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Print all possible words from phone digits

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  • Difficulty Level : Medium
  • Last Updated : 23 Jun, 2022

Before the advent of QWERTY keyboards, texts and numbers were placed on the same key. For example, 2 has “ABC” if we wanted to write anything starting with ‘A’ we need to type key 2 once. If we wanted to type ‘B’, press key 2 twice and thrice for typing ‘C’. Below is a picture of such a keypad.

Mobile-keypad

Given a keypad as shown in the diagram, and an n digit number, list all words which are possible by pressing these numbers.

Example: 

Input number: 234
Output:
adg adh adi aeg aeh aei afg afh 
afi bdg bdh bdi beg beh bei bfg 
bfh bfi cdg cdh cdi ceg ceh cei 
cfg cfh cfi

Explanation: All possible words which can be 
formed are (Alphabetical order):
adg adh adi aeg aeh aei afg afh 
afi bdg bdh bdi beg beh bei bfg 
bfh bfi cdg cdh cdi ceg ceh cei 
cfg cfh cfi
If 2 is pressed then the alphabet
can be a, b, c, 
Similarly, for 3, it can be 
d, e, f, and for 4 can be g, h, i. 

Input number: 5
Output: j k l
Explanation: All possible words which can be 
formed are (Alphabetical order):
j, k, l, only these three alphabets 
can be written with j, k, l.

Approach: It can be observed that each digit can represent 3 to 4 different alphabets (apart from 0 and 1). So the idea is to form a recursive function. Then map the number with its string of probable alphabets, i.e 2 with “abc”, 3 with “def” etc. Now the recursive function will try all the alphabets, mapped to the current digit in alphabetic order, and again call the recursive function for the next digit and will pass on the current output string.

Example: 

If the number is 23,

Then for 2, the alphabets are a, b, c
So 3 recursive function will be called 
with output string as a, b, c respectively 
and for 3 there are 3 alphabets d, e, f
So, the output will be ad, ae and af for 
the recursive function with output string.
Similarly, for b and c, the output will be: 
bd, be, bf and cd, ce, cf respectively.

Algorithm:  

  1. Map the number with its string of probable alphabets, i.e 2 with “abc”, 3 with “def” etc.
  2. Create a recursive function which takes the following parameters, output string, number array, current index, and length of number array
  3. If the current index is equal to the length of the number array then print the output string.
  4. Extract the string at digit[current_index] from the Map, where the digit is the input number array.
  5. Run a loop to traverse the string from start to end
  6. For every index again call the recursive function with the output string concatenated with the ith character of the string and the current_index + 1.

Implementation: Note that the input number is represented as an array to simplify the code. 

C++




// C++ program for the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find all possible combinations by
// replacing key's digits with characters of the 
// corresponding list
void findCombinations(vector<char> keypad[], 
                      int input[], string res, int index, int n)
{
    // If processed every digit of key, print result
    if (index == n) {
        cout << res << " ";
        return;
    }
  
    // Stores current digit
    int digit = input[index];
  
    // Size of the list corresponding to current digit
    int len = keypad[digit].size();
  
    // One by one replace the digit with each character in the
    // corresponding list and recur for next digit
    for (int i = 0; i < len; i++) {
        findCombinations(keypad, input, res + keypad[digit][i], index + 1, n);
    }
}
  
// Driver Code
int main()
{
    // Given mobile keypad
    vector<char> keypad[] =
    {
        {}, {},        // 0 and 1 digit don't have any characters associated
        { 'a', 'b', 'c' },
        { 'd', 'e', 'f' },
        { 'g', 'h', 'i' },
        { 'j', 'k', 'l' },
        { 'm', 'n', 'o' },
        { 'p', 'q', 'r', 's'},
        { 't', 'u', 'v' },
        { 'w', 'x', 'y', 'z'}
    };
  
    // Given input array
    int input[] = { 2, 3, 4 };
    
    // Size of the array
    int n = sizeof(input)/sizeof(input[0]);
  
    // Function call to find all combinations
    findCombinations(keypad, input, string(""), 0, n );
  
    return 0;
}

C




#include <stdio.h>
#include <string.h>
  
// hashTable[i] stores all characters that correspond to
// digit i in phone
const char hashTable[10][5]
    = { "",    "",    "abc""def", "ghi",
        "jkl", "mno", "pqrs", "tuv", "wxyz" };
  
// A recursive function to print all possible words that can
// be obtained by input number[] of size n.  The output
// words are one by one stored in output[]
void printWordsUtil(int number[], int curr_digit,
                    char output[], int n)
{
    // Base case, if current output word is prepared
    int i;
    if (curr_digit == n) {
        printf("%s ", output);
        return;
    }
  
    // Try all 3 possible characters for current digit in
    // number[] and recur for remaining digits
    for (i = 0; i < strlen(hashTable[number[curr_digit]]);
         i++) {
        output[curr_digit]
            = hashTable[number[curr_digit]][i];
        printWordsUtil(number, curr_digit + 1, output, n);
        if (number[curr_digit] == 0
            || number[curr_digit] == 1)
            return;
    }
}
  
// A wrapper over printWordsUtil().  It creates an output
// array and calls printWordsUtil()
void printWords(int number[], int n)
{
    char result[n + 1];
    result[n] = '\0';
    printWordsUtil(number, 0, result, n);
}
  
// Driver program
int main(void)
{
    int number[] = { 2, 3, 4 };
    int n = sizeof(number) / sizeof(number[0]);
    printWords(number, n);
    return 0;
}

Java




// Java program to implement the
// above approach
import java.util.*;
import java.lang.*;
import java.io.*;
class NumberPadString{
       
static Character[][] numberToCharMap;
  
private static List<String> printWords(int[] numbers, 
                                       int len, 
                                       int numIndex, 
                                       String s)
{
  if(len == numIndex)
  {
    return new ArrayList<>(Collections.singleton(s));
  }
    
  List<String> stringList = new ArrayList<>();
    
  for(int i = 0
          i < numberToCharMap[numbers[numIndex]].length; i++)
  {
    String sCopy = 
           String.copyValueOf(s.toCharArray());
    sCopy = sCopy.concat(
            numberToCharMap[numbers[numIndex]][i].toString());
    stringList.addAll(printWords(numbers, len, 
                                 numIndex + 1
                                 sCopy));
  }
  return stringList;
}
      
private static void printWords(int[] numbers)
{
  generateNumberToCharMap();
  List<String> stringList = 
       printWords(numbers, numbers.length, 0, "");
  stringList.stream().forEach(System.out :: println);
}
  
private static void generateNumberToCharMap()
{
  numberToCharMap = new Character[10][5];
  numberToCharMap[0] = new Character[]{'\0'};
  numberToCharMap[1] = new Character[]{'\0'};
  numberToCharMap[2] = new Character[]{'a','b','c'};
  numberToCharMap[3] = new Character[]{'d','e','f'};
  numberToCharMap[4] = new Character[]{'g','h','i'};
  numberToCharMap[5] = new Character[]{'j','k','l'};
  numberToCharMap[6] = new Character[]{'m','n','o'};
  numberToCharMap[7] = new Character[]{'p','q','r','s'};
  numberToCharMap[8] = new Character[]{'t','u','v'};
  numberToCharMap[9] = new Character[]{'w','x','y','z'};
}
  
// Driver code  
public static void main(String[] args) 
{
  int number[] = {2, 3, 4};
  printWords(number);
}
}
  
// This code is contributed by ankit pachori 1

Python3




# hashTable[i] stores all characters
# that correspond to digit i in phone
hashTable = ["", "", "abc", "def", "ghi", "jkl",
             "mno", "pqrs", "tuv", "wxyz"]
  
# A recursive function to print all
# possible words that can be obtained
# by input number[] of size n. The
# output words are one by one stored
# in output[]
  
  
def printWordsUtil(number, curr, output, n):
    if(curr == n):
        print(output)
        return
  
    # Try all 3 possible characters
    # for current digit in number[]
    # and recur for remaining digits
    for i in range(len(hashTable[number[curr]])):
        output.append(hashTable[number[curr]][i])
        printWordsUtil(number, curr + 1, output, n)
        output.pop()
        if(number[curr] == 0 or number[curr] == 1):
            return
  
# A wrapper over printWordsUtil().
# It creates an output array and
# calls printWordsUtil()
  
  
def printWords(number, n):
    printWordsUtil(number, 0, [], n)
  
  
# Driver function
if __name__ == '__main__':
    number = [2, 3, 4]
    n = len(number)
    printWords(number, n)
  
# This code is contributed by prajmsidc

C#




using System;
/*
  C# Program for
  Print all possible words from phone digits
*/
public class GFG {
    // Function to find all possible combinations by
    // replacing key's digits with characters of the
    // corresponding list
    static void findCombinations(String[] keypad,
                                 int[] input, String result,
                                 int index, int n)
    {
        // If processed every digit of key, print result
        if (index == n) {
            Console.Write(result + "\t");
            return;
        }
  
        // Stores current digit
        int digit = input[index];
        // Size of the list corresponding to current digit
        int len = keypad[digit].Length;
        // One by one replace the digit with each character
        // in the corresponding list and recur for next
        // digit
        for (int i = 0; i < len; i++) {
            findCombinations(keypad, input,
                             result + keypad[digit][i],
                             index + 1, n);
        }
    }
  
    public static void Main(String[] args)
    {
        // Keypad word of number of (1 to 9)
        // 0 and 1 digit don't have any characters
        // associated
        String[] keypad
            = { "",    "",    "abc""def", "ghi",
                "jkl", "mno", "pqrs", "tuv", "wxyz" };
        String result = "";
        // Given input array
        int[] input = { 2, 3, 4 };
        // Size of the array
        int n = input.Length;
        // Function call to find all combinations
        findCombinations(keypad, input, result, 0, n);
    }
}
// This code is contributed by Aarti_Rathi

Javascript




<script>
  
// Javascript program to implement the
// above approach
let hashTable = [ "", "", "abc", "def", "ghi", "jkl",
                  "mno", "pqrs", "tuv", "wxyz" ];
                    
// A recursive function to print all possible 
// words that can be obtained by input number[]
// of size n. The output words are one by one 
// stored in output[]
function printWordsUtil(number, curr, output, n)
{
      
    // Base case, if current output
    // word is prepared
    if (curr == n)
      {
        document.write(output.join("") + "<br>")
        return;
      }
        
      // Try all 3 possible characters for current
      // digit in number[] and recur for remaining digits
    for(let i = 0; 
            i < hashTable[number[curr]].length; 
            i++)
    {
        output.push(hashTable[number[curr]][i]);
        printWordsUtil(number, curr + 1, output, n);
          
        output.pop();
          
        if(number[curr] == 0 || number[curr] == 1)
            return
    }
}
  
// A wrapper over printWordsUtil(). It creates 
// an output array and calls printWordsUtil()
function printWords(numbers, n)
{
    printWordsUtil(number, 0, [], n);
}
  
// Driver code 
let number = [ 2, 3, 4 ];
let n = number.length;
  
printWords(number, n);
  
// This code is contributed by avanitrachhadiya2155
  
</script>

Output

adg adh adi aeg aeh aei afg afh afi bdg bdh bdi beg beh
 bei bfg bfh bfi cdg cdh cdi ceg ceh cei cfg cfh cfi 

Complexity Analysis:  

  • Time Complexity: O(4n), where n is a number of digits in the input number. 
    Each digit of a number has 3 or 4 alphabets, so it can be said that each digit has 4 alphabets as options. If there are n digits then there are 4 options for the first digit and for each alphabet of the first digit there are 4 options in the second digit, i.e for every recursion 4 more recursions are called (if it does not match the base case). So the time complexity is O(4n).
  • Space Complexity:O(1). 
    As no extra space is needed.

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