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

• Difficulty Level : Medium
• Last Updated : 04 May, 2021

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
Output: 4
ABCV - ABCD - EBCD - EBAD

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 the implementations of the above idea.

C++

// 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(start == target)
return 0;

// 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;
}

Java

// 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(start == target)
return 0;
// 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

// 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
}

// 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>();
String start = "toon";
String target = "plea";
System.out.print("Length of shortest chain is: "
+ shortestChainLen(start, target, D));
}
}

// This code is contributed by PrinciRaj1992

Python3

# Python3 program to find length of the
# shortest chain transformation from source
# to target
from collections import deque

# Returns length of shortest chain
# to reach 'target' from 'start'
# using minimum number of adjacent
# moves. D is dictionary
def shortestChainLen(start, target, D):

if start == target:
return 0
# If the target is not
# present in the dictionary
if target not in D:
return 0

# To store the current chain length
# and the length of the words
level, wordlength = 0, len(start)

# Push the starting word into the queue
Q =  deque()
Q.append(start)

# While the queue is non-empty
while (len(Q) > 0):

# Increment the chain length
level += 1

# Current size of the queue
sizeofQ = len(Q)

# 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 i in range(sizeofQ):

# Remove the first word from the queue
word = [j for j in Q.popleft()]
#Q.pop()

# For every character of the word
for pos in range(wordlength):

# Retain the original character
# at the current position
orig_char = word[pos]

# Replace the current character with
# every possible lowercase alphabet
for c in range(ord('a'), ord('z')+1):
word[pos] = chr(c)

# If the new word is equal
# to the target word
if ("".join(word) == target):
return level + 1

# Remove the word from the set
# if it is found in it
if ("".join(word) not in D):
continue

del D["".join(word)]

# And push the newly generated word
# which will be a part of the chain
Q.append("".join(word))

# Restore the original character
# at the current position
word[pos] = orig_char

return 0

# Driver code
if __name__ == '__main__':

# Make dictionary
D = {}
D["poon"] = 1
D["plee"] = 1
D["same"] = 1
D["poie"] = 1
D["plie"] = 1
D["poin"] = 1
D["plea"] = 1
start = "toon"
target = "plea"

print("Length of shortest chain is: ",
shortestChainLen(start, target, D))

# This code is contributed by mohit kumar 29

C#

// 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(start == target)
return 0;
// 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>();

// 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.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
}

// 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>();
String start = "toon";
String target = "plea";
Console.Write("Length of shortest chain is: "
+ shortestChainLen(start, target, D));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>
// Javascript program to find length
// of the shortest chain
// transformation from source
// to target

// Returns length of shortest chain
// to reach 'target' from 'start'
// using minimum number of adjacent moves.
// D is dictionary
function shortestChainLen(start,target,D)
{
if(start == target)
return 0;

// If the target String is not
// present in the dictionary
if (!D.has(target))
return 0;

// To store the current chain length
// and the length of the words
let level = 0, wordlength = start.length;

// Push the starting word into the queue
let Q = [];
Q.push(start);

// While the queue is non-empty
while (Q.length != 0)
{

// Increment the chain length
++level;

// Current size of the queue
let sizeofQ = Q.length;

// 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 (let i = 0; i < sizeofQ; ++i)
{

// Remove the first word from the queue
let word = Q.split("");
Q.shift();

// For every character of the word
for (let pos = 0; pos < wordlength; ++pos)
{

// Retain the original character
// at the current position
let orig_char = word[pos];

// Replace the current character with
// every possible lowercase alphabet
for (let c = 'a'.charCodeAt(0); c <= 'z'.charCodeAt(0); ++c)
{
word[pos] = String.fromCharCode(c);

// If the new word is equal
// to the target word
if (word.join("") == target)
return level + 1;

// Remove the word from the set
// if it is found in it
if (!D.has(word.join("")))
continue;
D.delete(word.join(""));

// And push the newly generated word
// which will be a part of the chain
Q.push(word.join(""));
}

// Restore the original character
// at the current position
word[pos] = orig_char;
}
}
}

return 0;
}

// Driver code
// make dictionary
let D = new Set();
let start = "toon";
let target = "plea";
document.write("Length of shortest chain is: "
+ shortestChainLen(start, target, D));

// This code is contributed by unknown2108
</script>
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++

// 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(start == target)
return 0;

// 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 code
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;
}
Output
Length of shortest chain is: 7

Thanks to Gaurav Ahirwar and Rajnish Kumar Jha for the above solution.