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Pandigital number in a given base

  • Difficulty Level : Easy
  • Last Updated : 16 Jun, 2021

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.

 

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;
}
 
// Driver 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;
}
 
// Driver Program
$b = 13;
$n = "1298450376ABC";
 
if(checkPandigital($b, $n))
    echo "Yes";
else
    echo "No";
                 
// This code is contributed by Sam007.
?>

Javascript




<script>
    // Javascript 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 (n.length < b)
        return 0;
 
    let hash = [];
     
    for(let i = 0; i< b; i++)
    hash[i] = 0;
     
     
    // Traversing each digit of the number.
    for (let i = 0; i < n.length; i++)
    {
        // If digit is integer
        if (n[i] >= '0' && n[i] <= '9')
            hash[n[i] - '0'] = 1;
 
        // If digit is alphabet
        else if (n.charCodeAt(i) - 'A'.charCodeAt(0) <= b - 11)
            hash[n.charCodeAt(i) - 'A'.charCodeAt(0) + 10] = 1;
    }
 
    // Checking hash array, if any index is
    // unmarked.
    for (let i = 0; i < b; i++)
        if (hash[i] == 0)
            return 0;
 
    return 1;
}
 
// Driver Program
let b = 13;
let n = "1298450376ABC";
 
if(checkPandigital(b, n))
    document.write("Yes");
else
    document.write("No");
                 
// This code is contributed by _saurabh_jaiswal.
</script>

Output: 
 

Yes

Reference: 
https://en.wikipedia.org/wiki/Pandigital_number
This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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