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Find all the possible mappings of characters in a sorted order
  • Difficulty Level : Expert
  • Last Updated : 11 Jun, 2021

Given a number find all the possible mappings of the characters in sorted order.
Examples: 
 

Input: 123
Output: ABC
         AW
         LC
Explanation:  
1 = A; 2 = B; 3 = C; 12 = L; 23 = W 

                 {1, 2, 3}                          
                /        \                          
               /          \                         
       "A"{2, 3}           "L"{3}                      
           /      \          /   \                  
          /        \        /     \
        "AB"{3}  "A"{23}  "LC"    null
        /  \        /  \
       /    \      /    \
      "ABC" null  "AW"  null 

Input : 2122
Output : BABB
         BAV
         BLB
         UBB
         UV

 

Approach :

  1. A recursive function which takes an input character array which contains the number stored in a form of characters, an output character array, and two variables(say i and j) that can be used to iterate over both the arrays is created.
  2. Using recursion, character for the ith digit in the input array is obtained and the corresponding mapped character with that digit is stored at the jth index in the output array.
  3. There will be two recursive calls. The first call will process a single-digit every time, whereas the second call will process two digits at a time.
  4. While processing two digits at a time, the number should be always less than 26 because after combining, the corresponding character has to lie between A to Z.
  5. At last, in the base case when the whole input string has been processed, the output array is printed.

Below is the implementation of the above approach. 
 

CPP




// C++ program to find all the possible
// string mappings of a given number
// in a sorted order
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the string mappings
void mapped(char inputarr[], char outputarr[],
            int i, int j)
{
 
    // Base case
    if (inputarr[i] == '\0') {
        outputarr[j] = '\0';
        cout << outputarr << endl;
        return;
    }
 
    // Convert the character to integer
    int digit = inputarr[i] - '0';
 
    // To store the characters corresponding
    // to the digits which are further
    // stored in outputarr[]
    char ch = digit + 'A' - 1;
    outputarr[j] = ch;
 
    // First recursive call taking one digit at a time
    mapped(inputarr, outputarr, i + 1, j + 1);
 
    if (inputarr[i + 1] != '\0') {
        int second_digit = inputarr[i + 1] - '0';
        int number = digit * 10 + second_digit;
 
        if (number <= 26) {
            ch = number + 'A' - 1;
            outputarr[j] = ch;
 
            // Second recursive call processing
            // two digits at a time
            mapped(inputarr, outputarr, i + 2, j + 1);
        }
    }
}
 
// Driver code
int main()
{
    char inputarr[] = { '1', '2', '3' };
    int m = pow(2, 3) - 1;
    int n = sizeof(m) / sizeof(int);
    char outputarr[n];
 
    mapped(inputarr, outputarr, 0, 0);
 
    return 0;
}

Java




// Java program to find all the possible
// string mappings of a given number
// in a sorted order
     
class GFG
{
     
    // Function to find the string mappings
    static void mapped(char inputarr[], char outputarr[],
                        int i, int j)
    {
     
        // Base case
        if (i >= inputarr.length)
        {
            String str = new String(outputarr);
            System.out.println(str.substring(0, j));
            return;
        }
     
        // Convert the character to integer
        int digit = inputarr[i] - '0';
     
        // To store the characters corresponding
        // to the digits which are further
        // stored in outputarr[]
        char ch = (char)(digit + (int)('A') - 1);
        outputarr[j] = ch;
     
        // First recursive call taking one digit at a time
        mapped(inputarr, outputarr, i + 1, j + 1);
     
        if (i + 1 < inputarr.length)
        {
            int second_digit = inputarr[i + 1] - '0';
            int number = digit * 10 + second_digit;
     
            if (number <= 26)
            {
                ch = (char)(number + (int)'A' - 1);
                outputarr[j] = ch;
     
                // Second recursive call processing
                // two digits at a time
                mapped(inputarr, outputarr, i + 2, j + 1);
            }
        }
    }
     
    // Driver code
    public static void main (String[] args)
    {
        char inputarr[] = { '1', '2', '3' };
        int m = (int)Math.pow(2, 3) - 1;
        int n = 1;
        char outputarr[] = new char[m];
     
        mapped(inputarr, outputarr, 0, 0);
    }
}
 
// This code is contributed by AnkitRai01

Python3




# Python3 program to find all the possible
# string mappings of a given number
# in a sorted order
 
# Function to find the string mappings
def mapped(inputarr, outputarr, i, j):
 
    # Base case
    if (i == len(inputarr)):
        print("".join(outputarr[:j]))
        return
 
    # Convert the character to integer
    digit = ord(inputarr[i]) - ord('0')
 
    # To store the characters corresponding
    # to the digits which are further
    # stored in outputarr[]
    ch = digit + ord('A') - 1
    outputarr[j] = chr(ch)
 
    # First recursive call taking one digit at a time
    mapped(inputarr, outputarr, i + 1, j + 1)
 
    if (i + 1 < len(inputarr) and [i + 1] != '\0'):
        second_digit = ord(inputarr[i + 1]) - ord('0')
        number = digit * 10 + second_digit
 
        if (number <= 26):
            ch = number + ord('A') - 1
            outputarr[j] = chr(ch)
 
            # Second recursive call processing
            # two digits at a time
            mapped(inputarr, outputarr, i + 2, j + 1)
 
# Driver code
inputarr = ['1', '2', '3']
m = pow(2, 3) - 1
n = 1
outputarr = ['0'] * m
 
mapped(inputarr, outputarr, 0, 0)
 
# This code is contributed by mohit kumar 29

C#




// C# program to find all the possible
// string mappings of a given number
// in a sorted order
using System;
 
class GFG
{
     
    // Function to find the string mappings
    static void mapped(char []inputarr, char []outputarr,
                        int i, int j)
    {
     
        // Base case
        if (i >= inputarr.Length)
        {
            string str = new string(outputarr);
            Console.WriteLine(str.Substring(0, j));
            return;
        }
     
        // Convert the character to integer
        int digit = inputarr[i] - '0';
     
        // To store the characters corresponding
        // to the digits which are further
        // stored in outputarr[]
        char ch = (char)(digit + (int)('A') - 1);
        outputarr[j] = ch;
     
        // First recursive call taking one digit at a time
        mapped(inputarr, outputarr, i + 1, j + 1);
     
        if (i + 1 < inputarr.Length)
        {
            int second_digit = inputarr[i + 1] - '0';
            int number = digit * 10 + second_digit;
     
            if (number <= 26)
            {
                ch = (char)(number + (int)'A' - 1);
                outputarr[j] = ch;
     
                // Second recursive call processing
                // two digits at a time
                mapped(inputarr, outputarr, i + 2, j + 1);
            }
        }
    }
     
    // Driver code
    public static void Main ()
    {
        char []inputarr = { '1', '2', '3' };
        int m = (int)Math.Pow(2, 3) - 1;
        int n = 1;
        char []outputarr = new char[m];
     
        mapped(inputarr, outputarr, 0, 0);
    }
}
 
// This code is contributed by AnkitRai01

Javascript




<script>
 
// JavaScript program to find all the possible
// string mappings of a given number
// in a sorted order
 
 // Function to find the string mappings
function mapped(inputarr,outputarr,i,j)
{
    // Base case
        if (i >= inputarr.length)
        {
            let str = (outputarr).join("");
            document.write(str.substring(0, j)+"<br>");
            return;
        }
       
        // Convert the character to integer
        let digit = inputarr[i].charCodeAt(0) -
        '0'.charCodeAt(0);
       
        // To store the characters corresponding
        // to the digits which are further
        // stored in outputarr[]
        let ch =
        String.fromCharCode(digit + ('A').charCodeAt(0) - 1);
        outputarr[j] = ch;
       
        // First recursive call taking one digit at a time
        mapped(inputarr, outputarr, i + 1, j + 1);
       
        if (i + 1 < inputarr.length)
        {
            let second_digit = inputarr[i + 1].charCodeAt(0) -
            '0'.charCodeAt(0);
            let number = digit * 10 + second_digit;
       
            if (number <= 26)
            {
                ch =
                String.fromCharCode(number + 'A'.charCodeAt(0) - 1);
                outputarr[j] = ch;
       
                // Second recursive call processing
                // two digits at a time
                mapped(inputarr, outputarr, i + 2, j + 1);
            }
        }
}
// Driver code
let inputarr=['1', '2', '3'];
let m = Math.pow(2, 3) - 1;
let n = 1;
let outputarr = new Array(m);
 mapped(inputarr, outputarr, 0, 0);
 
 
 
// This code is contributed by patel2127
 
</script>
Output: 
ABC
AW
LC

 

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