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:
C++
// 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 Nint 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 Codeint main(){ // Size of the // bigger square int N = 5; // Size of // smaller square int K = 3; cout << No_of_squares(N, K); return 0;} |
Java
// Java implementation of the// above approachimport java.util.*;class GFG{// Function to calculate the // number of squares of size // K in a square of size Nstatic 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 Codepublic 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 |
Python3
# Python3 implementation of the# above approach# Function to calculate the# number of squares of size# K in a square of size Ndef 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 Codeif __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 |
C#
// C# implementation of the// above approachusing System;class GFG{// Function to calculate the // number of squares of size // K in a square of size Nstatic 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 Codepublic 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 |
Javascript
<script>// JavaScript program for// the above approach// Function to calculate the// number of squares of size// K in a square of size Nfunction No_of_squares(N, K){ // Stores the number // of squares let no_of_squares = 0; no_of_squares = (N - K + 1) * (N - K + 1); return no_of_squares;}// Driver code // Size of the // bigger square let N = 5; // Size of // smaller square let K = 3; document.write(No_of_squares(N, K)); // This code is contributed by splevel62.</script> |
Output:
9
Time Complexity: O(1)
Auxiliary Space: O(1)
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