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:
- Write a recursive function that accepts key press string and returns all the words that can be formed in an Array list.
- If the length of the given string is 0 then return Arraylist containing empty string.
- 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
- Get the character set of the first character of original string, CSet
- 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.
- Return the arraylist, ans.
Implementation:
Java
// Java implementation of the approachimport 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)); }} |
[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.
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