Skip to content
Related Articles

Related Articles

Improve Article
Java ArrayList to print all possible words from phone digits
  • Difficulty Level : Hard
  • Last Updated : 27 Apr, 2020

Given a keypad of a mobile, and keys that need to be pressed, the task is to print all the words which are possible to generate by pressing these numbers.

Examples:

Input: str = "12" 
Output: [ad, bd, cd, ae, be, ce, af, bf, cf]
Explanation: The characters that can be formed
by pressing 1 is a, b, c and by pressing 2 characters
d, e, f can be formed.
So all the words will be a combination where first 
character belongs to a, b, c and 2nd character belongs
to d, e, f
 
Input: str = "4"
Output: [j, k, l]
Explanation: The characters that can be formed
by pressing 4 is j, k, l

Method 1: Another approach is discussed here Print all possible words from phone digits

Method 2:
Approach: The approach is slightly different from the approach in the other article. Suppose there are n keys which are pressed (a1 a2 a3 ..an). Find all the words that can be formed using (a2 a3 ..an). Suppose 3 characters can be generated by pressing a1 then for every character concatenate the character before all the words and insert them to the list.



For Example:

If the keypress is 12
The characters that can be formed by pressing 1 is a, b, c and by pressing 2 characters d, e, f can be formed.
So all the words that can be formed using 2 are [d, e, f]
So now concatenate ‘a’ with all words returned so, the list is [ad, ae, af] similarly concatenate b and c. So the list becomes [ad, ae, af, bd, be, bf, cd, ce, cf].

Algorithm:

  1. Write a recursive function that accepts key press string and returns all the words that can be formed in an Array list.
  2. If the length of the given string is 0 then return Arraylist containing empty string.
  3. Else recursively call the function with a string except the first character of original string, i.e string containing all the characters from index 1 to n-1. and store the arraylist returned, list and create a new arraylist ans
  4. Get the character set of the first character of original string, CSet
  5. For every word of the list run a loop through the Cset and concatenate the charcter of Cset infront of the word of list and insert them in the ans arraylist.
  6. Return the arraylist, ans.

Implementation:




// Java implementation of the approach
import java.util.ArrayList;
  
public class GFG {
  
    // String array to store keypad characters
    static final String codes[]
        = { " ", "abc", "def",
            "ghi", "jkl", "mno",
            "pqr", "stu", "vwx",
            "yz" };
  
    // Function that returns an Arraylist
    // which contains all the generated words
    public static ArrayList<String> printKeyWords(String str)
    {
  
        // If str is empty
        if (str.length() == 0) {
            ArrayList<String> baseRes = new ArrayList<>();
            baseRes.add("");
  
            // Return an Arraylist containing
            // empty string
            return baseRes;
        }
  
        // First character of str
        char ch = str.charAt(0);
  
        // Rest of the characters of str
        String restStr = str.substring(1);
  
        ArrayList<String> prevRes = printKeyWords(restStr);
        ArrayList<String> Res = new ArrayList<>();
  
        String code = codes[ch - '0'];
  
        for (String val : prevRes) {
  
            for (int i = 0; i < code.length(); i++) {
                Res.add(code.charAt(i) + val);
            }
        }
        return Res;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        String str = "23";
  
        // Print all the possible words
        System.out.println(printKeyWords(str));
    }
}
Output:
[dg, eg, fg, dh, eh, fh, di, ei, fi]

Complexity Analysis:

  • Time Complexity: O(3n).
    Though the recursive function runs n times. But the size of the arraylist grows exponentially. So there will be around 3n elements in the arraylist. Therefore, traversing them will take 3n time.
  • Space Complexity:O(3n).
    Space required to store all words is O(3n). As there will be around 3n words in the output.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live




My Personal Notes arrow_drop_up
Recommended Articles
Page :