Given a keypad as shown in the diagram, and an n digit number, list all words which are possible by pressing these numbers.
Before the advent of QWERTY keyboards, texts and numbers were placed on the same key. For example, 2 has “ABC”, so 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.
Examples:
Input: 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 cfiExplanation: 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: 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: To solve the problem follow the below idea:
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.
Illustration: 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: Follow the steps below to solve the problem:
- Map the number with its string of probable alphabets, i.e 2 with “abc”, 3 with “def” etc.
- Create a recursive function that 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.
Below is the implementation of the above approach:
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 listvoid 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 Codeint 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 findCombinations(keypad, input, string(""), 0, n); return 0;} |
C
// C program for the above approach#include <stdio.h>#include <string.h>// hashTable[i] stores all characters that correspond to// digit i in phoneconst 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 codeint main(void){ int number[] = { 2, 3, 4 }; int n = sizeof(number) / sizeof(number[0]); // Function call printWords(number, n); return 0;} |
Java
// Java program to implement the// above approachimport java.io.*;import java.lang.*;import java.util.*;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 }; // Function call printWords(number); }}// This code is contributed by ankit pachori 1 |
Python3
# Python3 program for the above approach# hashTable[i] stores all characters# that correspond to digit i in phonehashTable = ["", "", "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 functionif __name__ == '__main__': number = [2, 3, 4] n = len(number) # Function call 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 approachlet 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
Time Complexity: O(4n), where n is the 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).
Auxiliary Space: O(n)
- No extra data structure is explicitly used, so one may think there is no extra space required
- But in a recursive function, the space complexity depends on the maximum depth of the recursion tree
- Each function call creates a new stack frame that stores its local variables and parameters
- The recursion tree here has a depth of n, that’s how long a branch can be
- Therefore, complexity will be proportional to n, that is O(n)
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