# Count of alphabets whose ASCII values can be formed with the digits of N

Given an integer N. You can select any two digits from this number (the digits can be same but their positions should be different) and order them in any one of the two possible ways. For each of these ways, you create a two digit number from it (might contain leading zeros). Then, you will pick a character corresponding to the ASCII value equal to this number, i.e. the number 65 corresponds to ‘A’, 66 to ‘B’ and so on. The task is to count the number of english alphabets (lowercase or uppercase) that can be picked in this way.

Examples:

Input: N = 656
Output: 2
Only the characters ‘A’ (65) and ‘B’ (66) are possible.

Input: N = 1623455078
Output: 27

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

Approach: The idea is to observe that the total number of possible characters are (26 lowercase + 26 uppercase = 52). So, instead of generating all possible combinations of two digits from N, check the occurrences of these 52 characters.

Therefore, count the occurrences of each digit in N then for every character (lowercase or uppercase), find its ASCII value and check whether it can be picked from the given digits. Print the count of such alphabets in the end.

Below is the implementation of the above approach:

## Java

 // Java implementation of the approach class GFG {        // Function that returns true if num can be formed     // with the digits in digits[] array     static boolean canBePicked(int digits[], int num)     {         // Copy of the digits array         int copyDigits[] = digits.clone();         while (num > 0) {                // Get last digit             int digit = num % 10;                // If digit array doesn't contain current digit             if (copyDigits[digit] == 0)                 return false;                // One occurrence is used             else                 copyDigits[digit]--;                // Remove the last digit             num /= 10;         }            return true;     }        // Function to return the count of required alphabets     static int countAlphabets(int n)     {         int i, count = 0;            // To store the occurrences of digits (0 - 9)         int digits[] = new int[10];         while (n > 0) {                // Get last digit             int digit = n % 10;                // Update the occurrence of the digit             digits[digit]++;                // Remove the last digit             n /= 10;         }            // If any lowercase character can be picked          // from the current digits         for (i = 'a'; i <= 'z'; i++)             if (canBePicked(digits, i))                 count++;            // If any uppercase character can be picked          // from the current digits         for (i = 'A'; i <= 'Z'; i++)             if (canBePicked(digits, i))                 count++;            // Return the required count of alphabets         return count;     }        // Driver code     public static void main(String[] args)     {         int n = 1623455078;         System.out.println(countAlphabets(n));     } }

## Python3

 # Python3 implementation of the approach  import math    # Function that returns true if num  # can be formed with the digits in # digits[] array  def canBePicked(digits, num):         copyDigits = [];            # Copy of the digits array      for i in range(len(digits)):         copyDigits.append(digits[i]);                 while (num > 0):             # Get last digit          digit = num % 10;             # If digit array doesn't contain         # current digit          if (copyDigits[digit] == 0):              return False;             # One occurrence is used          else:             copyDigits[digit] -= 1;             # Remove the last digit          num = math.floor(num / 10);         return True;     # Function to return the count of # required alphabets  def countAlphabets(n):         count = 0;         # To store the occurrences of      # digits (0 - 9)      digits = [0] * 10;      while (n > 0):             # Get last digit          digit = n % 10;             # Update the occurrence of the digit          digits[digit] += 1;             # Remove the last digit          n = math.floor(n / 10);         # If any lowercase character can be      # picked from the current digits      for i in range(ord('a'), ord('z') + 1):          if (canBePicked(digits, i)):              count += 1;         # If any uppercase character can be      # picked from the current digits      for i in range(ord('A'), ord('Z') + 1):          if (canBePicked(digits, i)):              count += 1;         # Return the required count      # of alphabets      return count;     # Driver code  n = 1623455078;  print(countAlphabets(n));     # This code is contributed by mits

## C#

 // C# implementation of the approach using System;    class GFG  {        // Function that returns true     // if num can be formed with     // the digits in digits[] array     static bool canBePicked(int []digits, int num)     {         // Copy of the digits array         int []copyDigits = (int[]) digits.Clone();         while (num > 0)         {                // Get last digit             int digit = num % 10;                // If digit array doesn't              // contain current digit             if (copyDigits[digit] == 0)                 return false;                // One occurrence is used             else                 copyDigits[digit]--;                // Remove the last digit             num /= 10;         }            return true;     }        // Function to return the count      // of required alphabets     static int countAlphabets(int n)     {         int i, count = 0;            // To store the occurrences          // of digits (0 - 9)         int[] digits = new int[10];         while (n > 0)         {                // Get last digit             int digit = n % 10;                // Update the occurrence of the digit             digits[digit]++;                // Remove the last digit             n /= 10;         }            // If any lowercase character can be           // picked from the current digits         for (i = 'a'; i <= 'z'; i++)             if (canBePicked(digits, i))                 count++;            // If any uppercase character can be           // picked from the current digits         for (i = 'A'; i <= 'Z'; i++)             if (canBePicked(digits, i))                 count++;            // Return the required count of alphabets         return count;     }        // Driver code     public static void Main()     {         int n = 1623455078;         Console.WriteLine(countAlphabets(n));     } }    // This code is contributed by  // tufan_gupta2000

## PHP

 0)      {             // Get last digit          \$digit = \$num % 10;             // If digit array doesn't contain         // current digit          if (\$copyDigits[\$digit] == 0)              return false;             // One occurrence is used          else             \$copyDigits[\$digit]--;             // Remove the last digit          \$num = floor(\$num / 10);      }         return true;  }     // Function to return the count of // required alphabets  function countAlphabets(\$n)  {      \$count = 0;         // To store the occurrences of      // digits (0 - 9)      \$digits = array_fill(0, 10, 0);      while (\$n > 0)      {             // Get last digit          \$digit = \$n % 10;             // Update the occurrence of the digit          \$digits[\$digit]++;             // Remove the last digit          \$n = floor(\$n / 10);      }         // If any lowercase character can be      // picked from the current digits      for (\$i = ord('a'); \$i <= ord('z'); \$i++)          if (canBePicked(\$digits, \$i))              \$count++;         // If any uppercase character can be       // picked from the current digits      for (\$i = ord('A');           \$i <= ord('Z'); \$i++)          if (canBePicked(\$digits, \$i))              \$count++;         // Return the required count      // of alphabets      return \$count;  }     // Driver code  \$n = 1623455078;  echo countAlphabets(\$n);     // This code is contributed by Ryuga ?>

Output:

27

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