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

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

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// 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));
    }
}

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Python3














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# 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 



















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C#














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// 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 



















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PHP














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<?php 


// PHP implementation of the approach  


  


// Function that returns true if num  


// can be formed with the digits in 


// digits[] array  


function canBePicked($digits, $num)  


{  


    $copyDigits = array(); 


      


    // Copy of the digits array  


    for($i = 0; $i < sizeof($digits); $i++) 


        $copyDigits[$i] = $digits[$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]--;  


  


        // 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 


?> 



















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Output:


27






          
          
          
          

            

    

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