# Minimum number of points required to cover all blocks of a 2-D grid

Given two integers N and M. The task is to find the minimum number of points required to cover an N * M grid.

A point can cover two blocks in a 2-D grid when placed in any common line or sideline.

Examples:

Input: N = 5, M = 7
Output: 18

Input: N = 3, M = 8
Output: 12

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: This problem can be solved using Greedy Approach. The main idea is to observe that a single point placed on the common line or sideline covers two blocks. So the total number of points needed to cover all the blocks(say B blocks) is B/2 when B is even else B/2 + 1 when B is odd.

For a grid having N*M blocks, The total number of blocks will be (N*M)/2 when either one of them is even. Otherwise, it will require ((N*M)/2) + 1 points to cover all the blocks and one extra for last untouched block.

Below is the image to show how points can be used to cover block in a 2D-grid: Point ‘A’ covers two blocks and ‘B’ covers one block.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the minimum number ` `// of Points required to cover a grid ` `int` `minPoints(``int` `n, ``int` `m) ` `{ ` `    ``int` `ans = 0; ` ` `  `    ``// If number of block is even ` `    ``if` `((n % 2 != 0) ` `        ``&& (m % 2 != 0)) { ` `        ``ans = ((n * m) / 2) + 1; ` `    ``} ` `    ``else` `{ ` `        ``ans = (n * m) / 2; ` `    ``} ` ` `  `    ``// Return the minimum points ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Given size of grid ` `    ``int` `N = 5, M = 7; ` ` `  `    ``// Function Call ` `    ``cout << minPoints(N, M); ` `    ``return` `0; ` `} `

## Java

 `// Java program for the above approach ` `class` `GFG{ ` `     `  `// Function to find the minimum number ` `// of Points required to cover a grid ` `static` `int` `minPoints(``int` `n, ``int` `m) ` `{ ` `    ``int` `ans = ``0``; ` ` `  `    ``// If number of block is even ` `    ``if` `((n % ``2` `!= ``0``) && (m % ``2` `!= ``0``))  ` `    ``{ ` `        ``ans = ((n * m) / ``2``) + ``1``; ` `    ``} ` `    ``else`  `    ``{ ` `        ``ans = (n * m) / ``2``; ` `    ``} ` ` `  `    ``// Return the minimum points ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``// Given size of grid ` `    ``int` `N = ``5``, M = ``7``; ` ` `  `    ``// Function Call ` `    ``System.out.print(minPoints(N, M)); ` `} ` `} ` ` `  `// This code is contributed by Ritik Bansal `

## Python3

 `# Python3 program for the above approach ` ` `  `# Function to find the minimum number  ` `# of Points required to cover a grid  ` `def` `minPoints(n, m): ` ` `  `    ``ans ``=` `0` ` `  `    ``# If number of block is even ` `    ``if` `((n ``%` `2` `!``=` `0``) ``and` `(m ``%` `2` `!``=` `0``)): ` `        ``ans ``=` `((n ``*` `m) ``/``/` `2``) ``+` `1` ` `  `    ``else``: ` `        ``ans ``=` `(n ``*` `m) ``/``/` `2` ` `  `    ``# Return the minimum points ` `    ``return` `ans ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` ` `  `    ``# Given size of grid ` `    ``N ``=` `5` `    ``M ``=` `7` ` `  `    ``# Function call ` `    ``print``(minPoints(N, M)) ` ` `  `# This code is contributed by himanshu77 `

Output:

```18
```

Time Complexity: O(1)
Auxiliary Space: O(1)

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Improved By : btc_148, himanshu77