# 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.↵
↵
```cpp↵
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 opserations**↵
↵
↵
For **100k queries**:↵
↵
**1.3M × 100k = 1.3e11 operations → TLE, impossible**↵
↵
↵
## Method 2: Pattern indexing (per-query BFS usingot/ht/ho*)↵
----------------------------------------------------------* wildcards)↵
↵
```cpp↵
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 in C++ under 1-2s**↵
↵
↵
## 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.↵
↵
```cpp↵
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,and ↵
(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, wetry replacinge every character in the word with all 26 letters. ↵
Thisgiveyields up to **26·L possible** neighbors and avoids anyrequires no preprocessing.↵
Its c ↵
Complexity per query isroughly **O(N·L)**, which is fastine for a single query, but becomes prohibitive when Q issmall Q but disastrous for large Q.↵
↵
Forexample, with **N = 5000 and **, **Q = 100000**, the total work crosses several billion operations, which isoperations become several billions → not rfealistic for standard time limitssible.↵
↵
### 2. Pattern Indexing (“ot”, “ht”, “ho”)*↵
This is the classic editorial solution:↵
you↵
↵
We preprocess **N·L patterns, each mapping to the set of words that match that pattern.↵
Neighbor lookup becomes faster (L patterns + bucket lookups), but BFS still needs to examine** wildcard patterns. ↵
Lookup becomes faster, but BFS still touches a largeporfraction of the dictionary for each query.↵
Its total cost is about O(N·L + Q·N·L); much better constants.↵
↵
Total complexity: **O(N·L + Q·N·L)**. ↵
Better than method 1,but still too slow once Q grows into the tens of thousandsfor large Q.↵
↵
### 3. Full Adjacency Precomputation (O(N²·L) once, BFS per query thereafter)↵
When the dictionary is static, the graph structure does not change.↵
We can p↵
↵
Precompute all edges by checkingeach pairs of dictionary words and connecting thoswords differing in one tchat differ in exactly one position.↵
Thisracter.↵
↵
One-time preprocessing takes O(N²·L) time once, but every query afterward becomes a simple BFS on a ready-made adjacency list.↵
The per-query time drops to O(N), and the t: **O(N²·L)**. ↵
Then each query: simple BFS in **O(N)**.↵
↵
Totalbecomes plexity: **O(N²·L + Q·N), which is far more efficient than the previous method**. ↵
Dominates when **Q is large (for example, when Q ≥ N)**.↵
This approach leverages amortization: the heavy preprocessing cost is paid once, and repeated queries become cheap.↵
↵
↵
## In short:↵
↵
- For **small Q**, method (1) or (2) isperfectly fine.↵
↵
fine. ↵
- For **moderate Q**, pattern indexingscales better than neighbor generation.↵
↵
wins. ↵
- For **large Q**, especiallywhen **Q ≈ N or higher, **, **full adjacency precomputation is the superiorbest solution, as it minimizes repeated work across queries↵
↵
↵
↵
↵
↵
↵
↵
**.↵
↵
↵
Hello ,↵
↵
During learning graphs, I encountered a famous and classic problem
↵
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.↵
↵
```cpp↵
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 op
↵
↵
**1.3M × 100k = 1.3e11 operations
↵
↵
## Method 2: Pattern indexing (per-query BFS using
----------------------------------------------------------
↵
```cpp↵
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**↵
↵
↵
**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.↵
↵
```cpp↵
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
This
Its c
Complexity per query is
↵
For
↵
### 2. Pattern Indexing
This is the classic editorial solution:↵
you
↵
We preprocess **N·L
Neighbor lookup becomes faster (L patterns + bucket lookups), but BFS still needs to examine
Lookup becomes faster, but BFS still touches a large
Its total cost is about O(N·L + Q·N·L); much better constants
↵
Total complexity: **O(N·L + Q·N·L)**. ↵
Better than method 1,
↵
### 3. Full Adjacency Precomputation
When the dictionary is static, the graph structure does not change.↵
We can p
↵
Precompute all edges by checking
This
↵
One-time preprocessing
The per-query time drops to O(N), and the t
Then each query: simple BFS in **O(N)**.↵
↵
Total
Dominates when **Q is large (
↵
↵
## In short:↵
↵
- For **small Q**, method (1) or (2) is
↵
- For **moderate Q**, pattern indexing
↵
- For **large Q**, especially
↵
↵
↵
↵
↵
↵
↵
↵
