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 9Input : 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 Programint 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 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 in range(b): if (hash[i] == 0): return 0 return 1# Driver Codeb = 13n = "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 Programlet b = 13;let n = "1298450376ABC";if(checkPandigital(b, n)) document.write("Yes");else document.write("No"); // This code is contributed by _saurabh_jaiswal.</script> |
Yes
Time Complexity: O(b + strlen(n))
Auxiliary Space: O(b)
Reference:
https://en.wikipedia.org/wiki/Pandigital_number
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Using set in python:
Approach:
We can use set to check if a given number in a given base is pandigital or not. We create a set of all possible digits in the given base and check if the set of digits in the given number is equal to the set of all possible digits.
Create a set containing all the digits in the base.
Convert the given number to a string.
Convert the string to a set of characters.
Check if the two sets are equal.
Python3
def is_pandigital_set(n, b): digits = set(str(i) for i in range(b)) num_digits = set(str(n)) return digits == num_digits# test case 1n1 = "9651723480"b1 = 10print("Input:", "n =", n1, ", b =", b1)if is_pandigital_set(n1, b1): print("Output: Yes\nGiven number n has all digits from 0 to 9")else: print("Output: No")# test case 2n2 = "23456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"b2 = 36print("\nInput:", "n =", n2, ", b =", b2)if is_pandigital_set(n2, b2): print("Output: Yes\nGiven number n has all digits from 0 to 9")else: print("Output: No") |
Input: n = 9651723480 , b = 10 Output: Yes Given number n has all digits from 0 to 9 Input: n = 23456789ABCDEFGHIJKLMNOPQRSTUVWXYZ , b = 36 Output: No
Time complexity: O(n)
Space complexity: O(b)
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