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# 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:

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return a vector that contains``// all the generated letter combinations``vector letterCombinationsUtil(``const` `int` `number[],``                                      ``int` `n,``                                      ``const` `string table[])``{``    ``// To store the generated letter combinations``    ``vector list;` `    ``queue 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 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``    ``letterCombinationsUtil(``int``[] number, ``int` `n,``                           ``String[] table)``    ``{``        ``// To store the generated letter combinations``        ``ArrayList list = ``new` `ArrayList<>();` `        ``Queue 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 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``    ``letterCombinationsUtil(``int``[] number, ``int` `n,``                           ``String[] table)``    ``{``        ``// To store the generated letter combinations``        ``List list = ``new` `List();` `        ``Queue q = ``new` `Queue();``        ``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 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

 ```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`

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)