Check if a number is a perfect square having all its digits as a perfect square

Given an integer N, the task is to check if the given number is a perfect square having all its digits as a perfect square or not. If found to be true, then print “Yes”. Otherwise, print “No”.
Examples:

Input: N = 144 
Output: Yes 
Explanation: The number 144 is a perfect square and also the digits of the number {1(= 1², 4(= 2²} is also a perfect squares.

Input: N = 81 
Output: No

Approach:

The first step in the approach is to check if N is a perfect square. We can use the built-in sqrt() function to compute the square root of N, and then check if the square of this value is equal to N. If N is not a perfect square, we can immediately return “No” and exit the function.

If N is a perfect square, we can proceed to the next step, which is to check if each digit of N is a perfect square. To do this, we can convert N to a string using the to_string() function, which returns a string representation of N. We can then iterate over each character of the string using a for loop, and convert each character to an integer using the expression c – ‘0’. This expression subtracts the ASCII value of the character ‘0’ from the ASCII value of the current character c, resulting in an integer value corresponding to the current digit.

For each digit, we can use the isPerfectSquare() function to check if it is a perfect square. If the digit is not a perfect square, we can immediately return “No” and exit the function. If all digits are perfect squares, we can return “Yes” and exit the function.

C++

#include <bits/stdc++.h>
using namespace std;

bool isPerfectSquare(int n) {
    int root = sqrt(n);
    return root * root == n;
}

bool isFullSquare(long N)
{
    // Check if N is a perfect square
    if (!isPerfectSquare(N))
    {
        cout << "No";
        return false;
    }

    // Convert N to a string and check if each character is a perfect square
    string str = to_string(N);
    for (char c : str)
    {
        int digit = c - '0';
        if (!isPerfectSquare(digit))
        {
            cout << "No";
            return false;
        }
    }

    // All digits are perfect squares, so N is a full square
    cout << "Yes";
    return true;
}

int main()
{
    long N = 144;
    isFullSquare(N);
    return 0;
}

Java

import java.io.*;
import java.lang.*;
import java.util.*;

class Main {
  public static boolean isPerfectSquare(int n)
  {
    int root = (int)Math.sqrt(n);
    return root * root == n;
  }

  public static boolean isFullSquare(long N)
  {
    // Check if N is a perfect square
    if (!isPerfectSquare((int)N)) {
      System.out.println("No");
      return false;
    }

    // Convert N to a string and check if each character
    // is a perfect square
    String str = Long.toString(N);
    for (char c : str.toCharArray()) {
      int digit = c - '0';
      if (!isPerfectSquare(digit)) {
        System.out.println("No");
        return false;
      }
    }

    // All digits are perfect squares, so N is a full
    // square
    System.out.println("Yes");
    return true;
  }

  public static void main(String[] args)
    throws java.lang.Exception
  {
    long N = 144;
    isFullSquare(N);
  }
}

Python3

import math

# Function to check if a number is a perfect square
def isPerfectSquare(n):
    root = math.sqrt(n)
    return root * root == n

# Function to check if a number is a full square
def isFullSquare(N):
    # Check if N is a perfect square
    if not isPerfectSquare(N):
        print("No")
        return False

    # Convert N to a string and check if each character is a perfect square
    strN = str(N)
    for c in strN:
        digit = int(c)
        if not isPerfectSquare(digit):
            print("No")
            return False

    # All digits are perfect squares, so N is a full square
    print("Yes")
    return True

# Test the code with an example value of N
N = 144
isFullSquare(N)

C#

using System;

class Program {
    static bool IsPerfectSquare(int n)
    {
        int root = (int)Math.Sqrt(n);
        return root * root == n;
    }

    static bool IsFullSquare(long N)
    {
        // Check if N is a perfect square
        if (!IsPerfectSquare((int)N)) {
            Console.WriteLine("No");
            return false;
        }

        // Convert N to a string and check if each character
        // is a perfect square
        string str = N.ToString();
        foreach(char c in str)
        {
            int digit = c - '0';
            if (!IsPerfectSquare(digit)) {
                Console.WriteLine("No");
                return false;
            }
        }

        // All digits are perfect squares, so N is a full
        // square
        Console.WriteLine("Yes");
        return true;
    }

    static void Main(string[] args)
    {
        long N = 144;
        IsFullSquare(N);
    }
}

Javascript

// Function to check if a number is a perfect square
function isPerfectSquare(n) {
    let root = Math.sqrt(n);
    return root * root == n;
}

// Function to check if a number is a full square
function isFullSquare(N) {
    // Check if N is a perfect square
    if (!isPerfectSquare(N)) {
        console.log("No");
        return false;
    }

    // Convert N to a string and check if each character is a perfect square
    let str = N.toString();
    for (let c of str) {
        let digit = c - '0';
        if (!isPerfectSquare(digit)) {
            console.log("No");
            return false;
        }
    }

    // All digits are perfect squares, so N is a full square
    console.log("Yes");
    return true;
}

// Test the code with an example value of N
let N = 144;
isFullSquare(N);

Output: Yes

Time Complexity:  O(d), where d is the number of digits in the input number N.

Space Complexity: O(1)

Approach: The idea is to check if the given number N is a perfect square or not. If found to be true, check if all its digits are either 0, 1, 4, or 9. If found to be true, print “Yes”. Otherwise, print “No”.

Algorithm:

• Create a static method “check_digits” with a boolean return type that takes the long value “N” as the input parameter.
• Start a while loop with the condition “N” greater than 0.
  •  Extract the last digit of “N” and store it in “n”.
  • If “n” is not equal to 0, 1, 4, or 9, return false.
  • Divide “N” by 10 and store the value in “N”.
• If came out from the while loop then return true.
• Create a static method “is_perfect” with a boolean return type that takes the long value “N” as input.
  •  Now store the square root of “N” in variable “n” of double type.
  •  If the floor and ceiling of “n” are not equal, return false.
  • Otherwise, return true.
• Create a static method named “isFullSquare” with void return type which takes long as an input variable named as “N”.
  • If both “is_perfect” and “check_digits” methods return true for “N”, print “Yes”.
  • Otherwise, print “No”.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if digits of
// N is a perfect square or not
bool check_digits(long N)
{

    // Iterate over the digits
    while (N > 0) {

        // Extract the digit
        int n = N % 10;

        // Check if digit is a
        // perfect square or not
        if ((n != 0) && (n != 1)
            && (n != 4) && (n != 9)) {
            return 0;
        }

        // Divide N by 10
        N = N / 10;
    }

    // Return true
    return 1;
}

// Function to check if N is
// a perfect square or not
bool is_perfect(long N)
{
    long double n = sqrt(N);

    // If floor and ceil of n
    // is not same
    if (floor(n) != ceil(n)) {
        return 0;
    }
    return 1;
}

// Function to check if N satisfies
// the required conditions or not
void isFullSquare(long N)
{
    // If both the conditions
    // are satisfied
    if (is_perfect(N)
        && check_digits(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
}

// Driver Code
int main()
{
    long N = 144;

    // Function Call
    isFullSquare(N);

    return 0;
}

Java

// Java program for
// the above approach
import java.util.*;
class GFG{

// Function to check if digits of
// N is a perfect square or not
static boolean check_digits(long N)
{
  // Iterate over the digits
  while (N > 0)
  {
    // Extract the digit
    int n = (int) (N % 10);

    // Check if digit is a
    // perfect square or not
    if ((n != 0) && (n != 1) &&
        (n != 4) && (n != 9))
    {
      return false;
    }

    // Divide N by 10
    N = N / 10;
  }

  // Return true
  return true;
}

// Function to check if N is
// a perfect square or not
static boolean is_perfect(long N)
{
  double n = Math.sqrt(N);

  // If floor and ceil of n
  // is not same
  if (Math.floor(n) != Math.ceil(n))
  {
    return false;
  }
  return true;
}

// Function to check if N satisfies
// the required conditions or not
static void isFullSquare(long N)
{
  // If both the conditions
  // are satisfied
  if (is_perfect(N) &&
      check_digits(N))
  {
    System.out.print("Yes");
  }
  else
  {
    System.out.print("No");
  }
}

// Driver Code
public static void main(String[] args)
{
  long N = 144;

  // Function Call
  isFullSquare(N);
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program for the above approach
import math

# Function to check if digits of
# N is a perfect square or not
def check_digits(N):

    # Iterate over the digits
    while (N > 0):

        # Extract the digit
        n = N % 10

        # Check if digit is a
        # perfect square or not
        if ((n != 0) and (n != 1) and
            (n != 4) and (n != 9)):
            return 0

        # Divide N by 10
        N = N // 10

    # Return true
    return 1

# Function to check if N is
# a perfect square or not
def is_perfect(N):

    n = math.sqrt(N)

    # If floor and ceil of n
    # is not same
    if (math.floor(n) != math.ceil(n)):
        return 0

    return 1

# Function to check if N satisfies
# the required conditions or not
def isFullSquare(N):

    # If both the conditions
    # are satisfied
    if (is_perfect(N) and
      check_digits(N)):
        print("Yes")
    else:
        print("No")

# Driver Code
N = 144

# Function call
isFullSquare(N)

# This code is contributed by sanjoy_62

C#

// C# program for
// the above approach
using System;
class GFG{

// Function to check if digits of
// N is a perfect square or not
static bool check_digits(long N)
{
  // Iterate over the digits
  while (N > 0)
  {
    // Extract the digit
    int n = (int) (N % 10);

    // Check if digit is a
    // perfect square or not
    if ((n != 0) && (n != 1) &&
        (n != 4) && (n != 9))
    {
      return false;
    }

    // Divide N by 10
    N = N / 10;
  }

  // Return true
  return true;
}

// Function to check if N is
// a perfect square or not
static bool is_perfect(long N)
{
  double n = Math.Sqrt(N);

  // If floor and ceil of n
  // is not same
  if (Math.Floor(n) != Math.Ceiling(n))
  {
    return false;
  }
  return true;
}

// Function to check if N satisfies
// the required conditions or not
static void isFullSquare(long N)
{
  // If both the conditions
  // are satisfied
  if (is_perfect(N) &&
      check_digits(N))
  {
    Console.Write("Yes");
  }
  else
  {
    Console.Write("No");
  }
}

// Driver Code
public static void Main()
{
  long N = 144;

  // Function Call
  isFullSquare(N);
}
}

// This code is contributed by Chitranayal

Javascript

<script>

// JavaScript program for the above approach

// Function to check if digits of
// N is a perfect square or not
function check_digits(N)
{

    // Iterate over the digits
    while (N > 0) {

        // Extract the digit
        let n = N % 10;

        // Check if digit is a
        // perfect square or not
        if ((n != 0) && (n != 1)
            && (n != 4) && (n != 9)) {
            return 0;
        }

        // Divide N by 10
        N = Math.floor(N / 10);
    }

    // Return true
    return 1;
}

// Function to check if N is
// a perfect square or not
function is_perfect(N)
{
    let n = Math.sqrt(N);

    // If floor and ceil of n
    // is not same
    if (Math.floor(n) != Math.ceil(n)) {
        return 0;
    }
    return 1;
}

// Function to check if N satisfies
// the required conditions or not
function isFullSquare(N)
{
    // If both the conditions
    // are satisfied
    if (is_perfect(N)
        && check_digits(N)) {
        document.write("Yes");
    }
    else {
        document.write("No");
    }
}

// Driver Code

    let N = 144;

    // Function Call
    isFullSquare(N);


// This code is contributed by Surbhi Tyagi.

</script>

Output:

Yes

Time Complexity: O(log₁₀N), where N is the input variable
Auxiliary Space: O(1)