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# Count number of squares in a rectangle

Given a m x n rectangle, how many squares are there in it?

Examples :

`Input:  m = 2, n = 2Output: 5There are 4 squares of size 1x1 + 1 square of size 2x2.Input: m = 4, n = 3Output: 20There are 12 squares of size 1x1 +           6 squares of size 2x2 +           2 squares of size 3x3.` Recommended Practice

Let us first solve this problem for m = n, i.e., for a square:
For m = n = 1, output: 1
For m = n = 2, output: 4 + 1 [ 4 of size 1×1 + 1 of size 2×2 ]
For m = n = 3, output: 9 + 4 + 1 [ 9 of size 1×1 + 4 of size 2×2 + 1 of size 3×3 ]
For m = n = 4, output 16 + 9 + 4 + 1 [ 16 of size 1×1 + 9 of size 2×2 + 4 of size 3×3 + 1 of size 4×4 ]
In general, it seems to be n^2 + (n-1)^2 + … 1 = n(n+1)(2n+1)/6

Let us solve this problem when m may not be equal to n:
Let us assume that m <= n

From above explanation, we know that number of squares in a m x m matrix is m(m+1)(2m+1)/6

What happens when we add a column, i.e., what is the number of squares in m x (m+1) matrix?

When we add a column, number of squares increased is m + (m-1) + … + 3 + 2 + 1
[ m squares of size 1×1 + (m-1) squares of size 2×2 + … + 1 square of size m x m ]
Which is equal to m(m+1)/2

So when we add (n-m) columns, total number of squares increased is (n-m)*m(m+1)/2.
So total number of squares is m(m+1)(2m+1)/6 + (n-m)*m(m+1)/2.
Using same logic we can prove when n <= m.

So, in general,

`Total number of squares = m x (m+1) x (2m+1)/6 + (n-m) x m x (m+1)/2 when n is larger dimension`

Using above logic for rectangle, we can also prove that number of squares in a square is n(n+1)(2n+1)/6

Below is the implementation of above formula.

## C++

 `// C++ program to count squares``// in a rectangle of size m x n``#include``using` `namespace` `std;` `// Returns count of all squares``// in a rectangle of size m x n``int` `countSquares(``int` `m, ``int` `n)``{``// If n is smaller, swap m and n``if` `(n < m)``    ``swap(m, n);` `// Now n is greater dimension,``// apply formula``return` `m * (m + 1) * (2 * m + 1) /``     ``6 + (n - m) * m *(m + 1) / 2;``}` `// Driver Code``int` `main()``{``int` `m = 4, n = 3;``cout << ``"Count of squares is "``     ``<< countSquares(m, n);``}`

## C

 `// C program to count squares``// in a rectangle of size m x n` `#include ` `// Returns count of all squares``// in a rectangle of size m x n``int` `countSquares(``int` `m, ``int` `n)``{``  ``int` `temp;``// If n is smaller, swap m and n``  ``if` `(n < m)``  ``{``      ``temp=n;``      ``n=m;``      ``m=temp;``  ``}``  ``// Now n is greater dimension,``  ``// apply formula``  ``return` `m * (m + 1) * (2 * m + 1) /``      ``6 + (n - m) * m *(m + 1)/ 2;``}` `// Driver Code``int` `main()``{``    ``int` `m = 4, n = 3;``    ``printf``(``"Count of squares is %d"``,countSquares(m, n));``}` `// This code is contributed by Hemant Jain.`

## Java

 `// Java program to count squares``// in a rectangle of size m x n` `class` `GFG``{``    ``// Returns count of all squares``    ``// in a rectangle of size m x n``    ``static` `int` `countSquares(``int` `m, ``int` `n)``    ``{``    ``// If n is smaller, swap m and n``    ``if` `(n < m)``    ``{``        ``// swap(m, n)``        ``int` `temp = m;``        ``m = n;``        ``n = temp;``    ``}``        ` `    ` `    ``// Now n is greater dimension,``    ``// apply formula``    ``return` `m * (m + ``1``) * (``2` `* m + ``1``) /``        ``6` `+ (n - m) * m * (m + ``1``) / ``2``;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `m = ``4``, n = ``3``;``        ``System.out.println(``"Count of squares is "` `+``                            ``countSquares(m, n));``    ``}``}`

## Python3

 `# Python3 program to count squares``# in a rectangle of size m x n` `# Returns count of all squares``# in a rectangle of size m x n``def` `countSquares(m, n):``    ` `    ``# If n is smaller, swap m and n``    ``if``(n < m):``        ``temp ``=` `m``        ``m ``=` `n``        ``n ``=` `temp``        ` `    ``# Now n is greater dimension,``    ``# apply formula``    ``return` `((m ``*` `(m ``+` `1``) ``*` `(``2` `*` `m ``+` `1``) ``/``           ``6` `+` `(n ``-` `m) ``*` `m ``*` `(m ``+` `1``) ``/` `2``))` `# Driver Code``if` `__name__``=``=``'__main__'``:``    ``m ``=` `4``    ``n ``=` `3``    ``print``(``"Count of squares is "``         ``,countSquares(m, n))` `# This code is contributed by mits.`

## C#

 `// C# program to count squares in a rectangle``// of size m x n``using` `System;` `class` `GFG {``    ` `    ``// Returns count of all squares in a``    ``// rectangle of size m x n``    ``static` `int` `countSquares(``int` `m, ``int` `n)``    ``{``    ``// If n is smaller, swap m and n``    ``if` `(n < m)``    ``{``        ``// swap(m,n)``        ``int` `temp = m;``        ``m = n;``        ``n = temp;``    ``}``            ` `    ``// Now n is greater dimension, apply``    ``// formula``    ``return` `m * (m + 1) * (2 * m + 1) / 6 +``               ``(n - m) * m * (m + 1) / 2;``    ``}``    ` `    ``// Driver method``    ``public` `static` `void` `Main()``    ``{``        ``int` `m = 4, n = 3;``        ` `        ``Console.WriteLine(``"Count of squares is "``                          ``+ countSquares(m, n));``    ``}``}` `//This code is contributed by vt_m.`

## Javascript

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## PHP

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Output :

`Count of Squares is 20`

Time complexity: O(1)

Auxiliary Space: O(1)

Alternate Solution :

1. Let us take m = 2, n = 3;
2. The number of squares of side 1 will be 6 as there will be two cases one as squares of 1-unit sides along with the horizontal(2) and the second case as squares of 1-unit sides along the vertical(3). that give us 2*3 = 6 squares.
3. When the side is 2 units, one case will be as squares of the side of 2 units along only one place horizontally and the second case as two places vertically. So, the number of squares=2
4. So we can deduce that, Number of squares of size 1*1 will be m*n. The number of squares of size 2*2 will be (n-1)(m-1). So like this, the number of squares of size n will be 1*(m-n+1).

The final formula for the total number of squares will be n*(n+1)(3m-n+1)/6 .

## C++

 `// C++ program to count squares``// in a rectangle of size m x n``#include ``using` `namespace` `std;` `// Returns count of all squares``// in a rectangle of size m x n``int` `countSquares(``int` `m, ``int` `n)``{` `    ``// If n is smaller, swap m and n``    ``if` `(n < m) {``        ``int` `temp = m;``        ``m = n;``        ``n = temp;``    ``}` `    ``// Now n is greater dimension,``    ``// apply formula``    ``return` `n * (n + 1) * (3 * m - n + 1) / 6;``}` `// Driver Code``int` `main()``{``    ``int` `m = 4, n = 3;``    ``cout << ``"Count of squares is "` `<< countSquares(m, n);``}` `// This code is contributed by 29AjayKumar`

## C

 `// C program to count squares``// in a rectangle of size m x n` `#include ` `// Returns count of all squares``// in a rectangle of size m x n``int` `countSquares(``int` `m, ``int` `n)``{` `    ``// If n is smaller, swap m and n``    ``if` `(n < m)``    ``{``        ``int` `temp = m;``        ``m = n;``        ``n = temp;``    ``}` `    ``// Now n is greater dimension,``    ``// apply formula``    ``return` `n * (n + 1) * (3 * m - n + 1) / 6;``}` `// Driver Code``int` `main()``{``    ``int` `m = 4, n = 3;``    ``printf``(``"Count of squares is %d"``,countSquares(m, n));``}` `// This code is contributed by Hemant Jain`

## Java

 `// Java program to count squares``// in a rectangle of size m x n``import` `java.util.*;` `class` `GFG``{` `    ``// Returns count of all squares``    ``// in a rectangle of size m x n``    ``static` `int` `countSquares(``int` `m, ``int` `n)``    ``{` `        ``// If n is smaller, swap m and n``        ``if` `(n < m)``        ``{``            ``int` `temp = m;``            ``m = n;``            ``n = temp;``        ``}` `        ``// Now n is greater dimension,``        ``// apply formula``        ``return` `n * (n + ``1``) * (``3` `* m - n + ``1``) / ``6``;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `m = ``4``;``        ``int` `n = ``3``;``        ``System.out.print(``"Count of squares is "` `+``                             ``countSquares(m, n));``    ``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program to count squares``# in a rectangle of size m x n` `# Returns count of all squares``# in a rectangle of size m x n``def` `countSquares(m, n):``    ` `    ``# If n is smaller, swap m and n``    ``if``(n < m):``        ``temp ``=` `m``        ``m ``=` `n``        ``n ``=` `temp``        ` `    ``# Now n is greater dimension,``    ``# apply formula``    ``return` `n ``*` `(n ``+` `1``) ``*` `(``3` `*` `m ``-` `n ``+` `1``) ``/``/` `6` `# Driver Code``if` `__name__``=``=``'__main__'``:``    ``m ``=` `4``    ``n ``=` `3``    ``print``(``"Count of squares is"``,``           ``countSquares(m, n))` `# This code is contributed by AnkitRai01`

## C#

 `// C# program to count squares``// in a rectangle of size m x n``using` `System;` `class` `GFG``{` `    ``// Returns count of all squares``    ``// in a rectangle of size m x n``    ``static` `int` `countSquares(``int` `m, ``int` `n)``    ``{` `        ``// If n is smaller, swap m and n``        ``if` `(n < m)``        ``{``            ``int` `temp = m;``            ``m = n;``            ``n = temp;``        ``}` `        ``// Now n is greater dimension,``        ``// apply formula``        ``return` `n * (n + 1) * (3 * m - n + 1) / 6;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int` `m = 4;``        ``int` `n = 3;``        ``Console.Write(``"Count of squares is "` `+``                          ``countSquares(m, n));``    ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

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Output :

`Count of Squares is 20`

Time complexity: O(1)

Auxiliary Space: O(1)

Thanks to Pranav for providing this alternate solution.

#### Approach#3: Using loop

This approach counts the number of squares in a given rectangle of size m x n by iterating over all possible square sizes from 1 to the minimum of m and n, and adding up the number of squares of each size that can fit in the rectangle.

#### Algorithm

1. Initialize a counter variable to 0.
2. Iterate over all possible square sizes, i.e. from 1 to min(m, n).
3. For each square size, calculate the number of squares that can be formed.
4. Add this count to the counter variable.
5. Return the final count.

## C++

 `#include ``using` `namespace` `std;` `// Function to count the number of squares in an m x n grid``int` `count_squares(``int` `m, ``int` `n) {``    ``int` `count = 0;``    ``for` `(``int` `i = 1; i <= min(m, n); i++) {``        ``// Calculate the number of squares``        ``//with side length 'i' that fit in the grid``        ``count += (m - i + 1) * (n - i + 1);``    ``}``    ``return` `count;``}` `int` `main() {``    ``int` `m = 4;``    ``int` `n = 3;``    ``cout << count_squares(m, n) << endl;``    ``return` `0;``}`

## Java

 `public` `class` `Main {``    ``public` `static` `void` `main(String[] args) {``        ``int` `m = ``4``;``        ``int` `n = ``3``;``        ``System.out.println(countSquares(m, n));``    ``}` `    ``// Function to count the number of squares in an m x n grid``    ``public` `static` `int` `countSquares(``int` `m, ``int` `n) {``        ``int` `count = ``0``;``        ``for` `(``int` `i = ``1``; i <= Math.min(m, n); i++) {``            ``// Calculate the number of squares with``            ``// side length 'i' that fit in the grid``            ``count += (m - i + ``1``) * (n - i + ``1``);``        ``}``        ``return` `count;``    ``}``}`

## Python3

 `# Function to count the number of squares in an m x n grid``def` `count_squares(m, n):``    ``count ``=` `0``    ``for` `i ``in` `range``(``1``, ``min``(m, n) ``+` `1``):``        ``# Calculate the number of squares``        ``# with side length 'i' that fit in the grid``        ``count ``+``=` `(m ``-` `i ``+` `1``) ``*` `(n ``-` `i ``+` `1``)``    ``return` `count` `m ``=` `4``n ``=` `3``print``(count_squares(m, n))`

## C#

 `using` `System;` `public` `class` `CountSquaresInGrid``{``    ``public` `static` `int` `CountSquares(``int` `m, ``int` `n)``    ``{``        ``int` `count = 0;``        ` `        ``// Iterate through the side lengths of squares``        ``// from 1 to the minimum of m and n.``        ``for` `(``int` `i = 1; i <= Math.Min(m, n); i++)``        ``{``            ``// For each side length 'i', calculate``            ``// the number of squares that can fit in the grid.``            ``count += (m - i + 1) * (n - i + 1);``        ``}``        ` `        ``return` `count;``    ``}` `    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``int` `m = 4;``        ``int` `n = 3;``        ` `        ``// Call the function to count squares in the grid and print the result.``        ``int` `result = CountSquares(m, n);``        ``Console.WriteLine(result);``    ``}``}`

## Javascript

 `// Function to count the number of squares in an m x n grid``function` `count_squares(m, n) {``    ``let count = 0;``    ``// Iterate over all possible square sizes``    ``for` `(let i = 1; i <= Math.min(m, n); i++) {``        ``// Count the number of squares of size 'i' that can fit in the grid``        ``count += (m - i + 1) * (n - i + 1);``    ``}``    ``return` `count;``}` `let m = 4;``let n = 3;``console.log(count_squares(m, n)); ``// Output: 20`

Output

```20

```

Time Complexity: O(m * n * min(m, n))
Space Complexity: O(1)