# 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
Explanation: TOON - POON - POIN - POIE - PLIE - PLEE - PLEA```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to use BFS. We start from the given start word, traverse all words that adjacent (differ by one character) to it and keep doing so until we find the target word or we have traversed all words.

Below is C++ implementation of above idea.

## C

```// C++ program to find length of the shortest chain
// transformation from source to target
#include<bits/stdc++.h>
using namespace std;

// To check if strings differ by exactly one character
{
int count = 0;  // to store count of differences
int n = a.length();

// Iterate through all characters and return false
// if there are more than one mismatching characters
for (int i = 0; i < n; i++)
{
if (a[i] != b[i]) count++;
if (count > 1) return false;
}
return count == 1 ? true : false;
}

// A queue item to store word and minimum chain length
// to reach the word.
struct QItem
{
string word;
int len;
};

// 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)
{
// Create a queue for BFS and insert 'start' as source vertex
queue<QItem> Q;
QItem item = {start, 1};  // Chain length for start word is 1
Q.push(item);

// While queue is not empty
while (!Q.empty())
{
// Take the front word
QItem curr = Q.front();
Q.pop();

// Go through all words of dictionary
for (set<string>::iterator it = D.begin(); it != D.end(); it++)
{
// Process a dictionary word if it is adjacent to current
// word (or vertex) of BFS
string temp = *it;
{
// Add the dictionary word to Q
item.word = temp;
item.len = curr.len + 1;
Q.push(item);

// Remove from dictionary so that this word is not
// processed again.  This is like marking visited
D.erase(temp);

// If we reached target
if (temp == target)
return item.len;
}
}
}
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;
} ```

## Python

```# To check if strings differ by
# exactly one character

count = 0
n = len(a)

# Iterate through all characters and return false
# if there are more than one mismatching characters
for i in range(n):
if a[i] != b[i]:
count += 1
if count > 1:
break

return True if count == 1 else False

# A queue item to store word and minimum chain length
# to reach the word.
class QItem():

def __init__(self, word, len):
self.word = word
self.len = len

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

# Create a queue for BFS and insert
# 'start' as source vertex
Q = []
item = QItem(start, 1)
Q.append(item)

while( len(Q) > 0):

curr = Q.pop()

# Go through all words of dictionary
for it in D:

# Process a dictionary word if it is
# adjacent to current word (or vertex) of BFS
temp = it

# Add the dictionary word to Q
item.word = temp
item.len  = curr.len + 1
Q.append(item)

# Remove from dictionary so that this
# word is not processed again.  This is
# like marking visited
D.remove(temp)

# If we reached target
if temp == target:
return item.len

D = []
D.append("poon")
D.append("plee")
D.append("same")
D.append("poie")
D.append("plie")
D.append("poin")
D.append("plea")
start = "toon"
target = "plea"
print "Length of shortest chain is: %d" \
% shortestChainLen(start, target, D)

# This code is contributed by Divyanshu Mehta
```

Output:

`Length of shortest chain is: 7`

Time Complexity of the above code is O(n²m) where n is the number of entries originally in the dictionary and m is the size of the string

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

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