Leetcode Minimum Number of Arrows to Burst Balloons problem solution

In this Leetcode Minimum Number of Arrows to Burst Balloons problem solution 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.

Problem solution in Python.

class Solution(object):
    def findMinArrowShots(self, points):
        
        if not points: return 0

        points = sorted(points, key=lambda x:x[1])
        
        count = 1
        left = points[0][1]
        
        for i in points:
            if(i[0]>left):
                count += 1
                left = i[1]
        
        return count

Problem solution in Java.

class Solution {
    public int findMinArrowShots(int[][] points) {
        
        Arrays.sort(points, (a,b) -> Integer.compare(a[1], b[1]));
        int arrow = 1;
        int end = points[0][1];
        for(int i = 1;i < points.length;i++){
            if(points[i][0] > end){
                arrow++;
                end = points[i][1];
            }
        }
        return arrow;
    }
}

Problem solution in C++.

class Solution {
static bool comp(vector<int> a,vector<int> b){
    return a[1]<b[1];
}
public:
    int findMinArrowShots(vector<vector<int>>& points) {
        if(points.size()==0)
            return 0;
        sort(points.begin(),points.end(),comp);
        int count = 1;
        int prev = points[0][1];
        for(int i=0;i<points.size();i++){
            if(points[i][0]>prev){
                prev= points[i][1];
                count++;
            }
        }
        return count;
    }
};

Problem solution in C.

int comp(const void*a,const void*b){
    return ((int**)a)[0][1]-((int**)b)[0][1];
}
int findMinArrowShots(int** points, int pointsRowSize, int pointsColSize) {
    qsort(points,pointsRowSize,sizeof(points[0]),comp);
    int i;
    int end;
    int arr=0;
    for(i=0;i<pointsRowSize;){
        end=points[i][1];
        while(++i<pointsRowSize&&points[i][0]<=end);
        arr++;
    }
    return arr;
}