Skip to content
Related Articles

Related Articles

Improve Article
Number of rectangles in N*M grid
  • Difficulty Level : Hard
  • Last Updated : 30 Apr, 2021

We are given a N*M grid, print the number of rectangles in it.
Examples: 
 

Input  : N = 2, M = 2
Output : 9
There are 4 rectangles of size 1 x 1.
There are 2 rectangles of size 1 x 2
There are 2 rectangles of size 2 x 1
There is one rectangle of size 2 x 2.

Input  : N = 5, M = 4
Output : 150

Input :  N = 4, M = 3
Output: 60

 

We have discussed counting number of squares in a n x m grid,
Let us derive a formula for number of rectangles.
If the grid is 1×1, there is 1 rectangle. 
If the grid is 2×1, there will be 2 + 1 = 3 rectangles 
If it grid is 3×1, there will be 3 + 2 + 1 = 6 rectangles. 
we can say that for N*1 there will be N + (N-1) + (n-2) … + 1 = (N)(N+1)/2 rectangles
If we add one more column to N×1, firstly we will have as many rectangles in the 2nd column as the first, 
and then we have that same number of 2×M rectangles. 
So N×2 = 3 (N)(N+1)/2
After deducing this we can say 
For N*M we’ll have (M)(M+1)/2 (N)(N+1)/2 = M(M+1)(N)(N+1)/4
So the formula for total rectangles will be M(M+1)(N)(N+1)/4 

.

Combinatorial Logic:



N*M grid can be represented as (N+1) vertical lines and (M+1) horizontal lines.
In a rectangle, we need two distinct horizontal and two distinct verticals.
So going by the logic of Combinatorial Mathematics we can choose 2 vertical lines and 2 horizontal lines to form a rectangle. And total number of these combinations is the number of rectangles possible in the grid.

Total Number of Rectangles in N*M grid: N+1C2 * M+1C2 = (N*(N+1)/2!)*(M*(M+1)/2!) = N*(N+1)*M*(M+1)/4
 

C++




// C++ program to count number of rectangles
// in a n x m grid
#include <bits/stdc++.h>
using namespace std;
 
int rectCount(int n, int m)
{
    return (m * n * (n + 1) * (m + 1)) / 4;
}
 
/* driver code */
int main()
{
    int n = 5, m = 4;
    cout << rectCount(n, m);
    return 0;
}

Java




// JAVA Code to count number of
// rectangles in N*M grid
import java.util.*;
 
class GFG {
     
    public static long  rectCount(int n, int m)
    {
        return (m * n * (n + 1) * (m + 1)) / 4;
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int n = 5, m = 4;
       System.out.println(rectCount(n, m));
    }
}
 
// This code is contributed by Arnav Kr. Mandal.

Python3




# Python3 program to count number
# of rectangles in a n x m grid
 
def rectCount(n, m):
 
    return (m * n * (n + 1) * (m + 1)) // 4
 
# Driver code
n, m = 5, 4
print(rectCount(n, m))
 
# This code is contributed by Anant Agarwal.

C#




// C# Code to count number of
// rectangles in N*M grid
using System;
 
class GFG {
      
    public static long  rectCount(int n, int m)
    {
        return (m * n * (n + 1) * (m + 1)) / 4;
    }
      
    // Driver program
    public static void Main()
    {
        int n = 5, m = 4;
       Console.WriteLine(rectCount(n, m));
    }
}
  
// This code is contributed by Anant Agarwal.

PHP




<?php
// PHP program to count
// number of rectangles
// in a n x m grid
 
function rectCount($n, $m)
{
    return ($m * $n *
           ($n + 1) *
           ($m + 1)) / 4;
}
 
// Driver Code
$n = 5;
$m = 4;
echo rectCount($n, $m);
 
// This code is contributed
// by ajit
?>

Javascript




<script>
 
    // Javascript Code to count number
    // of rectangles in N*M grid
     
    function rectCount(n, m)
    {
        return parseInt((m * n * (n + 1) *
                        (m + 1)) / 4, 10);
    }
     
      let n = 5, m = 4;
      document.write(rectCount(n, m));
     
</script>

Output:  

150

This article is contributed by Pranav. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.




My Personal Notes arrow_drop_up
Recommended Articles
Page :