Given an integer N, the task is to find the count of squares of side 1 that can be drawn without lifting the pencil, starting at one corner of an N * N grid and never visiting an edge twice.
Input: N = 2
Output: 2
Input: N = 3
Output: 5
Approach: It can be observed that for the values of N = 1, 2, 3, …, a series will be formed as 1, 2, 5, 10, 17, 26, 37, 50, … whose Nth term is (N2 – (2 * N) + 2)
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the count of// squares that can be formedint countSquares(int n){ return (pow(n, 2) - (2 * n) + 2);}// Driver codeint main(){ int n = 2; cout << countSquares(n); return 0;} |
Java
// Java implementation of the approachclass GFG{// Function to return the count of// squares that can be formedstatic int countSquares(int n){ return (int) (Math.pow(n, 2) - (2 * n) + 2);}// Driver codepublic static void main(String []args){ int n = 2; System.out.println(countSquares(n));}}// This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach # Function to return the count of # squares that can be formed def countSquares(n) : return (pow(n, 2) - (2 * n) + 2); # Driver code if __name__ == "__main__" : n = 2; print(countSquares(n)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System; class GFG{ // Function to return the count of // squares that can be formed static int countSquares(int n) { return (int) (Math.Pow(n, 2) - (2 * n) + 2); } // Driver code public static void Main(String []args) { int n = 2; Console.WriteLine(countSquares(n)); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the approach// Function to return the count of// squares that can be formedfunction countSquares(n){ return (Math.pow(n, 2) - (2 * n) + 2);}// Driver codevar n = 2;document.write(countSquares(n));</script> |
Output:
2
Time Complexity: O(1)
Auxiliary Space: O(1)
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