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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++

 `// C++ program for the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Custom comparator function ` `bool` `compare(vector<``int``>& a, ` `             ``vector<``int``>& b) ` `{ ` `    ``return` `a < b; ` `} ` ` `  `// Function to find the minimum number of ` `// bullets required to penetrate all bricks ` `int` `findMinBulletShots(vector >& 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; ` ` `  `    ``// Iterate through all the points ` `    ``for` `(``int` `j = 1; j < points.size(); j++) { ` `        ``if` `(curr < points[j]) { ` ` `  `            ``// Increase the count ` `            ``cnt++; ` `            ``curr = points[j]; ` `        ``} ` `    ``} ` ` `  `    ``// Return the count ` `    ``return` `cnt; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Given coordinates of bricks ` `    ``vector > bricks{ { 5000, 900000 }, ` `                                 ``{ 1, 100 }, ` `                                 ``{ 150, 499 } }; ` ` `  `    ``// Function call ` `    ``cout << findMinBulletShots(bricks); ` ` `  `    ``return` `0; ` `} `

## Java

 `// 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 `

## Python3

 `# 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 `

Output:

```3
```

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

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