Optimized Prefix Repository — Problem Statement & Solution Guide
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.
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
An optimized approach involves using a Trie data structure to store the strings and then performing a depth-first search to find the lexicographically smallest prefix of a given length. This approach allows for efficient storage and retrieval of prefixes. The time complexity for adding a string and retrieving a prefix can be significantly improved.
Brute Force Approach
A brute force approach would involve storing all the strings in an array and then, for each retrieval operation, iterating over all strings to find the lexicographically smallest prefix of the given length. This approach would be inefficient and not scalable for large inputs. The time complexity would be high due to the nested loops. A more optimized approach is needed for better performance.
Verified 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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