Squarable Numbers

Given a positive integer N, the task is to check if N is a Squarable Number or not.

An integer "N" is said to be Squarable, if we can divide a square into N non-overlapping Squares (not necessarily of same size).

Examples:

Input: N = 1
Output: "YES, 1 is a squarable Number"
Explanation: Any Square satisfies this case.

Input: N=4
Output: "YES, 4 is a Squarable Number"
Explanation: A Square can be Divided into 4 Squares.

Input: N=5
Output: "NO, 5 is not a Squarable Number"
Explanation: Any Square cannot be Divided into 5 Squares.

Input: N=6 
Output: "YES, 6 is a Squarable Number"
Explanation: A Square can be divided into 6 Squares.

Approach: On the basis of below mentioned observations, it is possible to figure out that a number is squarable or not:

Observation:

The Trick to catch here is that every positive Integer N >= 6, will surely be Squarable.

• This can be proved by Inductive Hypothesis. Let's see how. 
• Before proceeding to Proof By Induction, let us see how numbers 7 and 8 are Squarable (Number 6 is Squarable which we have already seen in Examples above). Numbers 6, 7 and 8 will become the base Cases for Proving By Induction.

Number 7 is Squarable in Following way:

Number 8 is Squarable in Following way:

Base Cases:

We have already seen that N = 6, 7, 8 are Squarable and Hence our Base Case is Proved.

Proof By Induction:

Let us Assume that the Numbers "K", "K - 1" and "K - 2" are squarable where K >= 8. Now, Let us see how "K + 1" will also be squarable by Induction.

We Know,

 (K - 2) + 3 = (K + 1)

Therefore, If it is possible to form 3 more squares in "(K - 2)" which is a squarable Number, then we can say "K + 1" is also squarable. Forming 3 more squares in a square is easy. This can be achieved just by Dividing the Square into 4 small Squares from centre. 

Example Below:

One Square is Divided into 4 Squares, thus 3 new squares are formed. Hence, conclusively, it is proved that if "K - 2" is Squarable, then "K+1" is also Squarable.

Inductive Hypothesis:

Therefore, by Induction, we can say that our 3 base Cases (N = 6, 7, 8) are sufficient to prove the hypothesis, because for Proving any Number "X" Squarable, we will be having a Number "X - 3" which is Squarable, and inductively, "X" will also be Squarable (as 3 Squares can be Easily Formed in "X-3" to form "X" Squares).

Finally, we have 6, 7, 8 as Squarable numbers, which means

9 is Also Squarable (as 6 is Squarable, 6+3=9)
10 is Also Squarable (as 7 is Squarable, 7+3=10)
11 is Also Squarable (as 8 is Squarable, 8+3=11)
12 is Also Squarable (as 9 is Squarable, 9+3=12)......and so on

Hence, every N >= 6 is proved Squarable.

Below is the implementation for the above approach:

// C++ approach for the above problem:

#include <iostream>
using namespace std;

// Function to find a number is
// Squarable or not
void isSquarable(int N)
{
    if (N < 6) {
        if (N == 1 || N == 4)
            cout << "YES, " << N
          << " is a Squarable Number"
                 << endl;
        else
            cout << "NO, " << N 
          << " is not a "
                 << "Squarable Number" << endl;
    }
    else
        cout << "YES, " << N
      << " is a Squarable Number"
             << endl;
}

// Drivers code
int main()
{

    int N;

    isSquarable(1);
    isSquarable(4);
    isSquarable(5);
    isSquarable(6);
    isSquarable(100);

    return 0;
}

/*package whatever //do not write package name here */
import java.io.*;

class GFG {
  // Java approach for the above problem:

  // Function to find a number is
  // Squarable or not
  static void isSquarable(int N)
  {
    if (N < 6) {
      if (N == 1 || N == 4)
        System.out.println("YES, " + N + " is a Squarable Number");
      else
        System.out.println("NO, " + N +" is not a " + "Squarable Number");
    }
    else
      System.out.println("YES, " + N + " is a Squarable Number");
  }

  // Driver Code
  public static void main(String args[])
  {
    int N;

    isSquarable(1);
    isSquarable(4);
    isSquarable(5);
    isSquarable(6);
    isSquarable(100);
  }

}

// This code is contributed by shinjanpatra

# Python approach for the above problem:

# Function to find a number is
# Squarable or not
def isSquarable(N):
    if (N < 6):
        if (N == 1 or N == 4):
            print(f"YES {N} is a Squarable Number");
        else:
            print(f"NO {N} is not a Squarable Number");
    else:
        print(f"YES {N} is a Squarable Number");

# Drivers code
isSquarable(1);
isSquarable(4);
isSquarable(5);
isSquarable(6);
isSquarable(100);
    
# This code is contributed by Saurabh Jaiswal

// C# program to check if N is a Squarable Number or not.
using System;

public class GFG{

  // Function to find a number is
  // Squarable or not
  static void isSquarable(int N)
  {
    if (N < 6) {
      if (N == 1 || N == 4)
        Console.Write("YES, " + N + " is a Squarable Number\n");
      else
        Console.Write("NO, " + N +" is not a " + "Squarable Number\n");
    }
    else
      Console.Write("YES, " + N + " is a Squarable Number\n");
  }

  // Driver Code
  static public void Main (){

    isSquarable(1);
    isSquarable(4);
    isSquarable(5);
    isSquarable(6);
    isSquarable(100);
  }
}

// This code is contributed by shruti456rawal

// JavaScript approach for the above problem:

// Function to find a number is
// Squarable or not
function isSquarable(N)
{
    if (N < 6) {
        if (N == 1 || N == 4)
            console.log("YES, "+N+" is a Squarable Number");
        else
        console.log("NO, "+N+" is not a Squarable Number");
    }
    else
    console.log("YES, "+N+" is a Squarable Number");
}

// Drivers code
    isSquarable(1);
    isSquarable(4);
    isSquarable(5);
    isSquarable(6);
    isSquarable(100);
    
 // This code is contributed by Ishan Khandelwal

YES, 1 is a Squarable Number
YES, 4 is a Squarable Number
NO, 5 is not a Squarable Number
YES, 6 is a Squarable Number
YES, 100 is a Squarable Number

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