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:
C++
// C++ implementation of the approach #include<bits/stdc++.h> using namespace std; // Function that returns true if num // can be formed with the digits in // digits[] array bool canBePicked(int digits[], int num) { int copyDigits[10]; // Copy of the digits array for(int i =0 ; i < 10;i++) copyDigits[i]=digits[i]; 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] -= 1; // Remove the last digit num = floor(num / 10); } return true; } // Function to return the count of // required alphabets int countAlphabets(long n) { int count = 0; // To store the occurrences of // digits (0 - 9) int digits[10]= {0}; while (n > 0) { // Get last digit int digit = n % 10; // Update the occurrence of the digit digits[digit] += 1; // Remove the last digit n = floor(n / 10); } // If any lowercase character can be // picked from the current digits for(int i = 97; i <= 122 ;i ++) if (canBePicked(digits, i)) count += 1; // If any uppercase character can be // picked from the current digits for(int i = 65; i < 91;i++) if (canBePicked(digits, i)) count += 1; // Return the required count // of alphabets return count; } // Driver code int main() { long n = 1623455078; cout<<(countAlphabets(n)); } // This code is contributed by chitranayal |
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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)); } } |
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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 |
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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 |
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PHP
<?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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