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

  • Difficulty Level : Easy
  • Last Updated : 12 May, 2021

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 Polets 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 polets
    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)
 

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