LeetCode – Word Ladder
Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that only one letter can be changed at a time and each intermediate word must exist in the dictionary. For example, given:
start = "hit" end = "cog" dict = ["hot","dot","dog","lot","log"]
One shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog", the program should return its length 5.
Analysis
UPDATED on 06/07/2015.
So we quickly realize that this is a search problem, and breath-first search guarantees the optimal solution.
Java Solution
class WordNode{ String word; int numSteps; public WordNode(String word, int numSteps){ this.word = word; this.numSteps = numSteps; } } public class Solution { public int ladderLength(String beginWord, String endWord, Set<String> wordDict) { LinkedList<WordNode> queue = new LinkedList<WordNode>(); queue.add(new WordNode(beginWord, 1)); wordDict.add(endWord); while(!queue.isEmpty()){ WordNode top = queue.remove(); String word = top.word; if(word.equals(endWord)){ return top.numSteps; } char[] arr = word.toCharArray(); for(int i=0; i<arr.length; i++){ for(char c='a'; c<='z'; c++){ char temp = arr[i]; if(arr[i]!=c){ arr[i]=c; } String newWord = new String(arr); if(wordDict.contains(newWord)){ queue.add(new WordNode(newWord, top.numSteps+1)); wordDict.remove(newWord); } arr[i]=temp; } } } return 0; } } |
<pre><code> String foo = "bar"; </code></pre>
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The second algo has a bug. What the problem says is that “Each intermediate word must exist in the dictionary”. However, this solution assumes the last one(“cog”) also need to be in the dict, which is not necessary..
Also, “int result = 0;” This sentence seems redundant..
hey guess for the “a”, “c” dict = {a,b,c} example the correct ans should be 1 😉
That’s right. A sentence should be added “dict.add(end)”.
This should be a shortest path problem: if two words has common character, they are connected by edge. The weight of the edge is the number of character in common. So the problem becomes shortest path problem.
How can I print the world ladder answers, after getting the distance.? Thanks for the answers.
Nice!!!
The algorithm can’t guarantee find the shortest transformation sequence.
how it will accepted?? as it will not return as cog is not inside in dict, so currWord.equals(end) will false and return 0.
Priest solution is obviously wrong even if don’t consider if it’s optimally
I mean first
The second solution doesn’t work. Try with input: “hit”, “lag”, [“hot”,”hog”,”dot”,”dog”,”lot”,”log”]
Example sequence: hit > hot > hog > log > lag The answer should be 5. But, it shows 0
I think there should be a check on the currWord and the end to stop the search in the dict:
I think this solution is totally wrong.
Agree with most of the comments here:
First: the dictionary here needs to include the end word
Second: the second solution does not find the optimal shortest path, it only finds the first valid path.
I have come up with my own solution which does the following:
1. Instead of trying to iterate through all chars from a to z, it parses the dictionary and maintains a map between index and possible chars for that index in the dictionary, so flipping a character only flipped to possible dictionary words.
2. recursive Graph depth first search which keeps tracks of all sub path size and choose the shortest one.
Please kindly review and comment to help improve the algo
https://github.com/ifwonderland/leetcode/blob/master/src/main/java/leetcode/string/WordLadder.java
https://github.com/ifwonderland/leetcode/blob/master/src/main/java/leetcode/util/WordFlipperUtils.java
Why? And how does a correct one look like?
Hi! I think you were oversimplifying the problem. the edge cannot be weighted just based on the number of characters in common. And an edge cannot even be determined between two words just because they have common characters. You have to factor in what are included the dictionary. One simple example, ‘good’ and ‘doom’ have two common characters and suppose the dictionary only contains ‘good’ and ‘doom’…Apparently no edge of path could be established between the two words.
yeah second algo is wrong
We can add a condition that if (newword.equals(target)) return currDistance +1;
we stop at the first time we find the target word and return “length”. All the words that will be added in the stack after that will be > length as we push words in stack after length +1 .
shortest path problem with dictionary extended to include start and end and weight equals to 1 if distance(a, b) == 1, else weight equals infinite
Yes, you are right. Surprisingly, the seoncd one can be acceptted by leetcode.
You can create a separate list that stores words when found in the dictionary.
You just need to add end to dict. Then it will be correct.
Btw, thanks a lot for ur code:)
Thanks! Changed. And you are welcome:)
The problem is fixed.
Thanks a ton!! You rock!! Keep it up 🙂
You also don’t necessarily need two queues. You could create a WordNode class inside the method and have a word field and a distance field. Then you can create a queue of “WordNode”s. Also, I think there is no need to compare the distance to previously recorded one, since breadth search will reach the end word in the shortest path. Whenever you reach the end word, that is the shortest distance.
suboptimal solutions. Better would be 1-grams.
That’s nice idea to keep only the actual existing letters at each index.
Hi
I have solved this problem using Graph of word and then run on it BFS.
package Questions;
import java.util.HashMap;
import java.util.Queue;
import java.util.HashSet;
import java.util.LinkedList;
public class WordLadder {
//solution function
public static int ladderLength(String start, String end,
HashSet dict) {
// saving the graph nodes on hash map.
HashMap nodes = new HashMap();
// adding the start and the end words to the dict
dict.add(start);
dict.add(end);
for (String word : dict) {
nodes.put(word, new GraphNode(word));
}
// update each node’s adjacents according to one character different relation
Object[] dictArray = dict.toArray();
for (int i = 0; i < dictArray.length; i++) {
for (int j = i + 1; j < dictArray.length; j++) {
if (isNeighbor((String) dictArray[i], (String) dictArray[j])) {
nodes.get((String) dictArray[i]).childs.add(nodes
.get((String) dictArray[j]));
nodes.get((String) dictArray[j]).childs.add(nodes
.get((String) dictArray[i]));
}
}
}
// Run BFS on the Graph and take the dist generated as result
HashMap result = BFS(nodes, start);
// Return the distance of the end word node from the start word node
return result.get(end);
}
// BFS function
public static HashMap BFS(
HashMap nodes, String start) {
HashMap visited = new HashMap();
HashMap dist = new HashMap();
for (String key : nodes.keySet()) {
visited.put(key, 0);
dist.put(key, 0);
}
Queue q = new LinkedList();
q.add(start);
visited.put(start, 1);
while (!q.isEmpty()) {
String dequeued = q.remove();
GraphNode curNode = nodes.get(dequeued);
LinkedList currAdjs = curNode.childs;
for (int i = 0; i < currAdjs.size(); i++) {
GraphNode adj = (GraphNode) currAdjs.get(i);
if (visited.get(adj.word) == 0) {
visited.put(adj.word, 1);
dist.put(adj.word, dist.get(dequeued) + 1);
q.add(adj.word);
}
}
}
return dist;
}
// check if two words differ by one character
public static boolean isNeighbor(String a, String b) {
assert a.length() == b.length();
int differ = 0;
for (int i = 0; i 1)
return false;
}
return true;
}
public static void main(String[] args) {
// dict = [“hot”,”dot”,”dog”,”lot”,”log”] result 5;
HashSet dict = new HashSet();
dict.add(“hot”);
dict.add(“dot”);
dict.add(“dog”);
dict.add(“lot”);
dict.add(“log”);
System.out.println(ladderLength(“hit”, “cog”, dict));
}
}
class GraphNode {
String word;
LinkedList childs;
public GraphNode(String word) {
this.word = word;
childs = new LinkedList();
}
}
I would like to have feedback on my code ?
Thanks
BFS is a really good way of solving this problem
Too much code for an interview or competition problem.
Try to keep it as simple and small as possible 😉
If anyone interested for both BFS and DFS approach:
http://javabypatel.blogspot.in/2015/10/word-ladder-doublets-word-links-word-golf.html
Can be solved by Dijkstra.
Without using a queue. Please let me know if you see any issues with this?
public int getCount(String start, String end, Set dict){
// Basic edge cases
if(start==null || end==null || start.length()!=end.length()) return -1;
if(dict.isEmpty()) return -1;
dict.add(end);
int count = 0;
int i = 0;
while(i<start.length()){
boolean found = false;
char[] arr = start.toCharArray();
char tmp = arr[i];
for(int j=97;j<=122;j++){
arr[i] = (char)j;
String checker = new String(arr);
if(dict.contains(checker)){
count++;
dict.remove(checker);
if(checker.equals(end)) {
return count;
}
else {
start = checker;
found = true;
i = 0;
break;
}
}
}
if(!found){
arr[i] = tmp;
i++;
}
}
return -1;
}
Solution posted in the task is not correct.
Here is mine:
Test.java file:
import java.util.*;
public class Test {
public static void main(String argv[]){
String start = "hit";
String end = "cog";
String[] dict = {"cot","hot","dot","cit","dog","lot","log"};
// "hit" -> "cit" -> "cot" -> "cog"
// length = 4
int minDistance = getDistance(start, end);
List nodesToCheck = new ArrayList();
List newNodes = new ArrayList();
List winners = new ArrayList();
nodesToCheck.add(new TreeElem(start, null, minDistance, 1));
do {
for (TreeElem node : nodesToCheck) {
for (int i = 0; i 0);
if (winners.size() == 0) {
System.out.println("No solution found.");
return;
}
TreeElem winner = null;
int minLength = -1;
for (TreeElem e : winners) {
if (winner == null) {
winner = e;
minLength = e.distanceFromTop;
}
if (e.distanceFromTop = 0; i--) {
System.out.print("[" + steps[i] + "] -> ");
}
System.out.println("[" + end + "]");
}
public static int getDistance(String str1, String str2) {
int dist = 0;
for (int i = 0; i < str1.length(); i++) {
if (str1.charAt(i) != str2.charAt(i)) {
dist++;
}
}
return dist;
}
public static boolean isUsed(TreeElem node, String value) {
while (node.parent != null) {
if (node.value.equals(value)) {
return true;
}
node = node.parent;
}
return false;
}
}
TreeElem.java file:
import java.util.*;
public class TreeElem{
public String value;
public TreeElem parent;
List childs;
public int distanceToEnd;
public int distanceFromTop;
public TreeElem(String value, TreeElem parent, int distanceToEnd, int distanceFromTop) {
this.value = value;
this.parent = parent;
this.childs = new ArrayList();
this.distanceToEnd = distanceToEnd;
this.distanceFromTop = distanceFromTop;
}
public void addChild(TreeElem child) {
childs.add(child);
}
}
can you tell me what is the complexity of this method
Can you please tell me the complexity for both the methods?
Does this solution give length of shortest transformation sequence or length of first sequence found ?
Solution just using 2 HashMaps and 1 ArrayList
package leetcode;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
public class Solution {
private static int min = Integer.MAX_VALUE;
private static List visitedkeys = new ArrayList();
public void wordladder(String start, String end, String[] dict) {
//Word ladder
Map map2 = new HashMap();
Map<String, List> map = new HashMap();
int i=0;
map2.put(i++, start);
map.put(start, new ArrayList());
for (String d : dict) {
map2.put(i++, d);
map.put(d, new ArrayList());
}
map2.put(i, end);
map.put(end, new ArrayList());
int n = map2.size();
for (i=0; i < n; i++) {
for (int j=0; j < n; j++) {
if (isConvertible(map2.get(i), map2.get(j))) {
map.get(map2.get(i)).add(map2.get(j));
}
}
}
length(map, start, end, 1);
if (min != Integer.MAX_VALUE){
System.out.println(min);
}
else {
System.out.println("no path");
}
}
private boolean isConvertible(String s1, String s2) {
if (s1.length() != s2.length()) {
return false;
}
int diff = 0;
for (int i = 0; i < s1.length(); i++) {
if (s1.charAt(i) != s2.charAt(i)) {
diff += 1;
}
}
return (diff == 1) ? true : false;
}
private void length(Map<String, List> map, String key, String end, int len) {
if (key == end) {
if (len < min)
min = len;
return;
}
List lst = map.get(key);
for (String v : lst) {
if (!visitedkeys.contains(v)){
visitedkeys.add(v);
length(map, v, end, len+1);
visitedkeys.remove(visitedkeys.size()-1);
}
}
}
}
for the same input as yours with dictionary changed to below:
dict = [“hot”,”dot”,”dog”,”lot”,”aog”,”log”]
answer will change like : “hit” -> “hot” -> “dot” -> “dog” -> “aog”->”cog”,
but from dog to cog also we can go directly, how will you avoid that?
this is not working, returns 0
any c# code?
This can definitely be better. You should include visited words and make sure you are not adding visited words back to the graph.
I don’t understand why this code is posted, it has time limit exceeded issue on Leetcode.
I have Solved this problem in my way.Can please anyone locate the use case in which it will break
package com.cpa.examples;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
public class WordLadder {
private static String start = “hit”;
private static String end = “cog”;
private static List Dictionary = null;
private static ArrayList availablePathCntList = new ArrayList();
public static void main(String[] args) {
Dictionary = new ArrayList();
Dictionary.add(“hot”);
Dictionary.add(“dot”);
Dictionary.add(“dog”);
Dictionary.add(“lot”);
Dictionary.add(“log”);
Dictionary.add(“cog”);
HashSet used = new HashSet();
used.add(start);
int count = 0;
find(Dictionary, start, end, used, count);
System.out.println(availablePathCntList);
}
private static int find(List dictionary, String start, String end, HashSet used, int count) {
if (start == end) {
availablePathCntList.add(count);
return 0;
}
if (difference(start, end) == 1) {
availablePathCntList.add(count + 1);
return 0;
}
for (String word : dictionary) {
int diff = difference(start, word);
if (difference(end, word) == 0) {
availablePathCntList.add(1);
continue;
}
if (diff == 0) {
availablePathCntList.add(count);
continue;
}
if (diff <= 1 && used.add(word)) {
System.out.println("Your word is: " + start + " and matching word is " + word);
if (diff <= 1) {
if (difference(end, word) == 1) {
availablePathCntList.add(count + 1);
} else {
count++;
find(Dictionary, new String(word), new String(end), (HashSet) used.clone(), count);
}
}
}
}
return 0;
}
private static int difference(String first, String second) {
char[] firstArr = first.toCharArray();
char[] secondArr = second.toCharArray();
int mismatch = 0;
for (int i = 0; i < firstArr.length; i++) {
if (firstArr[i] != secondArr[i]) {
mismatch++;
}
}
return mismatch;
}
}
Uhhh… It’s breadth-first search
// “hit” -> “cit” -> “cot” -> “cog” => every intermediate word must exist in the dictionary, but “cit” doesn’t.
You have to check the newWord with the endWord
String newWord = new String(arr);
if (dict.contains(newWord)) {
System.out.println(newWord);
queue.add(new WordNode(newWord, top.numSteps + 1));
dict.remove(newWord);
// you have to add the next lines.
if (newWord.equals(end)) {
// System.out.println("equals!");
return top.numSteps;
}
}