# Word Ladder – Set 2 ( Bi-directional BFS )

• Difficulty Level : Expert
• Last Updated : 01 Dec, 2021

Given a dictionary, and two words start and target (both of the same length). Find length of the smallest chain from start to target if it exists, such that adjacent words in the chain only differ by one character and each word in the chain is a valid word i.e., it exists in the dictionary. It may be assumed that the target word exists in the dictionary and the lengths of all the dictionary words are equal.
Examples:

Input: Dictionary = {POON, PLEE, SAME, POIE, PLEA, PLIE, POIN}
start = “TOON”
target = “PLEA”
Output:
TOON -> POON –> POIN –> POIE –> PLIE –> PLEE –> PLEA

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Approach: This problem can be solved using the standard BFS approach as discussed here but its performance can be improved by using bi-directional BFS.
Bi-directional BFS doesn’t reduce the time complexity of the solution but it definitely optimizes the performance in many cases. This approach can also be used in many other shortest pathfinding problems where we have sufficient information about the source and the target node. The basic idea involved in bi-directional BFS is to start the search from both the ends of the path.
Therefore, two queues and two visited arrays are needed to be maintained to track both the paths. So, whenever a node (say A) is present in the source queue, encounters a node (say B) which is present in the target queue, then we can calculate the answer by adding the distance of A from source and the distance of B from target minus 1 (one node is common). This way we can calculate the answer in half the time as compared to the standard BFS approach. This method is also known as the meet-in-the-middle BFS approach.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Structure for queue``struct` `node {``    ``string word;``    ``int` `len;``};` `// Function that returns true if a and b``// differ in only a single character``bool` `isAdj(string a, string b)``{``    ``int` `count = 0;``    ``for` `(``int` `i = 0; i < a.length(); i++) {``        ``if` `(a[i] != b[i])``            ``count++;``    ``}``    ``if` `(count == 1)``        ``return` `true``;``    ``return` `false``;``}` `// Function to return the length of the shortest``// chain from ‘beginWord’ to ‘endWord’``int` `ladderLength(string beginWord, string endWord,``                 ``vector& wordList)``{` `    ``/* q1 is used to traverse the graph from beginWord``        ``and q2 is used to traverse the graph from endWord.``        ``vis1 and vis2 are used to keep track of the``        ``visited states from respective directions */``    ``queue q1;``    ``queue q2;``    ``unordered_map vis1;``    ``unordered_map vis2;` `    ``node start = { beginWord, 1 };``    ``node end = { endWord, 1 };` `    ``vis1[beginWord] = 1;``    ``q1.push(start);``    ``vis2[endWord] = 1;``    ``q2.push(end);` `    ``while` `(!q1.empty() && !q2.empty()) {` `        ``// Fetch the current node``        ``// from the source queue``        ``node curr1 = q1.front();``        ``q1.pop();` `        ``// Fetch the current node from``        ``// the destination queue``        ``node curr2 = q2.front();``        ``q2.pop();` `        ``// Check all the neighbors of curr1``        ``for` `(``auto` `it = wordList.begin(); it != wordList.end(); it++) {` `            ``// If any one of them is adjacent to curr1``            ``// and is not present in vis1``            ``// then push it in the queue``            ``if` `(isAdj(curr1.word, *it) && vis1.count(*it) == ``false``) {` `                ``node temp = { *it, curr1.len + 1 };``                ``q1.push(temp);``                ``vis1[*it] = curr1.len + 1;` `                ``// If temp is the destination node``                ``// then return the answer``                ``if` `(temp.word == endWord) {``                    ``return` `temp.len;``                ``}` `                ``// If temp is present in vis2 i.e. distance from``                ``// temp and the destination is already calculated``                ``if` `(vis2.count(temp.word)) {``                    ``return` `temp.len + vis2[temp.word] - 1;``                ``}``            ``}``        ``}` `        ``// Check all the neighbors of curr2``        ``for` `(``auto` `it = wordList.begin(); it != wordList.end(); it++) {` `            ``// If any one of them is adjacent to curr2``            ``// and is not present in vis1 then push it in the queue.``            ``if` `(isAdj(curr2.word, *it) && vis2.count(*it) == ``false``) {` `                ``node temp = { *it, curr2.len + 1 };``                ``q2.push(temp);``                ``vis2[*it] = curr2.len + 1;` `                ``// If temp is the destination node``                ``// then return the answer``                ``if` `(temp.word == beginWord) {``                    ``return` `temp.len;``                ``}` `                ``// If temp is present in vis1 i.e. distance from``                ``// temp and the source is already calculated``                ``if` `(vis1.count(temp.word)) {``                    ``return` `temp.len + vis1[temp.word] - 1;``                ``}``            ``}``        ``}``    ``}``    ``return` `0;``}` `// Driver code``int` `main()``{` `    ``vector wordList;``    ``wordList.push_back(``"poon"``);``    ``wordList.push_back(``"plee"``);``    ``wordList.push_back(``"same"``);``    ``wordList.push_back(``"poie"``);``    ``wordList.push_back(``"plie"``);``    ``wordList.push_back(``"poin"``);``    ``wordList.push_back(``"plea"``);` `    ``string start = ``"toon"``;``    ``string target = ``"plea"``;` `    ``cout << ladderLength(start, target, wordList);` `    ``return` `0;``}`

## Java

 `import` `java.util.*;``public` `class` `GFG``{``public` `static` `class` `node``{``    ``String word;``    ``int` `len;``    ``public` `node(String word, ``int` `len)``    ``{``        ``this``.word = word;``        ``this``.len = len;``    ``}``}` `public` `static` `boolean` `isAdj(String a, String b)``{``    ``int` `count = ``0``;``    ``for` `(``int` `i = ``0``; i < a.length(); i++)``    ``{``        ``if` `(a.charAt(i) != b.charAt(i))``            ``count++;``    ``}``    ``if` `(count == ``1``)``        ``return` `true``;``    ``return` `false``;``}` `// Function to return the length of the shortest``// chain from 'beginWord' to 'endWord'``public` `static` `int` `ladderLength(String beginWord, String endWord,``                               ``ArrayList wordList)``{` `    ``/* q1 is used to traverse the graph from beginWord``        ``and q2 is used to traverse the graph from endWord.``        ``vis1 and vis2 are used to keep track of the``        ``visited states from respective directions */``    ``Queue q1 = ``new` `LinkedList<>();``    ``Queue q2 = ``new` `LinkedList<>();``    ``HashMap vis1 = ``new` `HashMap<>();``    ``HashMap vis2 = ``new` `HashMap<>();` `    ``node start = ``new` `node(beginWord, ``1``);``    ``node end = ``new` `node(endWord, ``1``);` `    ``vis1.put(beginWord, ``1``);``    ``q1.add(start);``    ``vis2.put(endWord, ``1``);``    ``q2.add(end);` `    ``while` `(q1.size() > ``0` `&& q2.size() > ``0``)``    ``{` `        ``// Fetch the current node``        ``// from the source queue``        ``node curr1 = q1.remove();` `        ``// Fetch the current node from``        ``// the destination queue``        ``node curr2 = q2.remove();` `        ``// Check all the neighbors of curr1``        ``for` `(``int` `i = ``0``; i < wordList.size(); i++)``        ``{` `            ``// If any one of them is adjacent to curr1``            ``// and is not present in vis1``            ``// then push it in the queue``            ``if` `(isAdj(curr1.word,wordList.get(i)) &&``                ``vis1.containsKey(wordList.get(i)) == ``false``)``            ``{` `                ``node temp = ``new` `node(wordList.get(i),``                                      ``curr1.len + ``1``);``                ``q1.add(temp);``                ``vis1.put(wordList.get(i), curr1.len + ``1``);` `                ``// If temp is the destination node``                ``// then return the answer``                ``if` `(temp.word.equals(endWord))``                ``{``                    ``return` `temp.len;``                ``}` `                ``// If temp is present in vis2 i.e. distance from``                ``// temp and the destination is already calculated``                ``if` `(vis2.containsKey(temp.word))``                ``{``                    ``return` `temp.len + vis2.get(temp.word) - ``1``;``                ``}``            ``}``        ``}` `        ``// Check all the neighbors of curr2``        ``for` `(``int` `i = ``0``; i < wordList.size(); i++)``        ``{` `            ``// If any one of them is adjacent to curr2``            ``// and is not present in vis1 then push it in the queue.``            ``if` `(isAdj(curr2.word,wordList.get(i)) &&``                ``vis2.containsKey(wordList.get(i)) == ``false``)``            ``{` `                ``node temp = ``new` `node(wordList.get(i),``                                     ``curr2.len + ``1` `);``                ``q2.add(temp);``                ``vis2.put(wordList.get(i), curr2.len + ``1``);` `                ``// If temp is the destination node``                ``// then return the answer``                ``if` `(temp.word.equals(beginWord))``                ``{``                    ``return` `temp.len;``                ``}` `                ``// If temp is present in vis1 i.e. distance from``                ``// temp and the source is already calculated``                ``if` `(vis1.containsKey(temp.word))``                ``{``                    ``return` `temp.len + vis1.get(temp.word) - ``1``;``                ``}``            ``}``        ``}``    ``}``    ``return` `0``;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``ArrayList wordList = ``new` `ArrayList<>();``    ``wordList.add(``"poon"``);``    ``wordList.add(``"plee"``);``    ``wordList.add(``"same"``);``    ``wordList.add(``"poie"``);``    ``wordList.add(``"plie"``);``    ``wordList.add(``"poin"``);``    ``wordList.add(``"plea"``);` `    ``String start = ``"toon"``;``    ``String target = ``"plea"``;` `    ``System.out.println(ladderLength(start, target, wordList));``}``}` `// This code is contributed by Sambhav Jain`

## C#

 `// C# implementation of the approach   ``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``    ` `class` `node``{``    ``public` `String word;``    ``public` `int` `len;``    ``public` `node(String word, ``int` `len)``    ``{``        ``this``.word = word;``        ``this``.len = len;``    ``}``}` `public` `static` `bool` `isAdj(String a, String b)``{``    ``int` `count = 0;``    ``for` `(``int` `i = 0; i < a.Length; i++)``    ``{``        ``if` `(a[i] != b[i])``            ``count++;``    ``}``    ``if` `(count == 1)``        ``return` `true``;``    ``return` `false``;``}` `// Function to return the length of the shortest``// chain from 'beginWord' to 'endWord'``public` `static` `int` `ladderLength(String beginWord, String endWord,``                            ``List wordList)``{` `    ``/* q1 is used to traverse the graph from beginWord``        ``and q2 is used to traverse the graph from endWord.``        ``vis1 and vis2 are used to keep track of the``        ``visited states from respective directions */``    ``Queue q1 = ``new` `Queue();``    ``Queue q2 = ``new` `Queue();``    ``Dictionary vis1 = ``new` `Dictionary();``    ``Dictionary vis2 = ``new` `Dictionary();` `    ``node start = ``new` `node(beginWord, 1);``    ``node end = ``new` `node(endWord, 1);` `    ``vis1.Add(beginWord, 1);``    ``q1.Enqueue(start);``    ``vis2.Add(endWord, 1);``    ``q2.Enqueue(end);` `    ``while` `(q1.Count > 0 && q2.Count > 0)``    ``{` `        ``// Fetch the current node``        ``// from the source queue``        ``node curr1 = q1.Dequeue();` `        ``// Fetch the current node from``        ``// the destination queue``        ``node curr2 = q2.Dequeue();` `        ``// Check all the neighbors of curr1``        ``for` `(``int` `i = 0; i < wordList.Count; i++)``        ``{` `            ``// If any one of them is adjacent to curr1``            ``// and is not present in vis1``            ``// then push it in the queue``            ``if` `(isAdj(curr1.word,wordList[i]) &&``                ``vis1.ContainsKey(wordList[i]) == ``false``)``            ``{` `                ``node temp = ``new` `node(wordList[i],``                                    ``curr1.len + 1);``                ``q1.Enqueue(temp);``                ``vis1.Add(wordList[i], curr1.len + 1);` `                ``// If temp is the destination node``                ``// then return the answer``                ``if` `(temp.word.Equals(endWord))``                ``{``                    ``return` `temp.len;``                ``}` `                ``// If temp is present in vis2 i.e. distance from``                ``// temp and the destination is already calculated``                ``if` `(vis2.ContainsKey(temp.word))``                ``{``                    ``return` `temp.len + vis2[temp.word] - 1;``                ``}``            ``}``        ``}` `        ``// Check all the neighbors of curr2``        ``for` `(``int` `i = 0; i < wordList.Count; i++)``        ``{` `            ``// If any one of them is adjacent to curr2``            ``// and is not present in vis1 then push it in the queue.``            ``if` `(isAdj(curr2.word,wordList[i]) &&``                ``vis2.ContainsKey(wordList[i]) == ``false``)``            ``{` `                ``node temp = ``new` `node(wordList[i],``                                    ``curr2.len + 1 );``                ``q2.Enqueue(temp);``                ``vis2.Add(wordList[i], curr2.len + 1);` `                ``// If temp is the destination node``                ``// then return the answer``                ``if` `(temp.word.Equals(beginWord))``                ``{``                    ``return` `temp.len;``                ``}` `                ``// If temp is present in vis1 i.e. distance from``                ``// temp and the source is already calculated``                ``if` `(vis1.ContainsKey(temp.word))``                ``{``                    ``return` `temp.len + vis1[temp.word] - 1;``                ``}``            ``}``        ``}``    ``}``    ``return` `0;``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``List wordList = ``new` `List();``    ``wordList.Add(``"poon"``);``    ``wordList.Add(``"plee"``);``    ``wordList.Add(``"same"``);``    ``wordList.Add(``"poie"``);``    ``wordList.Add(``"plie"``);``    ``wordList.Add(``"poin"``);``    ``wordList.Add(``"plea"``);` `    ``String start = ``"toon"``;``    ``String target = ``"plea"``;` `    ``Console.WriteLine(ladderLength(start, target, wordList));``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python 3 implementation of the approach``from` `collections ``import` `deque as dq` `# class for queue``class` `node :``    ``def` `__init__(``self``, w, l):``        ``self``.word``=``w``        ``self``.``len``=``l`  `# Function that returns true if a and b``# differ in only a single character``def` `isAdj(a, b):``    ``count ``=` `0``    ``for` `i ``in` `range``(``len``(a)):``        ``if` `(a[i] !``=` `b[i]):``            ``count``+``=``1``    ` `    ``if` `(count ``=``=` `1``):``        ``return` `True``    ``return` `False`  `# Function to return the length of the shortest``# chain from ‘beginWord’ to ‘endWord’``def` `ladderLength(beginWord, endWord, wordList):` `    ``# q1 is used to traverse the graph from beginWord``    ``# and q2 is used to traverse the graph from endWord.``    ``# vis1 and vis2 are used to keep track of the``    ``# visited states from respective directions``    ``q1``=``dq([])``    ``q2``=``dq([])``    ``vis1``=``dict``()``    ``vis2``=``dict``()` `    ``start ``=` `node(beginWord, ``1``)``    ``end ``=` `node(endWord, ``1``)` `    ``vis1[beginWord] ``=` `1``    ``q1.append(start)``    ``vis2[endWord] ``=` `1``    ``q2.append(end)` `    ``while` `(q1 ``and` `q2):` `        ``# Fetch the current node``        ``# from the source queue``        ``curr1 ``=` `q1.popleft()` `        ``# Fetch the current node from``        ``# the destination queue``        ``curr2 ``=` `q2.popleft()` `        ``# Check all the neighbors of curr1``        ``for` `it ``in` `wordList:` `            ``# If any one of them is adjacent to curr1``            ``# and is not present in vis1``            ``# then push it in the queue``            ``if` `(isAdj(curr1.word, it) ``and` `it ``not` `in` `vis1) :` `                ``temp ``=` `node(it, curr1.``len` `+` `1``)``                ``q1.append(temp)``                ``vis1[it] ``=` `curr1.``len` `+` `1` `                ``# If temp is the destination node``                ``# then return the answer``                ``if` `(temp.word ``=``=` `endWord) :``                    ``return` `temp.``len``                `  `                ``# If temp is present in vis2 i.e. distance from``                ``# temp and the destination is already calculated``                ``if` `temp.word ``in` `vis2 :``                    ``return` `temp.``len` `+` `vis2[temp.word] ``-` `1``                ` `            ` `        `  `        ``# Check all the neighbors of curr2``        ``for` `it ``in` `wordList:` `            ``# If any one of them is adjacent to curr2``            ``# and is not present in vis1 then push it in the queue.``            ``if` `(isAdj(curr2.word, it) ``and` `it ``not` `in` `vis2) :` `                ``temp ``=` `node(it, curr2.``len` `+` `1``)``                ``q2.append(temp)``                ``vis2[it] ``=` `curr2.``len` `+` `1` `                ``# If temp is the destination node``                ``# then return the answer``                ``if` `(temp.word ``=``=` `beginWord) :``                    ``return` `temp.``len``                `  `                ``# If temp is present in vis1 i.e. distance from``                ``# temp and the source is already calculated``                ``if` `temp.word ``in` `vis1:``                    ``return` `temp.``len` `+` `vis1[temp.word] ``-` `1`          `    ``return` `0`  `# Driver code``if` `__name__``=``=``'__main__'``:` `    ``wordList``=``[]``    ``wordList.append(``"poon"``)``    ``wordList.append(``"plee"``)``    ``wordList.append(``"same"``)``    ``wordList.append(``"poie"``)``    ``wordList.append(``"plie"``)``    ``wordList.append(``"poin"``)``    ``wordList.append(``"plea"``)` `    ``start ``=` `"toon"``    ``target ``=` `"plea"` `    ``print``(ladderLength(start, target, wordList))`
Output:
`7`

Time Complexity: O(N^2), where N is the length of the string.
Auxiliary Space: O(N).

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