Related Articles

Related Articles

Print all possible shortest chains to reach a target word
  • Difficulty Level : Hard
  • Last Updated : 04 Sep, 2020

Given two strings start and target(both of the same length) and a list of strings str[], the task is to print all possible smallest sequences starting from start to target if it exists, such that adjacent words in the sequence only differ by a single character and each word in the sequence is present in the given list.
Note: It may be assumed that the target word is present in the list and the length of all the words is the same. If multiple sequences occur, print all of them.


Input: str[] = {poon, plee, same, poie, plea, plie, poin}, start = “toon”, target = “plea”
Output: [[toon, poon, poin, poie, plee, plea]]
Explanation: toon → poon → poin → poie → plee → plea

Input: str[] = { ted, tex, red, tax, tad, den, rex, pee}, start = “red”, target = “tax”
Output [[“red”, “ted”, “tad”, “tax”], [“red”, “ted”, “tex”, “tax”], [“red”, “rex”, “tex”, “tax”]]

Approach: The problem can be solved using BFS.The tricky part here is to do the BFS of the path instead of words. Follow the steps below to solve the problem:

  1. Initialize a variable, say res, to store all possible shortest paths.
  2. Create a Set to store all the visited words in current path and once the current path is completed, erase all the visited words.
  3. For each current word, find the possible next words present in str[] by changing each character from ‘a’ to ‘z’ and find all possible paths.
  4. Finally, print all the possible path.






// C++ Program to implememnt
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print all possible shortest
// sequences starting from start to target.
void displaypath(vector<vector<string> >& res)
    for (int i = 0; i < res.size(); i++) {
        cout << "[ ";
        for (int j = 0; j < res[0].size(); j++) {
            cout << res[i][j] << ", ";
        cout << " ]\n";
// Find words differing by a single
// character with word
vector<string> addWord(
    string word,
    unordered_set<string>& dict)
    vector<string> res;
    // Find next word in dict by changing
    // each element from 'a' to 'z'
    for (int i = 0; i < word.size(); i++) {
        char s = word[i];
        for (char c = 'a'; c <= 'z'; c++) {
            word[i] = c;
            if (dict.count(word))
        word[i] = s;
    return res;
// Function to get all the shortest possible
// sequences starting from 'start' to 'target'
vector<vector<string> > findLadders(
    vector<string>& Dict,
    string beginWord,
    string endWord)
    // Store all the shortest path.
    vector<vector<string> > res;
    // Store visited words in list
    unordered_set<string> visit;
    // Queue used to find the shortest path
    queue<vector<string> > q;
    // Stores the distinct words from given list
    unordered_set<string> dict(Dict.begin(),
    q.push({ beginWord });
    // Stores whether the shortest
    // path is found or not
    bool flag = false;
    while (!q.empty()) {
        int size = q.size();
        for (int i = 0; i < size; i++) {
            // Explore the next level
            vector<string> cur = q.front();
            vector<string> newadd;
            // Find words differing by a
            // single character
            newadd = addWord(cur.back(), dict);
            // Add words to the path.
            for (int j = 0; j < newadd.size(); j++) {
                vector<string> newline(cur.begin(),
                // Found the target
                if (newadd[j] == endWord) {
                    flag = true;
        // If already reached target
        if (flag) {
        // Erase all visited words.
        for (auto it : visit) {
    return res;
// Driver Code
int main()
    vector<string> str{ "ted", "tex", "red",
                        "tax", "tad", "den",
                        "rex", "pee" };
    string beginWord = "red";
    string endWord = "tax";
    vector<vector<string> > res
        = findLadders(str, beginWord, endWord);



[ red, ted, tad, tax,  ]
[ red, ted, tex, tax,  ]
[ red, rex, tex, tax,  ]

Time complexity: O(N²*M), where M is the number of strings in the given list and N is the length of each string.

Auxiliary Space:O(M*N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :