Iterative Letter Combinations of a Phone Number

Given an integer array containing digits from [0, 9], the task is to print all possible letter combinations that the numbers could represent. 

A mapping of digit to letters (just like on the telephone buttons) is being followed. Note that 0 and 1 do not map to any letters. All the mapping are shown in the image below: 

Example:

Input: arr[] = {2, 3} 
Output: ad ae af bd be bf cd ce cf

Input: arr[] = {9} 
Output: w x y z 

Approach: Now let us think how we would approach this problem without doing it in an iterative way. A recursive solution is intuitive and common. We keep adding each possible letter recursively and this will generate all the possible strings.
Let us think about how we can build an iterative solution using the recursive one. Recursion is possible through the use of a stack. So if we use a stack instead of a recursive function will that be an iterative solution? One could say so speaking technically but we then aren’t really doing anything different in terms of logic.

A Stack is a LIFO DS. Can we use another Data structure? What will be the difference if we use a FIFO DS? Let’s say a queue. Since BFS is done by queue and DFS by stack is there any difference between the two?
The difference between DFS and BFS is similar to this question. In DFS we will find each path possible in the tree one by one. It will perform all steps for a path first whereas BFS will build all paths together one step at a time.
So, a queue would work perfectly for this question. The only difference between the two algorithms using queue and stack will be the way in which they are formed. Stack will form all strings completely one by one whereas the queue will form all the strings together i.e. after x number of passes all the strings will have a length of x.

For example:

If the given number is "23", 
then using queue, the letter combinations 
obtained will be:
["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"] 
and using stack, the letter combinations obtained will 
be:
["cf","ce","cd","bf","be","bd","af","ae","ad"].

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return a vector that contains
// all the generated letter combinations
vector<string> letterCombinationsUtil(const int number[],
                                      int n,
                                      const string table[])
{
    // To store the generated letter combinations
    vector<string> list;

    queue<string> q;
    q.push("");

    while (!q.empty()) {
        string s = q.front();
        q.pop();

        // If complete word is generated
        // push it in the list
        if (s.length() == n)
            list.push_back(s);
        else

            // Try all possible letters for current digit
            // in number[]
            for (auto letter : table[number[s.length()]])
                q.push(s + letter);
    }

    // Return the generated list
    return list;
}

// Function that creates the mapping and
// calls letterCombinationsUtil
void letterCombinations(const int number[], int n)
{

    // table[i] stores all characters that
    // corresponds to ith digit in phone
    string table[10]
        = { "0",   "1",   "abc",  "def", "ghi",
            "jkl", "mno", "pqrs", "tuv", "wxyz" };

    vector<string> list
        = letterCombinationsUtil(number, n, table);

    // Print the contents of the vector
    for (auto word : list)
        cout << word << " ";

    return;
}

// Driver code
int main()
{
    int number[] = { 2, 3 };
    int n = sizeof(number) / sizeof(number[0]);

    // Function call
    letterCombinations(number, n);

    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
import java.util.*;

class GFG {
    // Function to return a vector that contains
    // all the generated letter combinations
    static ArrayList<String>
    letterCombinationsUtil(int[] number, int n,
                           String[] table)
    {
        // To store the generated letter combinations
        ArrayList<String> list = new ArrayList<>();

        Queue<String> q = new LinkedList<>();
        q.add("");

        while (!q.isEmpty()) {
            String s = q.remove();

            // If complete word is generated
            // push it in the list
            if (s.length() == n)
                list.add(s);
            else {
                String val = table[number[s.length()]];
                for (int i = 0; i < val.length(); i++)
                {
                    q.add(s + val.charAt(i));
                }
            }
        }
        return list;
    }

    // Function that creates the mapping and
    // calls letterCombinationsUtil
    static void letterCombinations(int[] number, int n)
    {
        // table[i] stores all characters that
        // corresponds to ith digit in phone
        String[] table
            = { "0",   "1",   "abc",  "def", "ghi",
                "jkl", "mno", "pqrs", "tuv", "wxyz" };

        ArrayList<String> list
            = letterCombinationsUtil(number, n, table);

        // Print the contents of the list
        for (int i = 0; i < list.size(); i++) {
            System.out.print(list.get(i) + " ");
        }
    }

    // Driver code
    public static void main(String args[])
    {
        int[] number = { 2, 3 };
        int n = number.length;

        // Function call
        letterCombinations(number, n);
    }
}

// This code is contributed by rachana soma

Python3

# Python3 implementation of the approach
from collections import deque

# Function to return a list that contains
# all the generated letter combinations


def letterCombinationsUtil(number, n, table):

    list = []
    q = deque()
    q.append("")

    while len(q) != 0:
        s = q.pop()

        # If complete word is generated
        # push it in the list
        if len(s) == n:
            list.append(s)
        else:

            # Try all possible letters for current digit
            # in number[]
            for letter in table[number[len(s)]]:
                q.append(s + letter)

    # Return the generated list
    return list


# Function that creates the mapping and
# calls letterCombinationsUtil
def letterCombinations(number, n):

    # table[i] stores all characters that
    # corresponds to ith digit in phone
    table = ["0", "1", "abc", "def", "ghi", "jkl",
             "mno", "pqrs", "tuv", "wxyz"]

    list = letterCombinationsUtil(number, n, table)

    s = ""
    for word in list:
        s += word + " "

    print(s)
    return


# Driver code
number = [2, 3]
n = len(number)

# Function call
letterCombinations(number, n)

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG {
    // Function to return a vector that contains
    // all the generated letter combinations
    static List<String>
    letterCombinationsUtil(int[] number, int n,
                           String[] table)
    {
        // To store the generated letter combinations
        List<String> list = new List<String>();

        Queue<String> q = new Queue<String>();
        q.Enqueue("");

        while (q.Count != 0) {
            String s = q.Dequeue();

            // If complete word is generated
            // push it in the list
            if (s.Length == n)
                list.Add(s);
            else {
                String val = table[number[s.Length]];
                for (int i = 0; i < val.Length; i++) {
                    q.Enqueue(s + val[i]);
                }
            }
        }
        return list;
    }

    // Function that creates the mapping and
    // calls letterCombinationsUtil
    static void letterCombinations(int[] number, int n)
    {
        // table[i] stores all characters that
        // corresponds to ith digit in phone
        String[] table
            = { "0",   "1",   "abc",  "def", "ghi",
                "jkl", "mno", "pqrs", "tuv", "wxyz" };

        List<String> list
            = letterCombinationsUtil(number, n, table);

        // Print the contents of the list
        for (int i = 0; i < list.Count; i++) {
            Console.Write(list[i] + " ");
        }
    }

    // Driver code
    public static void Main(String[] args)
    {
        int[] number = { 2, 3 };
        int n = number.Length;

        // Function call
        letterCombinations(number, n);
    }
}

// This code is contributed by Princi Singh

Javascript

</script>
function letterCombinations(digits) {

    if(digits == ""){
        return [];
    }
    let table = [ '0','1','abc','def','ghi','jkl','mno','pqrs','tuv','wxyz'];

    let res =[];
    let que = [''];

    while(que.length>0){
        let str = que[0];
        que.shift();

        if(str.length == digits.length){
            res.push(str); // if all digits are replaced with char push to result
        } else{
//             get the current number from the digits i.e if str.length = 2 , digits =123 s= 3
            let s= Number(digits.charAt(str.length));
            let val = table[s]; // get char from the table i.e def for s =3

            for(i=0;i<val.length;i++){
                que.push(str+val.charAt(i));
            }
        }
    }

    return res;

}

  // Driver code
    let str = "23";
    document.write(letterCombinations(str));
    // This code is contributed by Niloy Biswas(niloy1995)
</script>

Output

ad ae af bd be bf cd ce cf 

Time Complexity: O(4^n) as we get set of all possible numbers of length n. In worst case, for each number there can be 4 possibilities.
Auxiliary Space: O(4^n)