Given a matrix bricks[][] denoting the start and end coordinates of a brick’s breadth from an arrangement of rectangular bricks in a two-dimensional space. A bullet can be shot up exactly vertically from different points along the X-axis. A brick with coordinates Xstart and Xend is penetrated by the bullet which is shot from position X if Xstart ? X ? Xend. There is no limit on the number of bullets that can be shot and a bullet once shot keeps traveling along Y-axis infinitely. The task is to find the minimum number of bullets that must be shot to penetrate all the bricks.
Examples:
Input: bricks[][] = {{10, 16}, {2, 8}, {1, 6}, {7, 12}}
Output: 2
Explanation:
Shoot one bullet at X = 6, it penetrates the bricks places at {2, 8} and {1, 6}
Another bullet at X = 11, it penetrates the bricks places at {7, 12} and {10, 16}Input:bricks[][] = {{5000, 90000}, {150, 499}, {1, 100}}
Output: 3
Approach: The idea is to use Greedy technique. Follow the steps below to solve this problem:
- Initialize the required count of bullets with 0.
- Sort the Xstart and Xend on the basis of Xend in ascending order.
- Iterate over the sorted list of positions and check if the Xstart of the current brick is greater than or equal to the Xend of the previous brick. If so, then one more bullet is required. So increment the count by 1. Otherwise proceed.
- Return the count at the end.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Custom comparator functionbool compare(vector<int>& a, vector<int>& b){ return a[1] < b[1];}// Function to find the minimum number of// bullets required to penetrate all bricksint findMinBulletShots(vector<vector<int> >& points){ // Sort the points in ascending order sort(points.begin(), points.end(), compare); // Check if there are no points if (points.size() == 0) return 0; int cnt = 1; int curr = points[0][1]; // Iterate through all the points for (int j = 1; j < points.size(); j++) { if (curr < points[j][0]) { // Increase the count cnt++; curr = points[j][1]; } } // Return the count return cnt;}// Driver Codeint main(){ // Given coordinates of bricks vector<vector<int> > bricks{ { 5000, 900000 }, { 1, 100 }, { 150, 499 } }; // Function call cout << findMinBulletShots(bricks); return 0;} |
Java
// Java program for above approachimport java.util.*;class GFG{// Function to find the minimum number of // bullets required to penetrate all bricks static int findMinBulletShots(int[][] points) { // Sort the points in ascending order Arrays.sort(points, (a, b) -> a[1] - b[1]); // Check if there are no points if (points.length == 0) return 0; int cnt = 1; int curr = points[0][1]; // Iterate through all the points for(int j = 1; j < points.length; j++) { if (curr < points[j][0]) { // Increase the count cnt++; curr = points[j][1]; } } // Return the count return cnt; } // Driver codepublic static void main (String[] args){ // Given coordinates of bricks int[][] bricks = { { 5000, 900000 }, { 1, 100 }, { 150, 499 } }; // Function call System.out.print(findMinBulletShots(bricks)); }}// This code is contributed by offbeat |
Python3
# Python3 program for the above approach# Function to find the minimum number of# bullets required to penetrate all bricksdef findMinBulletShots(points): # Sort the points in ascending order for i in range(len(points)): points[i] = points[i][::-1] points = sorted(points) for i in range(len(points)): points[i] = points[i][::-1] # Check if there are no points if (len(points) == 0): return 0 cnt = 1 curr = points[0][1] # Iterate through all the points for j in range(1, len(points)): if (curr < points[j][0]): # Increase the count cnt += 1 curr = points[j][1] # Return the count return cnt# Driver Codeif __name__ == '__main__': # Given coordinates of bricks bricks = [ [ 5000, 900000 ], [ 1, 100 ], [ 150, 499 ] ] # Function call print(findMinBulletShots(bricks))# This code is contributed by mohit kumar 29 |
Javascript
<script>// JavaScript program for the above approach// Custom comparator functionfunction compare(a,b){ return a[1] - b[1];}// Function to find the minimum number of// bullets required to penetrate all bricksfunction findMinBulletShots(points){ // Sort the points in ascending order points.sort(compare); // Check if there are no points if (points.length == 0) return 0; let cnt = 1; let curr = points[0][1]; // Iterate through all the points for (let j = 1; j < points.length; j++) { if (curr < points[j][0]) { // Increase the count cnt++; curr = points[j][1]; } } // Return the count return cnt;}// Driver Code// Given coordinates of brickslet bricks = [ [ 5000, 900000 ], [ 1, 100 ], [ 150, 499 ] ];// Function calldocument.write(findMinBulletShots(bricks),"</br>");// This code is contributed by shinjanpatra</script> |
C#
using System;using System.Collections.Generic;public class PointComparer : IComparer<List<int>>{ public int Compare(List<int> a, List<int> b) { return a[1].CompareTo(b[1]); }}public class Program{ // Function to find the minimum number of // bullets required to penetrate all bricks public static int FindMinBulletShots(List<List<int>> points) { // Sort the points in ascending order points.Sort(new PointComparer()); // Check if there are no points if (points.Count == 0) return 0; int cnt = 1; int curr = points[0][1]; // Iterate through all the points for (int j = 1; j < points.Count; j++) { if (curr < points[j][0]) { // Increase the count cnt++; curr = points[j][1]; } } // Return the count return cnt; } // Driver Code public static void Main() { // Given coordinates of bricks List<List<int>> bricks = new List<List<int>>() { new List<int>(){ 5000, 900000 }, new List<int>(){ 1, 100 }, new List<int>(){ 150, 499 } }; // Function call Console.WriteLine(FindMinBulletShots(bricks)); }}// This code is contributed by Aditya Sharma |
Output:
3
Time Complexity: O(N * log N), where N is the number of bricks.
Auxiliary Space: O(1)
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