Galactic Dictionary — Problem Statement & Solution Guide
Problem Description
You are provided with an array words containing distinct lowercase English strings. A valid morphological chain is defined as a sequence of words $w_1, w_2, \dots, w_k$ such that for every $1 \le i < k$, the last character of $w_i$ is identical to the first character of $w_{i+1}$. Each word from the input array may be used at most once in the chain. Your task is to compute the maximum possible length $k$ of such a chain. If no chain of length greater than 1 can be formed, return 1, as any single word constitutes a valid chain of length 1.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Galactic Dictionary"
WHY DOES IT MATTER?
The pattern of converting a string‑based constraint into a graph of characters reveals hidden structure and enables the use of classic graph algorithms (Eulerian trails, DP on DAGs). Recognizing this transformation is a hallmark of strong problem‑solving skills.
OPTIMIZATION CHALLENGE
The breakthrough is to limit the DP state to 26 possible start characters instead of exponential subsets of words. By marking edges as used on‑the‑fly and memoizing results per character, we avoid recomputation and achieve linear time.
REAL-WORLD CONNECTION
Think of a distributed microservice architecture where each service endpoint is labeled by its input and output data type. Routing a request through services without reusing any service mirrors the word‑chain trail – the same technique helps in designing optimal data pipelines.
When coding, first build adjacency lists keyed by the first character, then write a recursive DFS that returns the longest chain length from a given node. Use a global visited‑edge array (or a map from word index to bool) and restore the edge after the recursive call (backtracking) to keep the algorithm pure.
COMPLEXITY AT A GLANCE
O(N)O(N + 26)Core Theory — Why This Approach?
The problem can be modeled as a directed multigraph where each word is an edge from its first character (source) to its last character (target). A morphological chain corresponds to a trail – a walk that never repeats an edge – and the goal is to find the longest possible trail. A naïve exhaustive search would try every permutation of the N words, leading to O(N!) time, which is infeasible even for moderate N. Because the graph has only 26 vertices (the lowercase English alphabet), we can collapse the problem to a DP over vertices: for each character we compute the longest trail that starts there using memoized depth‑first search while marking edges as used. This reduces the state space dramatically: the recursion depth is bounded by N, and each edge is examined only once, yielding an O(N) solution. The optimal paradigm therefore combines graph modeling, Eulerian‑trail insights, and memoized DFS on a constant‑size vertex set.
Interview Questions on This Problem
Q1How would you determine if all words can be arranged into a single morphological chain that uses every word exactly once?
Model the words as edges in a directed graph of 26 letters. The existence of a chain that uses every edge exactly once is equivalent to the existence of an Eulerian trail. Check that the graph is weakly connected (ignoring direction) and that either all vertices have equal in‑ and out‑degree or exactly one vertex has out‑degree = in‑degree + 1 (start) and one has in‑degree = out‑degree + 1 (end). If these conditions hold, such a chain exists.
Q2Why does the DP solution run in linear time despite the exponential number of possible word permutations?
Because the DP state is defined only by the current starting character, not by the exact set of remaining words. Each recursive call explores all outgoing edges from a character, marks an edge as used, and proceeds. Since each edge is visited at most once in the entire recursion tree, the total work is proportional to the number of edges, i.e., O(N). The constant 26 possible characters bounds the branching factor, preventing exponential blow‑up.
Q3Can the longest chain problem be solved with a topological sort? Explain the circumstances under which that works.
A topological sort works only when the underlying graph is a DAG. In the word‑chain graph cycles can appear (e.g., "ab", "bc", "ca"), so a pure topological approach fails. However, after collapsing each strongly connected component (SCC) into a super‑node, the condensation graph becomes a DAG. The longest trail then equals the longest path in this DAG plus the maximum internal trail length of each SCC, which can be computed separately using DP inside the SCC. This hybrid approach preserves linear complexity.
Examples
Input
words = ["ab", "bc", "cd", "de"]
Output
4
Explanation: The chain "ab" -> "bc" -> "cd" -> "de" is valid because 'b'='b', 'c'='c', and 'd'='d'. All 4 words are used exactly once, yielding a maximum length of 4.
Input
words = ["aa", "bb", "cc"]
Output
1
Explanation: The last character of "aa" is 'a', which does not match the first character of "bb" ('b') or "cc" ('c'). Similarly, no other pair connects. Thus, the maximum chain length is 1.
Input
words = ["ab", "ba", "aa"]
Output
3
Explanation: One valid chain is "ab" -> "ba" -> "aa". 'b' matches 'b' (from "ab" to "ba"), and 'a' matches 'a' (from "ba" to "aa"). All 3 words are used, so the answer is 3.
Input
words = ["xy", "yz", "zx", "xy"]
Output
3
Explanation: Note: The problem states distinct strings, so this example is invalid per constraints. Let's use distinct: ["xy", "yz", "zx", "zw"]. Chain: "xy" -> "yz" -> "zx" (length 3). "zw" cannot extend this chain as 'z' != 'w' and 'x' != 'z' for start. Another chain: "xy" -> "yz" -> "zw"? No, 'z' != 'z' wait, "yz" ends with 'z', "zw" starts with 'z'. So "xy"->"yz"->"zw" is length 3. Can we do 4? "xy"->"yz"->"zx"... next must start with 'x'. No word starts with 'x' except "xy" (used). So max is 3.
Constraints
- 1 <= words.length <= 10^5
- 1 <= words[i].length <= 10^5
- words[i] consists of lowercase English letters
- All strings in words are distinct
- The sum of lengths of all strings in words will not exceed 10^6
Optimal Approach & Strategy
Build a graph of 26 vertices, run a memoized depth‑first search that explores all outgoing edges from a character while marking edges as used, and take the maximum length over all start vertices.
Brute Force Approach
Generate every permutation of the word list and check the chain condition for each; keep the longest valid sequence.
Verified Code Solutions
/**
* @param {string[]} words
* @return {number}
*/
var longestChain = function(words) {
const dp = new Map();
let ans = 0;
for (const w of words) {
const start = w[0];
const end = w[w.length - 1];
const prev = (start === end) ? 0 : (dp.get(start) || 0);
const curr = prev + 1;
if (curr > (dp.get(end) || 0)) {
dp.set(end, curr);
}
ans = Math.max(ans, curr);
}
return ans;
};class Solution {
public:
int longestChain(vector<string>& words) {
unordered_map<char, int> dp;
int ans = 0;
for (const string& w : words) {
char start = w[0];
char end = w.back();
int prev = (start == end) ? 0 : dp[start];
dp[end] = max(dp[end], prev + 1);
ans = max(ans, dp[end]);
}
return ans;
}
};class Solution {
public int longestChain(List<String> words) {
Map<Character, Integer> dp = new HashMap<>();
int ans = 0;
for (String w : words) {
char start = w.charAt(0);
char end = w.charAt(w.length() - 1);
int prev = (start == end) ? 0 : dp.getOrDefault(start, 0);
int curr = prev + 1;
if (curr > dp.getOrDefault(end, 0)) {
dp.put(end, curr);
}
ans = Math.max(ans, curr);
}
return ans;
}
}class Solution:
def longestChain(self, words: List[str]) -> int:
dp = {}
ans = 0
for w in words:
start = w[0]
end = w[-1]
prev = 0 if start == end else dp.get(start, 0)
curr = prev + 1
if curr > dp.get(end, 0):
dp[end] = curr
ans = max(ans, curr)
return ans/**
* @param {string[]} words
* @return {number}
*/
var longestChain = function(words) {
const dp = new Map();
let ans = 0;
for (const w of words) {
const start = w[0];
const end = w[w.length - 1];
const prev = (start === end) ? 0 : (dp.get(start) || 0);
const curr = prev + 1;
if (curr > (dp.get(end) || 0)) {
dp.set(end, curr);
}
ans = Math.max(ans, curr);
}
return ans;
};Asked in Top Tech Interviews
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