Count number of squares in a rectangle - GeeksforGeeks

15 Apr, 2021
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: 1For 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
Java
Python3
C#
PHP
Javascript
// 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 nint countSquares(int m, int n){// If n is smaller, swap m and nif (n < m)    swap(m, n); // Now n is greater dimension,// apply formulareturn m * (m + 1) * (2 * m + 1) /     6 + (n - m) * m *(m + 1) / 2;} // Driver Codeint main(){int m = 4, n = 3;cout << "Count of squares is "     << countSquares(m, n);}
// 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 nint 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 Codeint main(){    int m = 4, n = 3;    printf("Count of squares is %d",countSquares(m, n));} // This code is contributed by Hemant Jain.
// 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 program to count squares# in a rectangle of size m x n # Returns count of all squares# in a rectangle of size m x ndef 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 Codeif __name__=='__main__':    m = 4    n = 3    print("Count of squares is "         ,countSquares(m, n)) # This code is contributed by mits.
// C# program to count squares in a rectangle// of size m x nusing 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 program to count squares// in a rectangle of size m x n // Returns count of all squares// in a rectangle of size m x nfunction 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.?>
<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 nfunction countSquares( m,  n){ // If n is smaller, swap m and nif (n < m)    [m, n] = [n, m]; // Now n is greater dimension,// apply formulareturn 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
Alternate Solution : 
Let us take m = 2, n = 3;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.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=2So 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).
Let us take m = 2, n = 3;
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.
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
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
Java
Python3
C#
Javascript
// 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 nint 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 Codeint main(){    int m = 4, n = 3;    cout << "Count of squares is " << countSquares(m, n);} // This code is contributed by 29AjayKumar
// 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 nint 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 Codeint main(){    int m = 4, n = 3;    printf("Count of squares is %d",countSquares(m, n));} // This code is contributed by Hemant Jain
// Java program to count squares// in a rectangle of size m x nimport 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 program to count squares# in a rectangle of size m x n # Returns count of all squares# in a rectangle of size m x ndef 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 Codeif __name__=='__main__':    m = 4    n = 3    print("Count of squares is",           countSquares(m, n)) # This code is contributed by AnkitRai01
// C# program to count squares// in a rectangle of size m x nusing 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
<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 nfunction 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 Codevar 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
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. 
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