452. Minimum Number of Arrows to Burst Balloons

Array

Greedy

Sorting

Description

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

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]

Intuition

There are several segments on the one-dimensional coordinate axis, find out the minimum number of arrows which are orthogonal to the coordinate axis so that each segment can be shot through by at least one arrow.

Approaches

Using Sorting

Given points

Sort the segments by the end

 sort(points.begin(), points.end(), [&](auto &x, auto &y) {
    return x[1] < y[1];
  });

Put an arrow at the end of the 1-st segment
From the 2-nd segment, we check whether the current arrow pass through the current segment, if not add an arrow, put it at the end of the current segment

Time Complexity

O(N * log(N)) where 'N' is points length

Space Complexity

O(1)

Code

cpp

class Solution {
public:
  int findMinArrowShots(vector<vector<int>>& points) {
    sort(points.begin(), points.end(), [&](auto &x, auto &y) {
      return x[1] < y[1];
    });
 
    int n = points.size(), ans = 1, max = points[0][1];
 
    for(int i = 1; i < n; i++) {
      if(points[i][0] > max) {
        ans++;
        max = points[i][1];
      }
    }
 
    return ans;
  }
};

Analysis