The O(n) Club: Minimum Number of Arrows to Burst Balloons Without Losing (Your Mind)

The O(n) Club: Minimum Number of Arrows to Burst Balloons Without Losing (Your Mind)

⚡ TL;DR

This is LeetCode 452, not the Macy’s parade: Each balloon is an interval. Sort them by end, fire your arrow at the end, and pack away your interval-merging fantasies. Java sketch below!

int findMinArrowShots(int[][] balloons) {
    Arrays.sort(balloons, Comparator.comparingInt(b -> b[1]));
    int arrows = 1, end = balloons[0][1];
    for (int i = 1; i < balloons.length; i++) {
        if (balloons[i][0] > end) {
            arrows++;
            end = balloons[i][1];
        }
    }
    return arrows;
}

🧠 How Devs Usually Mess This Up

Step 1: Spot an interval problem. Step 2: Merge intervals like your manager merges calendar invites. Classic! Sadly, that’s not the move here. Don’t fuse balloons together—our job is less Frankenstein, more Robin Hood. And yes, some folks fire their arrow at the start of a balloon. If only villains made it that easy.

🔍 What’s Actually Going On

Visualize balloons stretched between trees. You, armed with a bottomless quiver of vertical arrows, want to maximize bang-per-buck (or pop-per-arrow). If you shoot at the end of the leftmost “expiring” interval, you nail every balloon currently touching that end. Any balloon that starts after that? Well, congrats, it’s “ethical hunting” time again. It’s greedy, fast, and gives you maximum coverage with minimum work—just like any proper software developer wants.

🛠️ PseudoCode

• Sort all balloons (intervals) by their end coordinate ascending.
  

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

  Now you can sweep through easily.
• Initialize arrow count to 1, and set your first arrow to burst the end of the earliest-ending balloon.
  

  int arrows = 1;
  int end = balloons[0][1];

  Always gotta start somewhere.
• Sweep through the rest:
  

  for (int i = 1; i < balloons.length; i++) {
      if (balloons[i][0] > end) {
          arrows++;
          end = balloons[i][1];
      }
  }

  Any balloon that *starts* after where your arrow landed? Reload, shoot again at that new end.
• Return the total arrows used—nobody gets points for overkill.

💻 The Code

import java.util.*;
 public class BalloonArrows {
    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 end = points[0][1];
        for (int i = 1; i < points.length; i++) {
            if (points[i][0] > end) {
                arrows++;
                end = points[i][1];
            }
        }
        return arrows;
    }
}

⚠️ Pitfalls, Traps & Runtime Smacks

• Zero overlap? Every balloon needs its own arrow. (Performance: tragic, since O(n)).
• Total overlap? One arrow and you’re a hero.
• Sorting by start will break your code (and maybe your spirit)—always sort by end point.
• Balloons with same start/end: trust the sort, one arrow suffices.
• Complexity is O(n log n) for sorting, O(1) for cleverness. Even your CPU will thank you.

🧠 Interviewer Brain-Tattoo

“In the land of greedy algorithms, an arrow fired at the right end-point is mightier than a thousand merges. Stay sharp.”