Pandigital number in a given base

Given an integer n and its base b. The task is to check if given number is Pandigital Number in the given base or not. A Pandigital number is an integer that has each digit of its base at least once.

It may be assumed that base is smaller than or equal to 36. In base 36, digits are [0, 1, …9. A, B, …Z]

Examples :

Input : n = "9651723480", b = 10
Output : Yes
Given number n has all digits from 0 to 9

Input : n = "23456789ABCDEFGHIJKLMNOPQRSTUVWXYZ", 
        b = 36
Output : No
Given number n doesn't have all digits in base
36. For example 1 is missing.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Make a boolean hash array of size equal to base of the number and initialize it with false. Now, iterate each digit of the number mark its corresponding index value as true in the hash array. In the end, check whether all the value in hash array are marked or not, if marked print “Yes” i.e Pandigital number else print “No”.

Below is the implementation of this approach:

C++

// C++ program to check if a number is pandigital 
// in given base. 
#include<bits/stdc++.h> 
using namespace std; 
  
// Return true if n is pandigit else return false. 
bool checkPandigital(int b, char n[]) 
{ 
    // Checking length is less than base 
    if (strlen(n) < b) 
        return false; 
  
    bool hash[b]; 
    memset(hash, false, sizeof(hash)); 
  
    // Traversing each digit of the number. 
    for (int i = 0; i < strlen(n); i++) 
    { 
        // If digit is integer 
        if (n[i] >= '0' && n[i] <= '9') 
            hash[n[i] - '0'] = true; 
  
        // If digit is alphabet 
        else if (n[i] - 'A' <= b - 11) 
            hash[n[i] - 'A' + 10] = true; 
    } 
  
    // Checking hash array, if any index is 
    // unmarked. 
    for (int i = 0; i < b; i++) 
        if (hash[i] == false) 
            return false; 
  
    return true; 
} 
  
// Driven Program 
int main() 
{ 
    int b = 13; 
    char n[] = "1298450376ABC"; 
  
    (checkPandigital(b, n))? (cout << "Yes" << endl): 
                             (cout << "No" << endl); 
  
    return 0; 
} 

Java

// Java program to check if a number  
// is pandigital in given base. 
import java.util.*; 
  
class GFG { 
      
// Return true if n is pandigit 
// else return false. 
static boolean checkPandigital(int b, String n) { 
      
    // Checking length is less than base 
    if (n.length() < b) 
    return false; 
  
    boolean hash[] = new boolean[b]; 
    Arrays.fill(hash, false); 
  
    // Traversing each digit of the number. 
    for (int i = 0; i < n.length(); i++) { 
          
    // If digit is integer 
    if (n.charAt(i) >= '0' && n.charAt(i) <= '9') 
        hash[n.charAt(i) - '0'] = true; 
  
    // If digit is alphabet 
    else if (n.charAt(i) - 'A' <= b - 11) 
        hash[n.charAt(i) - 'A' + 10] = true; 
    } 
  
    // Checking hash array, if any  
    // index is unmarked. 
    for (int i = 0; i < b; i++) 
    if (hash[i] == false) 
        return false; 
  
    return true; 
} 
  
// Driver code 
public static void main(String[] args)  
{ 
    int b = 13; 
    String n = "1298450376ABC"; 
  
    if (checkPandigital(b, n)) 
    System.out.println("Yes"); 
    else
    System.out.println("No"); 
} 
} 
  
// This code is contributed by Anant Agarwal. 

Python3

# Python3 program to check if a number is  
# pandigital in given base.  
  
# Return true if n is pandigit else return false.  
def checkPandigital(b, n):  
  
    # Checking length is less than base  
    if (len(n) < b):  
        return 0;  
  
    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 0;  
  
    return 1; 
  
# Driver Code 
b = 13;  
n = "1298450376ABC";  
  
if(checkPandigital(b, n)):  
    print("Yes");  
else: 
    print("No");  
                  
# This code is contributed by mits 

C#

// C# program to check if a number  
// is pandigital in given base. 
using System; 
  
class GFG { 
      
// Return true if n is pandigit 
// else return false. 
static bool checkPandigital(int b, string n) { 
      
    // Checking length is less than base 
    if (n.Length < b) 
    return false; 
  
    bool []hash = new bool[b]; 
    for(int i = 0; i < b; i++) 
    hash[i] = false; 
  
  
    // Traversing each digit of the number. 
    for (int i = 0; i < n.Length; i++) { 
          
    // If digit is integer 
    if (n[i] >= '0' && n[i] <= '9') 
        hash[n[i] - '0'] = true; 
  
    // If digit is alphabet 
    else if (n[i] - 'A' <= b - 11) 
        hash[n[i] - 'A' + 10] = true; 
    } 
  
    // Checking hash array, if any  
    // index is unmarked. 
    for (int i = 0; i < b; i++) 
    if (hash[i] == false) 
        return false; 
  
    return true; 
} 
  
// Driver code 
public static void Main()  
{ 
    int b = 13; 
    String n = "1298450376ABC"; 
  
    if (checkPandigital(b, n)) 
    Console.Write("Yes"); 
    else
    Console.Write("No"); 
} 
} 
  
// This code is contributed by nitin mittal. 

PHP

<?php 
// php program to check if a number is pandigital 
// in given base. 
  
// Return true if n is pandigit else return false. 
function checkPandigital($b, $n) 
{ 
    // Checking length is less than base 
    if (strlen($n) < $b) 
        return 0; 
  
    $hash = array(); 
      
    for($i = 0; $i< $b; $i++) 
    $hash[$i] = 0; 
      
      
    // Traversing each digit of the number. 
    for ($i = 0; $i < strlen($n); $i++) 
    { 
        // If digit is integer 
        if ($n[$i] >= '0' && $n[$i] <= '9') 
            $hash[$n[$i] - '0'] = 1; 
  
        // If digit is alphabet 
        else if (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 = 0; $i < $b; $i++) 
        if ($hash[$i] == 0) 
            return 0; 
  
    return 1; 
} 
  
// Driven Program 
$b = 13; 
$n = "1298450376ABC"; 
  
if(checkPandigital($b, $n)) 
    echo "Yes"; 
else
    echo "No"; 
                  
// This code is contributed by Sam007. 
?> 

Output:

Yes

Reference:
https://en.wikipedia.org/wiki/Pandigital_number

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