Print all possible words from phone digits

Difficulty Level : Hard
Last Updated : 18 Nov, 2020

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:

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

#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 digir 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 digir 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

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
Process returned 0 (0x0)   execution time : 0.025 s
Press any key to continue.

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.

Reference:
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This article is contributed by Jitendra Sangar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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