Java ArrayList to print all possible words from phone digits

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 character of Cset infront of the word of list and insert them in the ans arraylist.
6. Return the arraylist, ans.

Implementation:

Java

// 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). 
  The space required to store all words is O(3ⁿ). As there will be around 3ⁿ words in the output.