LeetCode #126 — HARD

Word Ladder II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s₁ -> s₂ -> ... -> s_k such that:

Every adjacent pair of words differs by a single letter.
Every s_i for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
s_k == endWord

Given two words, beginWord and endWord, and a dictionary wordList, return all the shortest transformation sequences from beginWord to endWord, or an empty list if no such sequence exists. Each sequence should be returned as a list of the words [beginWord, s₁, s₂, ..., s_k].

Example 1:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: [["hit","hot","dot","dog","cog"],["hit","hot","lot","log","cog"]]
Explanation: There are 2 shortest transformation sequences:
"hit" -> "hot" -> "dot" -> "dog" -> "cog"
"hit" -> "hot" -> "lot" -> "log" -> "cog"

Example 2:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
Output: []
Explanation: The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.

Constraints:

1 <= beginWord.length <= 5
endWord.length == beginWord.length
1 <= wordList.length <= 500
wordList[i].length == beginWord.length
beginWord, endWord, and wordList[i] consist of lowercase English letters.
beginWord != endWord
All the words in wordList are unique.
The sum of all shortest transformation sequences does not exceed 10⁵.

Step 01

Brute Force Baseline

Problem summary: A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that: Every adjacent pair of words differs by a single letter. Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList. sk == endWord Given two words, beginWord and endWord, and a dictionary wordList, return all the shortest transformation sequences from beginWord to endWord, or an empty list if no such sequence exists. Each sequence should be returned as a list of the words [beginWord, s1, s2, ..., sk].

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Backtracking

Example 1

"hit"
"cog"
["hot","dot","dog","lot","log","cog"]

Example 2

"hit"
"cog"
["hot","dot","dog","lot","log"]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #126: Word Ladder II
class Solution {
    private List<List<String>> ans;
    private Map<String, Set<String>> prev;

    public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
        ans = new ArrayList<>();
        Set<String> words = new HashSet<>(wordList);
        if (!words.contains(endWord)) {
            return ans;
        }
        words.remove(beginWord);
        Map<String, Integer> dist = new HashMap<>();
        dist.put(beginWord, 0);
        prev = new HashMap<>();
        Queue<String> q = new ArrayDeque<>();
        q.offer(beginWord);
        boolean found = false;
        int step = 0;
        while (!q.isEmpty() && !found) {
            ++step;
            for (int i = q.size(); i > 0; --i) {
                String p = q.poll();
                char[] chars = p.toCharArray();
                for (int j = 0; j < chars.length; ++j) {
                    char ch = chars[j];
                    for (char k = 'a'; k <= 'z'; ++k) {
                        chars[j] = k;
                        String t = new String(chars);
                        if (dist.getOrDefault(t, 0) == step) {
                            prev.get(t).add(p);
                        }
                        if (!words.contains(t)) {
                            continue;
                        }
                        prev.computeIfAbsent(t, key -> new HashSet<>()).add(p);
                        words.remove(t);
                        q.offer(t);
                        dist.put(t, step);
                        if (endWord.equals(t)) {
                            found = true;
                        }
                    }
                    chars[j] = ch;
                }
            }
        }
        if (found) {
            Deque<String> path = new ArrayDeque<>();
            path.add(endWord);
            dfs(path, beginWord, endWord);
        }
        return ans;
    }

    private void dfs(Deque<String> path, String beginWord, String cur) {
        if (cur.equals(beginWord)) {
            ans.add(new ArrayList<>(path));
            return;
        }
        for (String precursor : prev.get(cur)) {
            path.addFirst(precursor);
            dfs(path, beginWord, precursor);
            path.removeFirst();
        }
    }
}

// Accepted solution for LeetCode #126: Word Ladder II
func findLadders(beginWord string, endWord string, wordList []string) [][]string {
	var ans [][]string
	words := make(map[string]bool)
	for _, word := range wordList {
		words[word] = true
	}
	if !words[endWord] {
		return ans
	}
	words[beginWord] = false
	dist := map[string]int{beginWord: 0}
	prev := map[string]map[string]bool{}
	q := []string{beginWord}
	found := false
	step := 0
	for len(q) > 0 && !found {
		step++
		for i := len(q); i > 0; i-- {
			p := q[0]
			q = q[1:]
			chars := []byte(p)
			for j := 0; j < len(chars); j++ {
				ch := chars[j]
				for k := 'a'; k <= 'z'; k++ {
					chars[j] = byte(k)
					t := string(chars)
					if v, ok := dist[t]; ok {
						if v == step {
							prev[t][p] = true
						}
					}
					if !words[t] {
						continue
					}
					if len(prev[t]) == 0 {
						prev[t] = make(map[string]bool)
					}
					prev[t][p] = true
					words[t] = false
					q = append(q, t)
					dist[t] = step
					if endWord == t {
						found = true
					}
				}
				chars[j] = ch
			}
		}
	}
	var dfs func(path []string, begin, cur string)
	dfs = func(path []string, begin, cur string) {
		if cur == beginWord {
			cp := make([]string, len(path))
			copy(cp, path)
			ans = append(ans, cp)
			return
		}
		for k := range prev[cur] {
			path = append([]string{k}, path...)
			dfs(path, beginWord, k)
			path = path[1:]
		}
	}
	if found {
		path := []string{endWord}
		dfs(path, beginWord, endWord)
	}
	return ans
}

# Accepted solution for LeetCode #126: Word Ladder II
class Solution:
    def findLadders(
        self, beginWord: str, endWord: str, wordList: List[str]
    ) -> List[List[str]]:
        def dfs(path, cur):
            if cur == beginWord:
                ans.append(path[::-1])
                return
            for precursor in prev[cur]:
                path.append(precursor)
                dfs(path, precursor)
                path.pop()

        ans = []
        words = set(wordList)
        if endWord not in words:
            return ans
        words.discard(beginWord)
        dist = {beginWord: 0}
        prev = defaultdict(set)
        q = deque([beginWord])
        found = False
        step = 0
        while q and not found:
            step += 1
            for i in range(len(q), 0, -1):
                p = q.popleft()
                s = list(p)
                for i in range(len(s)):
                    ch = s[i]
                    for j in range(26):
                        s[i] = chr(ord('a') + j)
                        t = ''.join(s)
                        if dist.get(t, 0) == step:
                            prev[t].add(p)
                        if t not in words:
                            continue
                        prev[t].add(p)
                        words.discard(t)
                        q.append(t)
                        dist[t] = step
                        if endWord == t:
                            found = True
                    s[i] = ch
        if found:
            path = [endWord]
            dfs(path, endWord)
        return ans

// Accepted solution for LeetCode #126: Word Ladder II
struct Solution;

use std::collections::HashMap;
use std::collections::HashSet;
use std::iter::FromIterator;

impl Solution {
    fn find_ladders(
        begin_word: String,
        end_word: String,
        word_list: Vec<String>,
    ) -> Vec<Vec<String>> {
        let mut dict: HashSet<String> = HashSet::new();
        for word in word_list {
            dict.insert(word);
        }
        if !dict.contains(&end_word) {
            return vec![];
        }
        let set1: HashSet<String> = HashSet::from_iter(vec![begin_word.to_string()]);
        let set2: HashSet<String> = HashSet::from_iter(vec![end_word.to_string()]);
        let mut map: HashMap<String, HashSet<String>> = HashMap::new();
        Self::bfs(set1, set2, false, &mut map, &mut dict);
        let mut path = vec![begin_word.to_string()];
        let mut res = vec![];
        Self::dfs(&begin_word, &mut path, &mut res, &map, &end_word);
        res
    }

    fn bfs(
        set1: HashSet<String>,
        set2: HashSet<String>,
        flipped: bool,
        map: &mut HashMap<String, HashSet<String>>,
        dict: &mut HashSet<String>,
    ) {
        if set1.is_empty() {
            return;
        }
        if set1.len() > set2.len() {
            Self::bfs(set2, set1, !flipped, map, dict);
            return;
        }
        for s in set1.iter() {
            dict.remove(s);
        }
        for s in set2.iter() {
            dict.remove(s);
        }
        let mut next: HashSet<String> = HashSet::new();
        let mut done = false;
        for word in set1.iter() {
            for next_word in Self::connected_words(&word.to_string()) {
                let key = if flipped {
                    next_word.to_string()
                } else {
                    word.to_string()
                };
                let value = if flipped {
                    word.to_string()
                } else {
                    next_word.to_string()
                };
                if set2.contains(&next_word) {
                    done = true;
                    map.entry(key).or_default().insert(value);
                } else if dict.contains(&next_word) {
                    next.insert(next_word);
                    map.entry(key).or_default().insert(value);
                }
            }
        }
        if !done {
            Self::bfs(set2, next, !flipped, map, dict);
        }
    }

    fn dfs(
        start: &str,
        path: &mut Vec<String>,
        all: &mut Vec<Vec<String>>,
        map: &HashMap<String, HashSet<String>>,
        end: &str,
    ) {
        if start == end {
            all.push(path.to_vec());
        } else {
            if let Some(nei) = map.get(start) {
                for next in nei.iter() {
                    path.push(next.to_string());
                    Self::dfs(next, path, all, map, end);
                    path.pop();
                }
            }
        }
    }

    fn connected_words(word: &str) -> Vec<String> {
        let n = word.len();
        let mut res = vec![];
        for i in 0..n {
            let mut s: Vec<char> = word.chars().collect();
            for j in 0..26 {
                let c = (b'a' + j as u8) as char;
                s[i] = c;
                let new_word: String = s.iter().collect();
                res.push(new_word);
            }
        }
        res
    }
}

#[test]
fn test() {
    let begin_word = "hit".to_string();
    let end_word = "cog".to_string();
    let word_list = vec_string!["hot", "dot", "dog", "lot", "log", "cog"];
    let mut res = vec_vec_string![
        ["hit", "hot", "dot", "dog", "cog"],
        ["hit", "hot", "lot", "log", "cog"]
    ];
    let mut ans = Solution::find_ladders(begin_word, end_word, word_list);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);

    let begin_word = "hit".to_string();
    let end_word = "cog".to_string();
    let word_list = vec_string!["hot", "dot", "dog", "lot", "log"];
    let res = vec_vec_string![];
    assert_eq!(Solution::find_ladders(begin_word, end_word, word_list), res);

    let begin_word = "a".to_string();
    let end_word = "c".to_string();
    let word_list = vec_string!["a", "b", "c"];
    let mut res = vec_vec_string![["a", "c"]];
    let mut ans = Solution::find_ladders(begin_word, end_word, word_list);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
}

// Accepted solution for LeetCode #126: Word Ladder II
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #126: Word Ladder II
// class Solution {
//     private List<List<String>> ans;
//     private Map<String, Set<String>> prev;
// 
//     public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
//         ans = new ArrayList<>();
//         Set<String> words = new HashSet<>(wordList);
//         if (!words.contains(endWord)) {
//             return ans;
//         }
//         words.remove(beginWord);
//         Map<String, Integer> dist = new HashMap<>();
//         dist.put(beginWord, 0);
//         prev = new HashMap<>();
//         Queue<String> q = new ArrayDeque<>();
//         q.offer(beginWord);
//         boolean found = false;
//         int step = 0;
//         while (!q.isEmpty() && !found) {
//             ++step;
//             for (int i = q.size(); i > 0; --i) {
//                 String p = q.poll();
//                 char[] chars = p.toCharArray();
//                 for (int j = 0; j < chars.length; ++j) {
//                     char ch = chars[j];
//                     for (char k = 'a'; k <= 'z'; ++k) {
//                         chars[j] = k;
//                         String t = new String(chars);
//                         if (dist.getOrDefault(t, 0) == step) {
//                             prev.get(t).add(p);
//                         }
//                         if (!words.contains(t)) {
//                             continue;
//                         }
//                         prev.computeIfAbsent(t, key -> new HashSet<>()).add(p);
//                         words.remove(t);
//                         q.offer(t);
//                         dist.put(t, step);
//                         if (endWord.equals(t)) {
//                             found = true;
//                         }
//                     }
//                     chars[j] = ch;
//                 }
//             }
//         }
//         if (found) {
//             Deque<String> path = new ArrayDeque<>();
//             path.add(endWord);
//             dfs(path, beginWord, endWord);
//         }
//         return ans;
//     }
// 
//     private void dfs(Deque<String> path, String beginWord, String cur) {
//         if (cur.equals(beginWord)) {
//             ans.add(new ArrayList<>(path));
//             return;
//         }
//         for (String precursor : prev.get(cur)) {
//             path.addFirst(precursor);
//             dfs(path, beginWord, precursor);
//             path.removeFirst();
//         }
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n!)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.