Check if a number is a K-persistent number

Given two numbers N and K, the task is to check if N is a K-Persistent Number or not. If N is a K-Persistent Number then print “Yes” else print “No”.

A number N is called K-Persistent Number if all of {n, 2n,…, kn} are Pandigital number in base 10.

Examples :

Input : N = 1023456789, K = 2
Output : Yes
Explanation :
1023456789 * 1 = 1023456789 = Pandigital number in base 10
1023456789 * 2 = 2046913578 = Pandigital number in base 10

Input : N = 561, K = 1
Output : No

Approach :

The idea is simple.We iterate through all numbers i from 1 to k and for every integer i in the loop, we check if i*n is a Pandigital number in base 10 or not. If at any i it’s i*n is not Pandigital number in base 10 the return false, otherwise return true at last.

Implementation :

# your code goes here
# python3 program for checking K-Persistent Number
import math 

# Return true if n is pandigital in base 10 else return false.  
def isPandigital(n):  
    
    n = str(n)
    b = 10
  
    hash = [0] * b; 
      
    # Traversing each digit of the number.  
    for i in range(len(n)):  
          
        # If digit is integer  
        if (n[i] >= '0' and n[i] <= '9'):  
            hash[ord(n[i]) - ord('0')] = 1;  
  
        # If digit is alphabet  
        elif (ord(n[i]) - ord('A') <= b - 11):  
            hash[ord(n[i]) - ord('A') + 10] = 1;  
  
    # Checking hash array, if any index is  
    # unmarked.  
    for i in range(b):  
        if (hash[i] == 0):  
            return False  
  
    return True 
    
# Returns true if N is a K-Persistent Number
def isPersistentNumber(N, K):
    for i in range(1, K+1):
        if(isPandigital(i*N)==False):
            return False
    return True
        
    

 
# Driver code
# Given Number N 
N = 1052674893
# Given Number K 
K = 3
# Function Call 
if (isPersistentNumber(N, K)):
    print("Yes")  
else:
    print("No")

Yes