Longest Prefix Sequence - Google Top Interview Questions


Problem Statement :


You are given a list of lowercase alphabet strings words. 

Return the length of the longest sequence of words where each previous word is the prefix of the next word and the next word has just one new character appended. 

You can rearrange words in any order.

Constraints

0 ≤ n ≤ 100,000 where n is the length of words

1 ≤ m ≤ 100,000 where m is the length of the longest string in words

Example 1

Input

words = ["abc", "ab", "x", "xy", "abcd"]

Output

3

Explanation

We can form the following sequence: ["ab", "abc", "abcd"].


Solution :



title-img 




                        Solution in C++ :

int solve(vector<string>& words) {
    sort(words.begin(), words.end());
    map<string, int> mp;
    int ret = 1;
    for (auto& word : words) {
        string tmp = word;
        tmp.pop_back();
        if (mp.count(tmp))
            mp[word] = mp[tmp] + 1;
        else
            mp[word] = 1;
        ret = max(ret, mp[word]);
    }
    return ret;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(String[] words) {
        Arrays.sort(words);
        Map<Integer, Integer> map = new HashMap<>();

        int res = 0;
        for (String w : words) {
            int prefixes = 0;

            if (w.length() > 1) {
                int prevHash = (w.length() - 1) * 100000 + w.charAt(w.length() - 2);
                if (map.containsKey(prevHash))
                    prefixes += map.get(prevHash);
            }

            int hash = w.length() * 100000 + w.charAt(w.length() - 1);
            map.put(hash, prefixes + 1);
            res = Math.max(res, map.get(hash));
        }

        return res;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, words):
        graph = {word: [] for word in words}

        for word in words:
            prev_word = word[: len(word) - 1]
            if prev_word in graph:
                graph[prev_word].append(word)

        @cache
        def dp(word):
            return 1 + max((dp(successor) for successor in graph[word]), default=0)

        return max((dp(word) for word in words), default=0)


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