✅

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: Scheduling tasks with deadlines. Each balloon represents a task with a start time and a deadline. Finding the minimum number of arrows is analogous to finding the minimum number of resources needed to complete all tasks by their deadlines.

🏦Real-world scenario 2: Interval covering problems in networking. Imagine each balloon represents a time interval during which a network resource is required. The problem becomes minimizing the number of servers needed to meet demands during these intervals.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: September 1, 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

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

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

❓

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 do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Minimum Number of Arrows to Burst Balloons"?

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

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