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: 7
TOON -> POON –> POIN –> POIE –> PLIE –> PLEE –> PLEA
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++ implementation of the approach #include <bits/stdc++.h> 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<string>& 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<node> q1;
queue<node> q2;
unordered_map<string, int > vis1;
unordered_map<string, int > 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<string> 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;
} |
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<String> 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<node> q1 = new LinkedList<>();
Queue<node> q2 = new LinkedList<>();
HashMap<String, Integer> vis1 = new HashMap<>();
HashMap<String, Integer> 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<String> 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# 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<String> 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<node> q1 = new Queue<node>();
Queue<node> q2 = new Queue<node>();
Dictionary<String, int > vis1 = new Dictionary<String, int >();
Dictionary<String, int > vis2 = new Dictionary<String, int >();
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<String> wordList = new List<String>();
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 |
# 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))
|
class Node { constructor(w, l) {
this .word = w;
this .len = l;
}
} // Function that returns true if a and b // differ in only a single character function isAdj(a, b) {
let count = 0;
for (let 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 function 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
const q1 = [];
const q2 = [];
const vis1 = {};
const vis2 = {};
const start = new Node(beginWord, 1);
const end = new Node(endWord, 1);
vis1[beginWord] = 1;
q1.push(start);
vis2[endWord] = 1;
q2.push(end);
while (q1.length > 0 && q2.length > 0) {
// Fetch the current node
// from the source queue
const curr1 = q1.shift();
const curr2 = q2.shift();
for (const it of wordList) {
if (isAdj(curr1.word, it) && !vis1[it]) {
const temp = new Node(it, curr1.len + 1);
q1.push(temp);
vis1[it] = curr1.len + 1;
if (temp.word === endWord) {
return temp.len;
}
if (temp.word in vis2) {
return temp.len + vis2[temp.word] - 1;
}
}
}
for (const it of wordList) {
if (isAdj(curr2.word, it) && !vis2[it]) {
const temp = new Node(it, curr2.len + 1);
q2.push(temp);
vis2[it] = curr2.len + 1;
if (temp.word === beginWord) {
return temp.len;
}
if (temp.word in vis1) {
return temp.len + vis1[temp.word] - 1;
}
}
}
}
return 0;
} const wordList = []; wordList.push( "poon" );
wordList.push( "plee" );
wordList.push( "same" );
wordList.push( "poie" );
wordList.push( "plie" );
wordList.push( "poin" );
wordList.push( "plea" );
const start = "toon" ;
const target = "plea" ;
console.log(ladderLength(start, target, wordList)); // This code is contributed by lokeshpotta20. |
7
Time Complexity: O(N^2), where N is the length of the string.
Auxiliary Space: O(N).
Approach 2:
[Printing the sequence itself]
- Perform BFS
- Now in the BFS, instead of storing the string and level
- We now store the sequence itself
- We delete the words used in every level at the start of next level.
- Hence we get the sequence
CODE:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the sequences itself vector<vector<string>> findSequences(string beginWord, string endWord, vector<string>& wordList) { vector<vector<string>> ans;
unordered_set<string> vis(wordList.begin(), wordList.end());
vector<string> UsedOnLvl;
queue<vector<string>> qu;
qu.push({beginWord});
int level = 0;
while (!qu.empty()){
vector<string> vec = qu.front();
qu.pop();
if (vec.size() > level){
level++;
//calc for next lvl is gonna happen
//lvl 0 1 2 3 4 5
//also del prev guys used
//or else infinite loop and not short len (duplication)
for ( auto str : UsedOnLvl){
vis.erase(str);
}
UsedOnLvl.clear(); // cleared
}
string last = vec.back();
if (last == endWord){
if (ans.size() == 0){
//1st time
ans.push_back(vec);
}
//we have one or more
else if (ans[0].size() == vec.size()){
//same len then add it
ans.push_back(vec);
}
}
for ( int i=0; i<last.size(); i++){
char org = last[i];
for ( char ch= 'a' ; ch<= 'z' ; ch++){
last[i] = ch;
if (vis.find(last)!=vis.end()){
//its there
vec.push_back(last);
qu.push(vec);
vec.pop_back();
UsedOnLvl.push_back(last); // we dont erase now, we do it before next lvl starts
}
}
//done so replace back to org, hit - ait...zit. Now zit to hit back
last[i] = org;
}
}
return ans;
} // Driver code int main()
{ vector<string> 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" ;
vector<vector<string>> ans = findSequences(start, target, wordList);
for ( auto a : ans){
for (string s : a){
cout<<s<< " " ;
}
cout<< "\n" << "Length of sequence is " <<a.size()<< "\n" ;
}
return 0;
} //Code done by Balakrishnan R (rbkraj000) |
import java.util.*;
public class Main {
// Function to return the sequences itself
static List<List<String>> findSequences(String beginWord, String endWord, List<String> wordList) {
List<List<String>> ans = new ArrayList<>();
Set<String> vis = new HashSet<>(wordList);
List<String> UsedOnLvl = new ArrayList<>();
Queue<List<String>> qu = new LinkedList<>();
List<String> temp = new ArrayList<>();
temp.add(beginWord);
qu.add(temp);
int level = 0 ;
while (!qu.isEmpty()) {
List<String> vec = qu.poll();
if (vec.size() > level) {
level++;
//calc for next lvl is gonna happen
//lvl 0 1 2 3 4 5
//also del prev guys used
//or else infinite loop and not short len (duplication)
for (String str : UsedOnLvl) {
vis.remove(str);
}
UsedOnLvl.clear(); // cleared
}
String last = vec.get(vec.size() - 1 );
if (last.equals(endWord)) {
if (ans.size() == 0 ) {
//1st time
ans.add(vec);
}
//we have one or more
else if (ans.get( 0 ).size() == vec.size()) {
//same len then add it
ans.add(vec);
}
}
for ( int i = 0 ; i < last.length(); i++) {
char [] arr = last.toCharArray();
char org = arr[i];
for ( char ch = 'a' ; ch <= 'z' ; ch++) {
arr[i] = ch;
String newStr = new String(arr);
if (vis.contains(newStr)) {
//its there
List<String> tempVec = new ArrayList<>(vec);
tempVec.add(newStr);
qu.add(tempVec);
UsedOnLvl.add(newStr); // we dont erase now, we do it before next lvl starts
}
}
//done so replace back to org, hit - ait...zit. Now zit to hit back
arr[i] = org;
last = new String(arr);
}
}
return ans;
}
// Driver code
public static void main(String[] args) {
List<String> 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" ;
List<List<String>> ans = findSequences(start, target, wordList);
for (List<String> a : ans) {
for (String s : a) {
System.out.print(s + " " );
}
System.out.println( "\nLength of sequence is " + a.size());
}
}
} |
from typing import List
from collections import deque
def findSequences(beginWord: str , endWord: str , wordList: List [ str ]) - > List [ List [ str ]]:
# Initialize the variables
ans = [] # list to hold the answer
vis = set (wordList) # set to keep track of visited words
usedOnLvl = [] # list to hold the words used on the current level
q = deque() # deque to implement the BFS
q.append([beginWord]) # start the BFS with the initial word
level = 0 # current level
# Implement BFS
while q:
vec = q.popleft() # get the first word in the deque
if len (vec) > level:
level + = 1
for str in usedOnLvl:
vis.remove( str ) # remove the words used on the current level from the visited set
usedOnLvl = [] # reset the list of words used on the current level
last = vec[ - 1 ] # get the last word in the sequence
if last = = endWord:
if not ans:
ans.append(vec) # if the answer is empty, add the sequence
elif len (ans[ 0 ]) = = len (vec):
ans.append(vec) # if the length of the sequence is equal to the first sequence in the answer, add the sequence
for i in range ( len (last)):
org = last[i] # store the original character
for c in 'abcdefghijklmnopqrstuvwxyz' :
last = last[:i] + c + last[i + 1 :] # replace the current character with a new character
if last in vis:
vec.append(last) # add the new word to the sequence
q.append(vec.copy()) # add the new sequence to the deque
vec.pop() # remove the new word from the sequence
usedOnLvl.append(last) # add the new word to the list of words used on the current level
last = last[:i] + org + last[i + 1 :] # restore the original character
return ans # return the answer
# Test the function with the given example wordList = [ "poon" , "plee" , "same" , "poie" , "plie" , "poin" , "plea" ]
start = "toon"
target = "plea"
ans = findSequences(start, target, wordList)
for a in ans:
print ( * a, f "\nLength of sequence is {len(a)}\n" )
|
using System;
using System.Collections.Generic;
class Program
{ // Function to return the sequences itself
static List<List< string >> findSequences( string beginWord,
string endWord,
List< string > wordList)
{
List<List< string >> ans = new List<List< string >>();
HashSet< string > vis = new HashSet< string >(wordList);
List< string > UsedOnLvl = new List< string >();
Queue<List< string >> qu = new Queue<List< string >>();
qu.Enqueue( new List< string > { beginWord });
int level = 0;
while (qu.Count > 0)
{
List< string > vec = qu.Dequeue();
if (vec.Count > level)
{
level++;
//calc for next lvl is gonna happen
//lvl 0 1 2 3 4 5
//also del prev guys used
//or else infinite loop and not short len (duplication)
for ( int i = 0; i < UsedOnLvl.Count; i++)
{
vis.Remove(UsedOnLvl[i]);
}
UsedOnLvl.Clear();
}
string last = vec[vec.Count - 1];
if (last == endWord)
{
if (ans.Count == 0)
{
ans.Add(vec);
}
else if (ans[0].Count == vec.Count)
{
ans.Add(vec);
}
}
for ( int i = 0; i < last.Length; i++)
{
char org = last[i];
for ( char ch = 'a' ; ch <= 'z' ; ch++)
{
last = last.Substring(0, i) + ch + last.Substring(i + 1);
if (vis.Contains(last))
{
vec.Add(last);
qu.Enqueue( new List< string >(vec));
vec.RemoveAt(vec.Count - 1);
UsedOnLvl.Add(last);
}
}
last = last.Substring(0, i) + org + last.Substring(i + 1);
}
}
return ans;
}
static void Main( string [] args)
{
List< string > wordList = new List< string >();
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" ;
List<List< string >> ans = findSequences(start, target, wordList);
foreach (List< string > a in ans)
{
foreach ( string s in a)
{
Console.Write(s + " " );
}
Console.WriteLine( "\nLength of sequence is " + a.Count);
}
}
} |
// Javascript program for the above approach function findSequences(beginWord, endWord, wordList) {
let ans = []; // list to hold the answer
let vis = new Set(wordList); // set to keep track of visited words
let usedOnLvl = []; // list to hold the words used on the current level
let q = []; // array to implement the BFS
q.push([beginWord]); // start the BFS with the initial word
let level = 0; // current level
// Implement BFS
while (q.length) {
let vec = q.shift(); // get the first word in the array
if (vec.length > level) {
level += 1;
for (let str of usedOnLvl) {
vis. delete (str); // remove the words used on the current level from the visited set
}
usedOnLvl = []; // reset the list of words used on the current level
}
let last = vec[vec.length - 1]; // get the last word in the sequence
if (last == endWord) {
if (!ans.length) {
ans.push(vec); // if the answer is empty, add the sequence
} else if (ans[0].length == vec.length) {
ans.push(vec); // if the length of the sequence is equal to the first sequence in the answer, add the sequence
}
}
for (let i = 0; i < last.length; i++) {
let org = last[i]; // store the original character
for (let c of 'abcdefghijklmnopqrstuvwxyz' ) {
last = last.slice(0, i) + c + last.slice(i + 1); // replace the current character with a new character
if (vis.has(last)) {
vec.push(last); // add the new word to the sequence
q.push(vec.slice()); // add the new sequence to the array
vec.pop(); // remove the new word from the sequence
usedOnLvl.push(last); // add the new word to the list of words used on the current level
}
}
last = last.slice(0, i) + org + last.slice(i + 1); // restore the original character
}
}
return ans; // return the answer
} // Test the function with the given example let wordList = [ "poon" , "plee" , "same" , "poie" , "plie" , "poin" , "plea" ];
let start = "toon" ;
let target = "plea" ;
let ans = findSequences(start, target, wordList); for (let a of ans) {
console.log(...a, `\nLength of sequence is ${a.length}\n`);
} // This code is contributed by princekumaras |
toon poon poin poie plie plee plea Length of sequence is 7
Time Complexity: O(N*M*26)
Auxiliary Space: O(N*M)
where N = length of wordList and M = | wordList[i] |