algoadvance

Leetcode 452. Minimum Number of Arrows to Burst Balloons

Problem Statement

You are given a set of n balloons, each represented as an interval [start, end], where start is the start position and end is the end position of the balloon. The balloons can be burst by arrows. An arrow can be shot from any point along the coordinate axis, and it bursts all balloons that it touches. You need to find the minimum number of arrows required to burst all the balloons.

Example:

Input: [[10,16], [2,8], [1,6], [7,12]]
Output: 2
Explanation: 
One way is to shoot one arrow for example at x = 6 (bursting balloons [2,8], [1,6]) and another arrow at x = 11 (bursting [10,16], [7,12]).

Clarifying Questions

  1. Will there always be at least one balloon in the input?
    • Yes, there will always be at least one balloon.
  2. Can balloons have negative start and end values?
    • Yes, the intervals can have negative values.
  3. Can intervals be overlapping?
    • Yes, intervals can be overlapping.

Strategy

The key idea is to use a greedy algorithm to determine the minimum number of arrows required. Here’s the step-by-step strategy:

  1. Sort the Intervals:
    • First, sort the balloon intervals based on their end coordinates. This will help us to always target the balloon that finishes the earliest.
  2. Initialize Counters:
    • Use a variable to keep track of the number of arrows and another to track the position at which the last arrow was shot.
  3. Iterate Through the Sorted Intervals:
    • For each balloon, if the start of the current balloon is greater than the position of the last shot arrow, we need a new arrow.
    • Update the arrow position to the end of the current balloon.

By following this approach, we minimize the need for arrows by always greedily shooting an arrow at the end of the balloon that finishes the earliest and covering as many subsequent overlapping balloons as possible.

Code

Here is the C++ code implementing the above strategy:

#include <vector>
#include <algorithm>

int findMinArrowShots(std::vector<std::vector<int>>& points) {
    if (points.empty()) return 0;
    
    // Sort the intervals by their end points
    std::sort(points.begin(), points.end(), [](const std::vector<int>& a, const std::vector<int>& b) {
        return a[1] < b[1];
    });
    
    int arrows = 1; // Start with one arrow
    int end = points[0][1]; // Shoot the first arrow at the end of the first balloon
    
    for(const auto& balloon : points) {
        // If the current balloon starts after the position of the last arrow
        if (balloon[0] > end) {
            arrows++; // Need another arrow
            end = balloon[1]; // Update the end to the current balloon's end
        }
    }
    
    return arrows;
}

Time Complexity

Hence, the overall time complexity is (O(n \log n)), dominated by the sorting step. This approach ensures that we achieve the desired efficiency.

Cut your prep time in half and DOMINATE your interview with AlgoAdvance AI