LeetCode #127 — HARD

Word Ladder

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 the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if no such sequence exists.

Example 1:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: 5
Explanation: One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long.

Example 2:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
Output: 0
Explanation: The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.

Constraints:

1 <= beginWord.length <= 10
endWord.length == beginWord.length
1 <= wordList.length <= 5000
wordList[i].length == beginWord.length
beginWord, endWord, and wordList[i] consist of lowercase English letters.
beginWord != endWord
All the words in wordList are unique.

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 the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if no such sequence exists.

Baseline thinking

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

Pattern signal: Hash Map

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 #127: Word Ladder
class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        Set<String> words = new HashSet<>(wordList);
        Queue<String> q = new ArrayDeque<>();
        q.offer(beginWord);
        int ans = 1;
        while (!q.isEmpty()) {
            ++ans;
            for (int i = q.size(); i > 0; --i) {
                String s = q.poll();
                char[] chars = s.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 (!words.contains(t)) {
                            continue;
                        }
                        if (endWord.equals(t)) {
                            return ans;
                        }
                        q.offer(t);
                        words.remove(t);
                    }
                    chars[j] = ch;
                }
            }
        }
        return 0;
    }
}

// Accepted solution for LeetCode #127: Word Ladder
func ladderLength(beginWord string, endWord string, wordList []string) int {
	words := make(map[string]bool)
	for _, word := range wordList {
		words[word] = true
	}
	q := []string{beginWord}
	ans := 1
	for len(q) > 0 {
		ans++
		for i := len(q); i > 0; i-- {
			s := q[0]
			q = q[1:]
			chars := []byte(s)
			for j := 0; j < len(chars); j++ {
				ch := chars[j]
				for k := 'a'; k <= 'z'; k++ {
					chars[j] = byte(k)
					t := string(chars)
					if !words[t] {
						continue
					}
					if t == endWord {
						return ans
					}
					q = append(q, t)
					words[t] = false
				}
				chars[j] = ch
			}
		}
	}
	return 0
}

# Accepted solution for LeetCode #127: Word Ladder
class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        words = set(wordList)
        q = deque([beginWord])
        ans = 1
        while q:
            ans += 1
            for _ in range(len(q)):
                s = q.popleft()
                s = list(s)
                for i in range(len(s)):
                    ch = s[i]
                    for j in range(26):
                        s[i] = chr(ord('a') + j)
                        t = ''.join(s)
                        if t not in words:
                            continue
                        if t == endWord:
                            return ans
                        q.append(t)
                        words.remove(t)
                    s[i] = ch
        return 0

// Accepted solution for LeetCode #127: Word Ladder
pub fn encode(word : String) -> u64 {
    word.into_bytes().into_iter().fold(0, |mut acc, c|{
        acc <<= 5;
        acc | (c - b'a' + 1) as u64
    })
}

use std::collections::HashSet;
use std::collections::VecDeque;

impl Solution {
    pub fn ladder_length(begin_word: String, end_word: String, word_list: Vec<String>) -> i32 {
        let word_len = begin_word.len();
        let begin_word = encode(begin_word);
        let end_word = encode(end_word);
        
        let mut word_list : HashSet<u64> = word_list.into_iter().map(|word| encode(word)).collect();
        let mut frontier : VecDeque<u64> = VecDeque::with_capacity(5000);
        let mut seen : HashSet<u64> = HashSet::new();
        let mut res = 1;
        
        frontier.push_back(begin_word);
        while !frontier.is_empty() {
            let len = frontier.len();
            for _ in 0..len {
                let curr = frontier.pop_front().unwrap();
                if curr == end_word {
                    return res;
                }
                
                for i in 0..word_len {
                    let filter = !(0b11111 << (i * 5));
                    for character in 1..=26 {
                        let neighbour = ((curr & filter) | (character << (i * 5)));
                        if word_list.contains(&neighbour) && !seen.contains(&neighbour) {
                            frontier.push_back(neighbour);
                            seen.insert(neighbour);
                        }
                    }
                }
                
            }
            res += 1;
        }
        0
    }
}

// Accepted solution for LeetCode #127: Word Ladder
function ladderLength(beginWord: string, endWord: string, wordList: string[]): number {
    if (!wordList.includes(endWord)) return 0;

    const replace = (s: string, i: number, ch: string) => s.slice(0, i) + ch + s.slice(i + 1);
    const { length } = beginWord;
    const words: Record<string, string[]> = {};
    const g: Record<string, string[]> = {};

    for (const w of [beginWord, ...wordList]) {
        const derivatives: string[] = [];

        for (let i = 0; i < length; i++) {
            const nextW = replace(w, i, '*');
            derivatives.push(nextW);

            g[nextW] ??= [];
            g[nextW].push(w);
        }

        words[w] = derivatives;
    }

    let ans = 0;
    let q = words[beginWord];
    const vis = new Set<string>([beginWord]);

    while (q.length) {
        const nextQ: string[] = [];
        ans++;

        for (const variant of q) {
            for (const w of g[variant]) {
                if (w === endWord) return ans + 1;

                if (vis.has(w)) continue;
                vis.add(w);

                nextQ.push(...words[w]);
            }
        }

        q = nextQ;
    }

    return 0;
}

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

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

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.