Word Ladder With 1e5 Queries — Efficient Approaches and Analysis
Hello ,
During learning graphs, I encountered a famous and classic problem Word Ladder problem. Given a beginWord and endWord, find the minimum number of single-character transformations, where each intermediate word must be from a given list. Given all words are of the same length.
This shows up on LeetCode, GFG, interviews, everywhere.
But almost every version of the problem is single-query.
You solve it once and move on.
Now I want to discuss for a scenario where the dictionary is fixed and you need to answer queries in range of 1e5.
N ≤ 5000 words
L ≤ 10 (word length)
Q ≤ 100000 queries
This completely changes the problem.
The typical BFS solution (where we generate neighbours (26*L) at each stage and check whether they are in the dictionary or not and proceed) became far too slow.
This article explores that space in depth and shows efficient techniques depending on the ratio of N vs Q.
To understand the original classical problem better, you may refer to:
https://www.geeksforgeeks.org/problems/word-ladder/1
Method 1: "Generate 26·L neighbors per BFS" (per query BFS)
This is the classic solution.
int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
unordered_set<string> dict(wordList.begin(), wordList.end());
if (!dict.count(endWord)) return 0;
queue<pair<string,int>> q;
q.push({beginWord, 1});
unordered_set<string> vis;
vis.insert(beginWord);
while (!q.empty()) {
auto [w, d] = q.front(); q.pop();
if (w == endWord) return d;
for (int i = 0; i < w.size(); i++) {
char orig = w[i];
for (char c = 'a'; c <= 'z'; c++) {
if (c == orig) continue;
w[i] = c;
if (dict.count(w) && !vis.count(w)) {
vis.insert(w);
q.push({w, d + 1});
}
}
w[i] = orig;
}
}
return 0;
}
Per query BFS:
O(N × 26 × L) ≈ 5000 × 260 = 1.3 million operations
For 100k queries:
1.3M × 100k = 1.3e11 operations → TLE, impossible
Method 2: Pattern indexing (per-query BFS using * wildcards)
int ladder(string beginWord, string endWord, vector<string> wordList) {
int L = beginWord.size();
unordered_set<string> dict(wordList.begin(), wordList.end());
if (!dict.count(endWord)) return 0;
unordered_map<string, vector<string>> buckets;
for (auto &w : dict) {
for (int i = 0; i < L; i++) {
string p = w;
p[i] = '*';
buckets[p].push_back(w);
}
}
queue<pair<string,int>> q;
unordered_set<string> vis;
q.push({beginWord, 1});
vis.insert(beginWord);
while (!q.empty()) {
auto [w, d] = q.front(); q.pop();
if (w == endWord) return d;
for (int i = 0; i < L; i++) {
string p = w;
p[i] = '*';
for (auto &nei : buckets[p]) {
if (!vis.count(nei)) {
vis.insert(nei);
q.push({nei, d+1});
}
}
buckets[p].clear(); // important optimization
}
}
return 0;
}
Per query BFS:
O(N × L) = 5000 × 10 = 50k operations
For 100k queries:
50k × 100k = 5e9 → borderline slow, likely TLE
Method 3: Precomputed adjacency (O(N²·L) cost once) + BFS per query
Solving this question, I first came up with a logic where words in the dictionary can be considered as nodes and those words having only one character different have an edge joining them. So in this way I developed an adjacency list but this came out to be O(N²·L) which was not good for the original problem but for this variation it is efficient.
bool differ1(const string &a, const string &b) {
int c = 0;
for (int i = 0; i < a.size(); i++)
c += (a[i] != b[i]);
return c == 1;
}
vector<vector<int>> buildAdj(const vector<string>& dict) {
int n = dict.size();
vector<vector<int>> adj(n);
for (int i = 0; i < n; i++) {
for (int j = i+1; j < n; j++) {
if (differ1(dict[i], dict[j])) {
adj[i].push_back(j);
adj[j].push_back(i);
}
}
}
return adj;
}
int bfsAdj(int start, int goal, const vector<vector<int>>& adj) {
if (start == -1 || goal == -1) return 0;
int n = adj.size();
vector<int> dist(n, -1);
queue<int> q;
dist[start] = 1;
q.push(start);
while (!q.empty()) {
int u = q.front(); q.pop();
if (u == goal) return dist[u];
for (int v : adj[u]) {
if (dist[v] == -1) {
dist[v] = dist[u] + 1;
q.push(v);
}
}
}
return 0;
}
Analysis of the Three Main Approaches
There are three reasonable strategies for solving multiple Word-Ladder queries when the dictionary is fixed:
(1) On-the-fly neighbor generation,
(2) pattern-indexed BFS,
(3) full adjacency precomputation.
Each comes with different trade-offs depending on the number of queries Q compared to the dictionary size N.
1. Neighbor Generation (26·L BFS)
For each BFS step, we replace every character in the word with all 26 letters.
This yields up to 26·L neighbors and requires no preprocessing.
Complexity per query is O(N·L), which is fine for small Q but disastrous for large Q.
For N = 5000, Q = 100000, total operations become several billions → not feasible.
2. Pattern Indexing
We preprocess N·L wildcard patterns.
Lookup becomes faster, but BFS still touches a large fraction of the dictionary.
Total complexity: O(N·L + Q·N·L).
Better than method 1, still too slow for large Q.
3. Full Adjacency Precomputation
Precompute all edges by checking pairs of words differing in one character.
One-time preprocessing: O(N²·L).
Then each query: simple BFS in O(N).
Total complexity: O(N²·L + Q·N).
Dominates when Q is large (Q ≥ N).
In short:
- For small Q, method (1) or (2) is fine.
- For moderate Q, pattern indexing wins.
- For large Q, especially Q ≈ N or higher, full adjacency precomputation is the best solution.
