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 = 2
Output: 5
There are 4 squares of size 1x1 + 1 square of size 2x2.

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

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<iostream>
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 <stdio.h>

// 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.

PHP

<?php
// PHP 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
function countSquares($m, $n)
{
    // If n is smaller, swap m and n
    if ($n < $m)
        list($m, $n) = array($n, $m);

    // Now n is greater dimension,
    // apply formula
    return $m * ($m + 1) * (2 * $m + 1) /
       6 + ($n - $m) * $m * ($m + 1) / 2;
}

// Driver Code
$m = 4; $n = 3;
echo("Count of squares is " . countSquares($m, $n));

// This code is contributed by Ajit.
?>

Javascript

<script>

// javascript 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
function countSquares( m,  n)
{

// If n is smaller, swap m and n
if (n < m)
    [m, n] = [n, m];

// Now n is greater dimension,
// apply formula
return m * (m + 1) * (2 * m + 1) /
    6 + (n - m) * m *(m + 1) / 2;
}

// Driver Code
    let m = 4;
    let n = 3;
document.write("Count of squares is "+countSquares(n, m));

// This code is contributed by jana_sayantan.
</script>

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 <iostream>
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 <stdio.h>

// 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

<script>

// Javascript 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
function countSquares(m , n)
{

    // If n is smaller, swap m and n
    if (n < m)
    {
        var temp = m;
        m = n;
        n = temp;
    }

    // Now n is greater dimension,
    // apply formula
    return n * (n + 1) * (3 * m - n + 1) / 6;
}

// Driver Code
var m = 4;
var n = 3;

document.write("Count of squares is " +
               countSquares(m, n));

// This code is contributed by shikhasingrajput

</script>

Output :

Count of Squares is 20

Time complexity: O(1)

Auxiliary Space: O(1)

Thanks to Pranav for providing this alternate solution.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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 <iostream>
using namespace std;

int count_squares(int m, int n) {
    int count = 0;
    for (int i = 1; i <= min(m, n); i++) {
        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;
}

Python3

def count_squares(m, n):
    count = 0
    for i in range(1, min(m, n) + 1):
        count += (m - i + 1) * (n - i + 1)
    return count

m=4
n=3
print(count_squares(m, n))

Java

public class Main {
    public static void main(String[] args) {
        int m = 4;
        int n = 3;
        System.out.println(countSquares(m, n));
    }

    public static int countSquares(int m, int n) {
        int count = 0;
        for (int i = 1; i <= Math.min(m, n); i++) {
            count += (m - i + 1) * (n - i + 1);
        }
        return count;
    }
}

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)