Check if a number is a Pangram or not

Last Updated : 08 Apr, 2021

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:

Java

// Java implementation of above approach
import java.math.BigInteger;
import java.util.HashSet;
 
class GFG{
     
// Function to check if N
// is a Pangram or not
static 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 code
public 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 not
def 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 Code
 
N = 10239876540022
print(numberPangram(N))

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:

Python3

# Python implementation of above approach
 
from collections import Counter
 
# Function to check if
# N is a Pangram or not
def 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 Code
 
N =10239876540022
print(numberPangram(N))

Output:

True

Time Complexity: O(log₁₀N * log(log₁₀N))
Auxiliary Space: O(1)

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