 GeeksforGeeks App
Open App Browser
Continue

# Number of rectangles in N*M grid

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```
Recommended Practice

Brute Force Approach :

• Iterate over all possible pairs of horizontal lines.
• Iterate over all possible pairs of vertical lines.
• count the number of rectangles that can be formed using these lines.

Below is the code for the above approach :

## C++

 `#include ``using` `namespace` `std;` `int` `rectCount(``int` `n, ``int` `m)``{``    ``int` `count = 0;``    ``for``(``int` `i=1; i<=n; i++) ``// iterating over all possible pairs of horizontal lines``    ``{``        ``for``(``int` `j=1; j<=m; j++) ``// iterating over all possible pairs of vertical lines``        ``{``            ``count += (n-i+1)*(m-j+1); ``// counting the number of rectangles that can be formed using these lines``        ``}``    ``}``    ``return` `count;``}` `/* driver code */``int` `main()``{``    ``int` `n = 5, m = 4;``    ``cout << rectCount(n, m);``    ``return` `0;``}`

Output

`150`

Time Complexity : O(N^2)
Space Complexity : O(1)

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 ``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

 ``

## Javascript

 ``

Output

`150`

Time complexity: O(1)
Auxiliary Space: O(1), since no extra space has been taken.

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 review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.