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:

Below is the implementation of the above approach:

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

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

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

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.




Recommended Posts:


Check out this Author's contributed articles.

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.



Article Tags :
Practice Tags :