# Find the number of squares inside the given square grid

Given a grid of side N * N, the task is to find the total number of squares that exist inside it. All squares selected can be of any length.

Examples:

Input: N = 1
Output: 1 Input: N = 2
Output: 5 Input: N = 4
Output: 30

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach 1: Taking a few examples, it can be observed that for a grid on size N * N, the number of squares inside it will be 12 + 22 + 32 + … + N2

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach  #include  using namespace std;     // Function to return the number  // of squares inside an n*n grid  int cntSquares(int n)  {      int squares = 0;      for (int i = 1; i <= n; i++) {          squares += pow(i, 2);      }      return squares;  }     // Driver code  int main()  {      int n = 4;         cout << cntSquares(4);         return 0;  }

## Java

 // Java implementation of the approach  class GFG {         // Function to return the number      // of squares inside an n*n grid      static int cntSquares(int n)      {          int squares = 0;          for (int i = 1; i <= n; i++) {              squares += Math.pow(i, 2);          }          return squares;      }         // Driver code      public static void main(String args[])      {          int n = 4;             System.out.print(cntSquares(4));      }  }

## Python3

 # Python3 implementation of the approach      # Function to return the number   # of squares inside an n*n grid   def cntSquares(n) :          squares = 0;       for i in range(1, n + 1) :          squares += i ** 2;          return squares;      # Driver code   if __name__ == "__main__" :          n = 4;          print(cntSquares(4));      # This code is contributed by AnkitRai01

## C#

 // C# implementation of the approach  using System;     class GFG   {         // Function to return the number      // of squares inside an n*n grid      static int cntSquares(int n)      {          int squares = 0;          for (int i = 1; i <= n; i++)           {              squares += (int)Math.Pow(i, 2);          }          return squares;      }         // Driver code      public static void Main(String []args)      {          int n = 4;             Console.Write(cntSquares(n));      }  }     // This code is contributed by 29AjayKumar

Output:

30


Approach 2: By the use of direct formula.
However, the sum has the closed form (direct formula) . Hence, we can employ this to calculate the sum in time.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach  #include     using namespace std;     int cnt_squares (int n)  {      /* Function to return the number       of squares inside an n*n grid */        return n * (n + 1) * (2 * n + 1) / 6;  }     // Driver code  int main()  {      cout << cnt_squares (4) << endl;         return 0;  }

## Java

 // Java implementation of the approach  class GFG {      static int cntSquares (int n) {          /* Function to return the number          of squares inside an n*n grid */                return n * (n + 1) * (2 * n + 1) / 6;      }         // Driver code      public static void main(String args[]) {          System.out.println (cntSquares(4));      }  }

## Python3

 # Python3 implementation of the approach      """  Function to return the number   of squares inside an n*n grid   """    def cntSquares(n) :       return int (n * (n + 1) * (2 * n + 1) / 6)     # Driver code   if __name__ == "__main__" :       print (cntSquares (4));

## C#

 // C# implementation of the approach  using System;     class GFG   {         /* Function to return the number       of squares inside an n*n grid */     static int cntSquares (int n)      {          return n * (n + 1) * (2 * n + 1) / 6;      }         // Driver code      public static void Main (String[] args)      {          Console.Write (cntSquares (4));      }  }

Output:

30


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