Word Ladder (Length of shortest chain to reach a target word)

Given a dictionary, and two words ‘start’ and ‘target’ (both of 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 dictionary and length of all dictionary words is same. 
Example: 

Input: Dictionary = {POON, PLEE, SAME, POIE, PLEA, PLIE, POIN}
       start = TOON
       target = PLEA
Output: 7
TOON - POON - POIN - POIE - PLIE - PLEE - PLEA

Input: Dictionary = {ABCD, EBAD, EBCD, XYZA}
       Start = ABCV
       End = EBAD
Output: 4
ABCV - ABCD - EBCD - EBAD

Solution: The idea is to use BFS
Approach: 

  1. Start from the given start word.
  2. Push the word in the queue
  3. Run a loop until the queue is empty
  4. Traverse all words that adjacent (differ by one character) to it and push the word in a queue (for BFS)
  5. Keep doing so until we find the target word or we have traversed all words.

Below are implementation of above idea.

C++

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// C++ program to find length
// of the shortest chain
// transformation from source
// to target
#include <bits/stdc++.h>
using namespace std;
 
// Returns length of shortest chain
// to reach 'target' from 'start'
// using minimum number of adjacent
// moves.  D is dictionary
int shortestChainLen(
string start, string target,
set<string>& D)
{
 
    // If the target string is not
    // present in the dictionary
    if (D.find(target) == D.end())
        return 0;
 
    // To store the current chain length
    // and the length of the words
    int level = 0, wordlength = start.size();
 
    // Push the starting word into the queue
    queue<string> Q;
    Q.push(start);
 
    // While the queue is non-empty
    while (!Q.empty()) {
 
        // Increment the chain length
        ++level;
 
        // Current size of the queue
        int sizeofQ = Q.size();
 
        // Since the queue is being updated while
        // it is being traversed so only the
        // elements which were already present
        // in the queue before the start of this
        // loop will be traversed for now
        for (int i = 0; i < sizeofQ; ++i) {
 
            // Remove the first word from the queue
            string word = Q.front();
            Q.pop();
 
            // For every character of the word
            for (int pos = 0; pos < wordlength; ++pos) {
 
                // Retain the original character
                // at the current position
                char orig_char = word[pos];
 
                // Replace the current character with
                // every possible lowercase alphabet
                for (char c = 'a'; c <= 'z'; ++c) {
                    word[pos] = c;
 
                    // If the new word is equal
                    // to the target word
                    if (word == target)
                        return level + 1;
 
                    // Remove the word from the set
                    // if it is found in it
                    if (D.find(word) == D.end())
                        continue;
                    D.erase(word);
 
                    // And push the newly generated word
                    // which will be a part of the chain
                    Q.push(word);
                }
 
                // Restore the original character
                // at the current position
                word[pos] = orig_char;
            }
        }
    }
 
    return 0;
}
 
// Driver program
int main()
{
    // make dictionary
    set<string> D;
    D.insert("poon");
    D.insert("plee");
    D.insert("same");
    D.insert("poie");
    D.insert("plie");
    D.insert("poin");
    D.insert("plea");
    string start = "toon";
    string target = "plea";
    cout << "Length of shortest chain is: "
         << shortestChainLen(start, target, D);
    return 0;
}

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Java

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// Java program to find length
// of the shortest chain
// transformation from source
// to target
import java.util.*;
 
class GFG
{
 
// Returns length of shortest chain
// to reach 'target' from 'start'
// using minimum number of adjacent moves.
// D is dictionary
static int shortestChainLen(String start,
                            String target,
                            Set<String> D)
{
 
    // If the target String is not
    // present in the dictionary
    if (!D.contains(target))
        return 0;
 
    // To store the current chain length
    // and the length of the words
    int level = 0, wordlength = start.length();
 
    // Push the starting word into the queue
    Queue<String> Q = new LinkedList<>();
    Q.add(start);
 
    // While the queue is non-empty
    while (!Q.isEmpty())
    {
 
        // Increment the chain length
        ++level;
 
        // Current size of the queue
        int sizeofQ = Q.size();
 
        // Since the queue is being updated while
        // it is being traversed so only the
        // elements which were already present
        // in the queue before the start of this
        // loop will be traversed for now
        for (int i = 0; i < sizeofQ; ++i)
        {
 
            // Remove the first word from the queue
            char []word = Q.peek().toCharArray();
            Q.remove();
 
            // For every character of the word
            for (int pos = 0; pos < wordlength; ++pos)
            {
 
                // Retain the original character
                // at the current position
                char orig_char = word[pos];
 
                // Replace the current character with
                // every possible lowercase alphabet
                for (char c = 'a'; c <= 'z'; ++c)
                {
                    word[pos] = c;
 
                    // If the new word is equal
                    // to the target word
                    if (String.valueOf(word).equals(target))
                        return level + 1;
 
                    // Remove the word from the set
                    // if it is found in it
                    if (!D.contains(String.valueOf(word)))
                        continue;
                    D.remove(String.valueOf(word));
 
                    // And push the newly generated word
                    // which will be a part of the chain
                    Q.add(String.valueOf(word));
                }
 
                // Restore the original character
                // at the current position
                word[pos] = orig_char;
            }
        }
    }
 
    return 0;
}
 
// Driver code
public static void main(String[] args)
{
    // make dictionary
    Set<String> D = new HashSet<String>();
    D.add("poon");
    D.add("plee");
    D.add("same");
    D.add("poie");
    D.add("plie");
    D.add("poin");
    D.add("plea");
    String start = "toon";
    String target = "plea";
    System.out.print("Length of shortest chain is: "
        + shortestChainLen(start, target, D));
}
}
 
// This code is contributed by PrinciRaj1992

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C#

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// C# program to find length of the shortest chain
// transformation from source to target
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Returns length of shortest chain
// to reach 'target' from 'start'
// using minimum number of adjacent moves.
// D is dictionary
static int shortestChainLen(String start,
                            String target,
                            HashSet<String> D)
{
 
    // If the target String is not
    // present in the dictionary
    if (!D.Contains(target))
        return 0;
 
    // To store the current chain length
    // and the length of the words
    int level = 0, wordlength = start.Length;
 
    // Push the starting word into the queue
    List<String> Q = new List<String>();
    Q.Add(start);
 
    // While the queue is non-empty
    while (Q.Count != 0)
    {
 
        // Increment the chain length
        ++level;
 
        // Current size of the queue
        int sizeofQ = Q.Count;
 
        // Since the queue is being updated while
        // it is being traversed so only the
        // elements which were already present
        // in the queue before the start of this
        // loop will be traversed for now
        for (int i = 0; i < sizeofQ; ++i)
        {
 
            // Remove the first word from the queue
            char []word = Q[0].ToCharArray();
            Q.RemoveAt(0);
 
            // For every character of the word
            for (int pos = 0; pos < wordlength; ++pos)
            {
 
                // Retain the original character
                // at the current position
                char orig_char = word[pos];
 
                // Replace the current character with
                // every possible lowercase alphabet
                for (char c = 'a'; c <= 'z'; ++c)
                {
                    word[pos] = c;
 
                    // If the new word is equal
                    // to the target word
                    if (String.Join("", word).Equals(target))
                        return level + 1;
 
                    // Remove the word from the set
                    // if it is found in it
                    if (!D.Contains(String.Join("", word)))
                        continue;
                    D.Remove(String.Join("", word));
 
                    // And push the newly generated word
                    // which will be a part of the chain
                    Q.Add(String.Join("", word));
                }
 
                // Restore the original character
                // at the current position
                word[pos] = orig_char;
            }
        }
    }
    return 0;
}
 
// Driver code
public static void Main(String[] args)
{
    // make dictionary
    HashSet<String> D = new HashSet<String>();
    D.Add("poon");
    D.Add("plee");
    D.Add("same");
    D.Add("poie");
    D.Add("plie");
    D.Add("poin");
    D.Add("plea");
    String start = "toon";
    String target = "plea";
    Console.Write("Length of shortest chain is: "
        + shortestChainLen(start, target, D));
}
}
 
// This code is contributed by PrinciRaj1992

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Output

Length of shortest chain is: 7

Complexity Analysis: 



  • Time complexity: O(n²m), where m is the number of entries originally in the dictionary and n is the size of the string. 
     
  • Auxiliary Space:O(m*n), where m are the strings are stored in queue. 
    So the space Complexity is O(m*n).

Alternate Implementation: (Maintaining the mapping of the intermediate words and the original word):

Below is an alternative implementation to the above approach. 

Here, in this approach, we find out all the intermediate words of the start word and the words in the given list of dictionary and maintain a map of the intermediate word and a vector of the original word (map<string, vector<string>>). For instance, for the word “POON”, the intermediate words are “*OON” , “P*ON”, “PO*N”, “POO*”. Then, we perform BFS traversal starting with the start word and push a pair of start word and the distance (pair(word, distance)) to the queue until we reach the target word. Then, the distance is our answer.

C++

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// C++ program to find length
// of the shortest chain
// transformation from source
// to target
#include <bits/stdc++.h>
using namespace std;
 
// Returns length of shortest chain
// to reach 'target' from 'start'
// using minimum number of adjacent
// moves.  D is dictionary
int shortestChainLen(
string start, string target,
set<string>& D)
{
   
  // Map of intermediate words and
  // the list of original words
  map<string, vector<string>> umap;
 
  // Find all the intermediate
  // words for the start word
  for(int i = 0; i < start.size(); i++)
  {
    string str = start.substr(0,i) + "*" +
                        start.substr(i+1);
    umap[str].push_back(start);
  }
 
  // Find all the intermediate words for
  // the words in the given Set
  for(auto it = D.begin(); it != D.end(); it++)
  {
    string word = *it;
    for(int j = 0; j < word.size(); j++)
    {
      string str = word.substr(0,j) + "*" +
                          word.substr(j+1);
      umap[str].push_back(word);
    }
  }
 
  // Perform BFS and push (word, distance)
  queue<pair<string, int>> q;
 
  map<string, int> visited;
 
  q.push(make_pair(start,1));
  visited[start] = 1;
 
  // Traverse until queue is empty
  while(!q.empty())
  {
    pair<string, int> p = q.front();
    q.pop();
 
    string word = p.first;
    int dist = p.second;
 
    // If target word is found
    if(word == target)
    {
      return dist;
    }
 
    // Finding intermediate words for
    // the word in front of queue
    for(int i = 0; i < word.size(); i++)
    {
      string str = word.substr(0,i) + "*" +
                           word.substr(i+1);
 
      vector<string> vect = umap[str];
      for(int j = 0; j < vect.size(); j++)
      {
        // If the word is not visited
        if(visited[vect[j]] == 0)
        {
          visited[vect[j]] = 1;
          q.push(make_pair(vect[j], dist + 1));
        }
      }
    }
 
  }
 
    return 0;
}
 
// Driver program
int main()
{
    // make dictionary
    set<string> D;
    D.insert("poon");
    D.insert("plee");
    D.insert("same");
    D.insert("poie");
    D.insert("plie");
    D.insert("poin");
    D.insert("plea");
    string start = "toon";
    string target = "plea";
    cout << "Length of shortest chain is: "
         << shortestChainLen(start, target, D);
    return 0;
}

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Output

Length of shortest chain is: 7

https://www.youtube.com/watch?v=6pIC20wCm20
Thanks to Gaurav Ahirwar and Rajnish Kumar Jha for the above solution.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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