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:

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

Output: 

Yes

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