Count squares of size K inscribed in a square of size N

Given two integers N and K, the task is to find the number of squares of size K that is inscribed in a square of size N.

Examples:

Input: N = 4, K = 2
Output: 9
Explanation:
There are 9 squares of size 2 inscribed in a square of size 4.

Input: N = 5, K = 3
Output: 9
Explanation:
There are 9 squares of size 3 inscribed in a square of size 5.



Approach: The key observation to solve the problem is that the total number of squares in a square of size N is (N * (N + 1)* (2*N + 1)) / 6. Therefore, the total number of squares of size K possible from a square of size N are:

(N - K + 1) * (N - K + 1)

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the
// above approach
 
#include <iostream>
using namespace std;
 
// Function to calculate the number
// of squares of size K in a square
// of size N
int No_of_squares(int N, int K)
{
    // Stores the number of squares
    int no_of_squares = 0;
 
    no_of_squares
        = (N - K + 1) * (N - K + 1);
 
    return no_of_squares;
}
 
// Driver Code
int main()
{
    // Size of the
    // bigger square
    int N = 5;
 
    // Size of
    // smaller square
    int K = 3;
    cout << No_of_squares(N, K);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the
// above approach
import java.util.*;
class GFG{
 
// Function to calculate the
// number of squares of size
// K in a square of size N
static int No_of_squares(int N,
                         int K)
{
  // Stores the number
  // of squares
  int no_of_squares = 0;
 
  no_of_squares = (N - K + 1) *
                  (N - K + 1);
 
  return no_of_squares;
}
 
// Driver Code
public static void main(String[] args)
{
  // Size of the
  // bigger square
  int N = 5;
 
  // Size of
  // smaller square
  int K = 3;
  System.out.print(No_of_squares(N, K));
}
}
 
// This code is contributed by Princi Singh

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the
# above approach
 
# Function to calculate the
# number of squares of size
# K in a square of size N
def No_of_squares(N, K):
   
    # Stores the number
    # of squares
    no_of_squares = 0;
    no_of_squares = (N - K + 1) * (N - K + 1);
    return no_of_squares;
 
# Driver Code
if __name__ == '__main__':
   
    # Size of the
    # bigger square
    N = 5;
 
    # Size of
    # smaller square
    K = 3;
    print(No_of_squares(N, K));
 
# This code is contributed by 29AjayKumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the
// above approach
using System;
 
class GFG{
 
// Function to calculate the
// number of squares of size
// K in a square of size N
static int No_of_squares(int N, int K)
{
     
    // Stores the number
    // of squares
    int no_of_squares = 0;
     
    no_of_squares = (N - K + 1) *
                    (N - K + 1);
     
    return no_of_squares;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Size of the
    // bigger square
    int N = 5;
     
    // Size of
    // smaller square
    int K = 3;
     
    Console.Write(No_of_squares(N, K));
}
}
 
// This code is contributed by Amit Katiyar

chevron_right


Output: 

9



Time Complexity: O(1) 
Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Recommended Posts:



If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.