Given an integer N, the task is to check whether the given number is a pangram or not.
Note: A Pangram Number contains every digit [0- 9] at least once.
Examples:
Input : N = 10239876540022
Output : Yes
Explanation: N contains all the digits from 0 to 9. Therefore, it is a pangram.Input : N = 234567890
Output : No
Explanation: N doesn’t contain the digit 1. Therefore, it is not a pangram.
Set-based Approach: The idea is to use Sets to store the count of distinct digits present in N. Follow the steps below to solve the problem:
- Convert the number N to equivalent string.
- Convert the string to a set.
- If the size of the Set is 10, then it contains all the distinct possible digits [0 – 9]. Therefore, print “Yes”.
- Otherwise, print “No”
Below is the implementation of the above approach:
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Function to check if N// is a Pangram or notstring numberPangram(string N){ // Add all characters pf arrNum to set set<char> setNum; for (int i = 0; i < N.length(); i++) { setNum.insert(N[i]); } // If the length of set is 10 // The number is a Pangram if (setNum.size() == 10) return "True"; else return "False";}// Driver codeint main(){ string N = "10239876540022"; cout << (numberPangram(N));}// This code is contributed by ukasp. |
Java
// Java implementation of above approachimport java.math.BigInteger;import java.util.HashSet;class GFG{ // Function to check if N// is a Pangram or notstatic String numberPangram(BigInteger N){ // Stores equivalent string // representation of N String num = N.toString(); // Convert the string to Character array char[] arrNum = num.toCharArray(); // Add all characters pf arrNum to set HashSet<Character> setNum = new HashSet<Character>(); for(char ch : arrNum) { setNum.add(ch); } // If the length of set is 10 // The number is a Pangram if (setNum.size() == 10) return "True"; else return "False";}// Driver codepublic static void main(String[] args){ BigInteger N = new BigInteger("10239876540022"); System.out.print(numberPangram(N));}}// This code is contributed by abhinavjain194 |
Python3
# Python3 implementation of above approach# Function to check if N# is a Pangram or notdef numberPangram(N): # Stores equivalent string # representation of N num = str(N) # Convert the string to set setnum = set(num) # If the length of set is 10 if(len(setnum) == 10): # The number is a Pangram return True else: return False# Driver CodeN = 10239876540022print(numberPangram(N)) |
C#
// C# implementation of above approachusing System;using System.Globalization;using System.Numerics;using System.Collections.Generic;class GFG{ // Function to check if N// is a Pangram or notstatic String numberPangram(ulong N){ // Stores equivalent string // representation of N string num = N.ToString(); // Convert the string to Character array char[] arrNum = num.ToCharArray(); // Add all characters pf arrNum to set HashSet<char> setNum = new HashSet<char>(); foreach(char ch in arrNum) { setNum.Add(ch); } // If the length of set is 10 // The number is a Pangram if (setNum.Count == 10) return "True"; else return "False";}// Driver Code static void Main() { ulong N = 10239876540022; Console.Write(numberPangram(N)); }}// This code is contributed by SoumikMondal |
Javascript
<script>// Javascript implementation of above approach// Function to check if N// is a Pangram or notfunction numberPangram(N){ // Add all characters pf arrNum to set var setNum = new Set(); for (var i = 0; i < N.length; i++) { setNum.add(N[i]); } // If the length of set is 10 // The number is a Pangram if (setNum.size == 10) return "True"; else return "False";}// Driver codevar N = "10239876540022";document.write(numberPangram(N));</script> |
Output:
True
Time Complexity: O(log10N * log(log10N))
Auxiliary Space: O(1)
Hashing-based Approach: Follow the steps to solve the problem:
- Convert N to its equivalent string.
- Calculate frequencies of all characters in this string. Counter() function can be used for this purpose in Python.
- If the length of the Dictionary / HashMap storing the frequencies is 10, the number contains all possible distinct digits. Therefore, print “Yes“.
- Otherwise, print “No”.
Below is the implementation of the above approach:
C++
// C++ code to implement the approach#include <iostream>#include <string>#include <map>bool numberPangram(long N) { // Stores equivalent string representation of N std::string num = std::to_string(N); // Count frequencies of digits present in N std::map<char, int> frequency; for (int i = 0; i < num.length(); i++) { char digit = num[i]; frequency[digit]++; } // If the size of the map frequency is 10 if (frequency.size() == 10) { // The number is a Pangram return true; } else { return false; }}// Driver codeint main() { long N = 10239876540022; std::cout << (numberPangram(N) ? "true" : "false") << std::endl; return 0;}// This code is contributed by phasing17 |
Java
import java.util.*;class GFG { // Driver code public static void main(String[] args) { long N = 10239876540022L; boolean result = numberPangram(N); System.out.println(result); } public static boolean numberPangram(long N) { // Stores equivalent string representation of N String num = Long.toString(N); // Count frequencies of digits present in N Map<Character, Integer> frequency = new HashMap<>(); for (int i = 0; i < num.length(); i++) { char digit = num.charAt(i); if (frequency.containsKey(digit)) { frequency.put(digit, frequency.get(digit) + 1); } else { frequency.put(digit, 1); } } // If the length of the dictionary frequency is 10 if (frequency.size() == 10) { // The number is a Pangram return true; } else { return false; } }}// This code is contributed by phasing17. |
Python3
# Python implementation of above approachfrom collections import Counter# Function to check if # N is a Pangram or notdef numberPangram(N): # Stores equivalent string # representation of N num = str(N) # Count frequencies of # digits present in N frequency = Counter(num) # If the length of the # dictionary frequency is 10 if(len(frequency) == 10): # The number is a Pangram return True else: return False# Driver CodeN =10239876540022print(numberPangram(N)) |
C#
// C# code to implement the approachusing System;using System.Collections.Generic;class GFG { // Driver Code static void Main(string[] args) { long N = 10239876540022L; // Function call bool result = NumberPangram(N); Console.WriteLine(result); } static bool NumberPangram(long N) { // Stores equivalent string representation of N string num = N.ToString(); // Count frequencies of digits present in N Dictionary<char, int> frequency = new Dictionary<char, int>(); for (int i = 0; i < num.Length; i++) { char digit = num[i]; if (frequency.ContainsKey(digit)) { frequency[digit]++; } else { frequency.Add(digit, 1); } } // If the length of the dictionary frequency is 10 if (frequency.Count == 10) { // The number is a Pangram return true; } else { return false; } }}// This code is contributed by phasing17. |
Javascript
// JavaScript implementation of above approach// Function to check if // N is a Pangram or notfunction numberPangram(N) { // Stores equivalent string // representation of N let num = N.toString(); // Count frequencies of // digits present in N let frequency = {}; for (let i = 0; i < num.length; i++) { let digit = num[i]; if (frequency[digit]) { frequency[digit] += 1; } else { frequency[digit] = 1; } } // If the length of the // dictionary frequency is 10 if (Object.keys(frequency).length === 10) { // The number is a Pangram return True; } else { return False; }}// Driver Codelet N = 10239876540022;console.log(numberPangram(N)); |
Output:
True
Time Complexity: O(log10N * log(log10N))
Auxiliary Space: O(1)
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