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 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”.
Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output: 

Yes






Time Complexity: O(log10N) 
Auxiliary Space: O(1)

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