LeetCode – Word Ladder

So my thought, if the dictionary isn’t that big, provide all options of the dictionary words with one letter missing.
This way, we can find the matching “1-letter changed” quickly – without traversing through all the abc.
Such a dictionary from the example would be:
[-ot, h-t, ho-, -ot, d-t, do-, -og, d-g, do-, -ot, l-t, lo-, -og, l-g, lo-].
We still need to save the original word so we’ll add it to the dictionary.
If we have the space, and can partition the dictionary based on string length, the iteration over the abc would be faster – O(size-of-word) instead of O(abc…z)=O(26).

What do you think?

LOL SERIOUSLY 😀

breadth not breath ..

defining distance between words to be number of letters which are different then we start like this

add initial into processing queue
add end to dictionary
dequeuing from processing queue
compute distance from item to each in the dictionary O(Dict)
add to queue the ones with distance one and remove them from dict so that you don’t loop

If the dictionary can be prepressed then you can organize it as a graph with arches connecting elements which have dinstance 1.

Then is just a matter of breath-first search and display the result.

// “hit” -> “cit” -> “cot” -> “cog” => every intermediate word must exist in the dictionary, but “cit” doesn’t.

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;
}

}

This can definitely be better. You should include visited words and make sure you are not adding visited words back to the graph.

this is not working, returns 0

Solution just using 2 HashMaps and 1 ArrayList

Can you please tell me the complexity for both the methods?

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;

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 be solved by Dijkstra.

Too much code for an interview or competition problem.
Try to keep it as simple and small as possible 😉

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 GraphNode(String word) {
this.word = word;
childs = new LinkedList();
}
}
I would like to have feedback on my code ?
Thanks

suboptimal solutions. Better would be 1-grams.

Thanks a ton!! You rock!! Keep it up 🙂

Thanks! Changed. And you are welcome:)

You can create a separate list that stores words when found in the dictionary.

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

We can add a condition that if (newword.equals(target)) return currDistance +1;

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.

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

I think there should be a check on the currWord and the end to stop the search in the dict:

 
/*if(currWord.equals(end)){
   return currDistance;
 }*/
if(isLastWord(currWord,end)){
                break;
 }
where isLastWord is something like this:

private boolean isLastWord(String current, String end){
        int count = 0;
        boolean isLastWord = true;
        for(int i=0; i1){
                    isLastWord = false;
                    break;
                }
            }
        }
        return isLastWord;
    }

I mean first

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.

Nice!!!

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.

hey guess for the “a”, “c” dict = {a,b,c} example the correct ans should be 1 😉

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..