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Minimum Number of Bullets required to penetrate all bricks
  • Last Updated : 22 Sep, 2020

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:
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:



  1. Initialize the required count of bullets with 0.
  2. Sort the Xstart and Xend on the basis of Xend in ascending order.
  3. 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.
  4. Return the count at the end.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Custom comparator function
bool compare(vector<int>& a,
             vector<int>& b)
{
    return a[1] < b[1];
}
  
// Function to find the minimum number of
// bullets required to penetrate all bricks
int 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 Code
int main()
{
    // Given coordinates of bricks
    vector<vector<int> > bricks{ { 5000, 900000 },
                                 { 1, 100 },
                                 { 150, 499 } };
  
    // Function call
    cout << findMinBulletShots(bricks);
  
    return 0;
}

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Java

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// Java program for above approach
import 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 code
public 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

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Python3

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# Python3 program for the above approach
  
# Function to find the minimum number of
# bullets required to penetrate all bricks
def 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 Code
if __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

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Output: 

3

Time Complexity: O(N * log N), where N is the number of bricks. 
Auxiliary Space: O(1)

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