452.FindMinArrowShots.cs

// 452. Minimum Number of Arrows to Burst Balloons
// 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] = [xstart, xend] denotes a balloon whose horizontal diameter stretches between xstart and xend. 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 xstart and xend is burst by an arrow shot at x if xstart <= x <= xend. There is no limit to the number of arrows that can be shot. A shot arrow keeps traveling up infinitely, bursting any balloons in its path.
// Given the array points, return the minimum number of arrows that must be shot to burst all balloons.
// Example 1:
// Input: points = [[10,16],[2,8],[1,6],[7,12]]
// Output: 2
// Explanation: The balloons can be burst by 2 arrows:
// - Shoot an arrow at x = 6, bursting the balloons [2,8] and [1,6].
// - Shoot an arrow at x = 11, bursting the balloons [10,16] and [7,12].
// Example 2:
// Input: points = [[1,2],[3,4],[5,6],[7,8]]
// Output: 4
// Explanation: One arrow needs to be shot for each balloon for a total of 4 arrows.
// Example 3:
// Input: points = [[1,2],[2,3],[3,4],[4,5]]
// Output: 2
// Explanation: The balloons can be burst by 2 arrows:
// - Shoot an arrow at x = 2, bursting the balloons [1,2] and [2,3].
// - Shoot an arrow at x = 4, bursting the balloons [3,4] and [4,5].
// Constraints:
// 1 <= points.length <= 105
// points[i].length == 2
// -231 <= xstart < xend <= 231 - 1


//The intuition behind the solution is to sort the intervals (balloons) by their end points. 
// Then, iterate through the sorted list and shoot an arrow at the end point of the current balloon. 
// This arrow will also burst all other balloons whose start point is less than or equal to the current
// balloon's end point. 
// This is because the sorted order ensures that all such balloons will overlap with the current balloon.
public class Solution {
    public int FindMinArrowShots(int[][] points) {
        points = points.OrderBy(x => x[1]).ThenBy(x => x[0]).ToArray();
        int count = 0;
        for(int i = 0; i < points.Length; i++){
            count++;
            int j = i + 1;
            while(j < points.Length && points[i][1] >= points[j][0])
                j++;
            i = j-1;
        }
        return count;
    }
}