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 

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 <bits/stdc++.h>
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

C#

// C# program for the above approach
using System;
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
    Console.Write(minPoints(N, M));
}
}

// This code is contributed by sapnasingh4991

Javascript

<script>
// Javascript implementation for the above approach

// Function to find the minimum number
// of Points required to cover a grid
function minPolets(n, m)
{
    let ans = 0;

    // If number of block is even
    if ((n % 2 != 0) && (m % 2 != 0))
    {
        ans = Math.floor((n * m) / 2) + 1;
    }
    else
    {
        ans = Math.floor((n * m) / 2);
    }

    // Return the minimum points
    return ans;
}

    // Driver Code

    // Given size of grid
    let N = 5, M = 7;

    // Function Call
    document.write(minPolets(N, M));

</script>

Output:

18

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