Optimized Prefix Repository — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Stack with O(1) minimum retrieval and Trie operations
Insertion O(|s|), Query O(L)O(total distinct prefix characters) ≈ O(Σ|s_i|)Problem Description
Design a data structure to store and manage a collection of strings, supporting two primary operations: adding a string to the repository and retrieving the lexicographically smallest prefix of a given length that exists in the repository. If no such prefix exists for the specified length, the retrieval operation should return an empty string.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimized Prefix Repository"
WHY DOES IT MATTER?
Prefix‑based queries appear in autocomplete, DNS resolution, and security rule matching; delivering the smallest lexicographic match guarantees deterministic user experience and can be used to enforce ordering constraints without extra sorting.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that you only need existence information at each node, not the full list of strings, allowing a greedy walk that never backtracks. Storing children in lexical order eliminates the need for a priority queue during queries.
REAL-WORLD CONNECTION
Think of a telephone exchange routing table: each node (digit) forwards the call to the smallest viable next digit that leads to an active line. The Trie mirrors this hierarchical routing, instantly finding the shortest valid dialing prefix.
During implementation, use a fixed-size array (e.g., 26 for lowercase English) for children; this gives O(1) child access and natural ordering. If the alphabet is larger, a sorted map works but adds a log factor—still far better than scanning all strings.
COMPLEXITY AT A GLANCE
Insertion O(|s|), Query O(L)O(total distinct prefix characters) ≈ O(Σ|s_i|)Core Theory — Why This Approach?
A Trie (prefix tree) is the canonical data structure for handling collections of strings where prefix queries are frequent. Each node represents a character and stores a map to its children; traversing from the root to a node yields the prefix formed by the visited characters. By augmenting each node with a boolean flag indicating whether any inserted string passes through that node, we can answer the query "lexicographically smallest prefix of length L" in O(L) time: starting at the root, we greedily select the smallest child (according to alphabetical order) that is marked as reachable, repeat until we have consumed L characters, and return the accumulated path. Naïve solutions—such as storing all strings in a hash set and scanning every possible prefix for each query—degenerate to O(N·L) per query, where N is the number of stored strings, and quickly become infeasible for large repositories (10⁵+ strings, each up to 10⁵ characters). The optimal paradigm leverages the hierarchical nature of a Trie, turning a potentially linear scan over all strings into a deterministic walk bounded by the requested prefix length, independent of the total number of strings.
The key to optimality lies in two lightweight augmentations: (1) a counter at each node that records how many strings have this node on their path (or simply a boolean if existence suffices), and (2) storing children in a fixed-size array (or ordered map) that respects lexical order. These allow constant‑time selection of the smallest viable child during a query. Insertion updates the counters along the path, guaranteeing that the query never explores dead branches. Consequently, both operations run in O(|s|) for insertion and O(L) for retrieval, with overall space proportional to the total number of characters across all distinct prefixes, i.e., O(total characters).
Interview Questions on This Problem
Q1How would you modify a standard Trie to support "lexicographically smallest prefix of length L" queries in O(L) time?
Add a boolean (or count) flag to each node indicating whether any inserted string passes through it, and store children in an array indexed by character so they are naturally ordered. During a query, start at the root and at each depth choose the smallest indexed child whose flag is true, building the prefix until length L is reached. If no such child exists at any step, return an empty string.
Q2Why does a hash‑set based solution become O(N·L) for each query, and how does the Trie avoid this cost?
A hash set can only answer exact‑match lookups; to find the smallest prefix of length L you would have to generate all possible prefixes of length L from every stored string (O(N·L) total) and compare them. A Trie encodes all prefixes implicitly in its structure, so the query walks a single path of length L, inspecting only the relevant nodes, yielding O(L) regardless of N.
Q3In a distributed system where strings are sharded across multiple nodes, how could you maintain the "smallest prefix" property efficiently?
Each shard maintains its own local Trie with the same augmentations. For a global query, each node returns its local smallest prefix of length L (or a sentinel if none). A lightweight coordinator then picks the overall smallest among the returned candidates. Because each local query is O(L) and the merge step is O(k) where k is the number of shards, the total cost remains near‑optimal.
Examples
Input
repository = { 'app', 'apple', 'application' }, length = 3Output
app
Explanation: Step 1: We start by iterating over each string in the repository. We find the first string 'app' that has a length of 3 or more. Step 2: We then return the lexicographically smallest prefix of 'app' of length 3, which is 'app'.
Input
repository = { 'app', 'apple', 'application' }, length = 4Output
Explanation: Step 1: We start by iterating over each string in the repository. We find that there is no string with a length of 4 or more. Step 2: We then return an empty string because the problem statement asks for the lexicographically smallest prefix of length 4, but there is no such prefix.
Constraints
- The total number of strings added to the repository will not exceed 10^5.
- The total length of all strings added will not exceed 10^6 characters.
- Each retrieval operation should be performed in reasonable time complexity, e.g., O(n) or better where n is the length of the strings or the given length.
- The repository is initially empty.
Optimal Approach & Strategy
Insert strings into a Trie with a reachability flag at each node; during a query, greedily traverse the smallest marked child at each depth up to L, building the answer in O(L) time.
Brute Force Approach
Store all strings in a hash set and, for each query, generate every possible prefix of length L from every string, then pick the smallest lexicographically.
Code Solutions
class TrieNode {
constructor() {
this.children = new Map();
this.isEndOfWord = false;
}
}
class Trie {
constructor() {
this.root = new TrieNode();
}
addWord(word) {
let node = this.root;
for (let char of word) {
if (!node.children.has(char)) {
node.children.set(char, new TrieNode());
}
node = node.children.get(char);
}
node.isEndOfWord = true;
}
getSmallestPrefix(length) {
let node = this.root;
let prefix = '';
for (let i = 0; i < length; i++) {
if (node.children.size === 0) {
return prefix;
}
let chars = Array.from(node.children.keys()).sort();
prefix += chars[0];
node = node.children.get(chars[0]);
}
return prefix;
}
}
function solution(words, length) {
let trie = new Trie();
for (let word of words) {
trie.addWord(word);
}
return trie.getSmallestPrefix(length);
}class TrieNode {
public:
std::map<char, TrieNode*> children;
bool isEndOfWord;
TrieNode() : isEndOfWord(false) {}
};
class Trie {
public:
TrieNode* root;
Trie() : root(new TrieNode()) {}
void addWord(const std::string& word) {
TrieNode* node = root;
for (char c : word) {
if (node->children.find(c) == node->children.end()) {
node->children[c] = new TrieNode();
}
node = node->children[c];
}
node->isEndOfWord = true;
}
std::string getSmallestPrefix(int length) {
TrieNode* node = root;
std::string prefix;
for (int i = 0; i < length; i++) {
if (node->children.empty()) {
return prefix;
}
std::vector<char> chars(node->children.begin()->first, node->children.end()->first);
std::sort(chars.begin(), chars.end());
prefix += chars[0];
node = node->children[chars[0]];
}
return prefix;
}
};
class Solution {
public:
std::string solution(const std::string words[], int length) {
Trie trie;
for (const std::string& word : words) {
trie.addWord(word);
}
return trie.getSmallestPrefix(length);
}
};class TrieNode {
Map<Character, TrieNode> children;
boolean isEndOfWord;
public TrieNode() {
children = new HashMap<>();
isEndOfWord = false;
}
}
class Trie {
TrieNode root;
public Trie() {
root = new TrieNode();
}
public void addWord(String word) {
TrieNode node = root;
for (char c : word.toCharArray()) {
if (!node.children.containsKey(c)) {
node.children.put(c, new TrieNode());
}
node = node.children.get(c);
}
node.isEndOfWord = true;
}
public String getSmallestPrefix(int length) {
TrieNode node = root;
StringBuilder prefix = new StringBuilder();
for (int i = 0; i < length; i++) {
if (node.children.isEmpty()) {
return prefix.toString();
}
char[] chars = node.children.keySet().toArray(new Character[0]);
Arrays.sort(chars);
prefix.append(chars[0]);
node = node.children.get(chars[0]);
}
return prefix.toString();
}
}
public class Solution {
public String solution(String[] words, int length) {
Trie trie = new Trie();
for (String word : words) {
trie.addWord(word);
}
return trie.getSmallestPrefix(length);
}
}class TrieNode:
def __init__(self):
self.children = {}
self.isEndOfWord = False
class Trie:
def __init__(self):
self.root = TrieNode()
def addWord(self, word):
node = self.root
for char in word:
if char not in node.children:
node.children[char] = TrieNode()
node = node.children[char]
node.isEndOfWord = True
def getSmallestPrefix(self, length):
node = self.root
prefix = ''
for i in range(length):
if not node.children:
return prefix
chars = sorted(node.children.keys())
prefix += chars[0]
node = node.children[chars[0]]
return prefix
def solution(words, length):
trie = Trie()
for word in words:
trie.addWord(word)
return trie.getSmallestPrefix(length)class TrieNode {
constructor() {
this.children = new Map();
this.isEndOfWord = false;
}
}
class Trie {
constructor() {
this.root = new TrieNode();
}
addWord(word) {
let node = this.root;
for (let char of word) {
if (!node.children.has(char)) {
node.children.set(char, new TrieNode());
}
node = node.children.get(char);
}
node.isEndOfWord = true;
}
getSmallestPrefix(length) {
let node = this.root;
let prefix = '';
for (let i = 0; i < length; i++) {
if (node.children.size === 0) {
return prefix;
}
let chars = Array.from(node.children.keys()).sort();
prefix += chars[0];
node = node.children.get(chars[0]);
}
return prefix;
}
}
function solution(words, length) {
let trie = new Trie();
for (let word of words) {
trie.addWord(word);
}
return trie.getSmallestPrefix(length);
}Solve in Interative Editor
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