269.cpp

/*
269. Alien Dictionary
Hard

1765

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There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of non-empty words from the dictionary, where words are sorted lexicographically by the rules of this new language. Derive the order of letters in this language.

Example 1:

Input:
[
  "wrt",
  "wrf",
  "er",
  "ett",
  "rftt"
]

Output: "wertf"
Example 2:

Input:
[
  "z",
  "x"
]

Output: "zx"
Example 3:

Input:
[
  "z",
  "x",
  "z"
] 

Output: "" 

Explanation: The order is invalid, so return "".
Note:

You may assume all letters are in lowercase.
If the order is invalid, return an empty string.
There may be multiple valid order of letters, return any one of them is fine.
*/


class Solution {
public:
    string alienOrder(vector<string>& alien) {
        unordered_set<char> ust;
        for(auto& word: alien)
            for(auto& ch: word)
                ust.insert(ch);
        int n = alien.size();
        vector<int> indegree(26,0);
        vector<unordered_set<int>> graph(26);
        for(int i=0; i<n-1; i++){
            auto prev = alien[i], curr = alien[i+1];
            for(int j=0; j<min(prev.size(), curr.size()); j++){
                if(prev.size()>curr.size() && prev.substr(0,curr.size())==curr)
                    return "";// case when abc->ab which is not valid case as orderly ab->abc
                if(prev[j]!=curr[j]){// only considering first distinctive charachters
                    if(graph[prev[j]-'a'].insert(curr[j]-'a').second)//if not present previously
                        indegree[curr[j]-'a']++;
                    break;
                }
            }
        }
        
        // indegree topsort bfs;
        queue<int> q;
        for(auto ch: ust){
            if(indegree[ch-'a']==0){
                q.push(ch-'a');
                indegree[ch-'a']--;
            }
        }
        
        string res = "";
        while(!q.empty()){
            auto curr =q.front();q.pop();
            res += (curr+'a');
            for(auto v: graph[curr]){
                indegree[v]--;
                if(indegree[v]==0){
                    q.push(v);
                    indegree[v]--;
                }
            }
        }
        if(res.size()==ust.size())//check is size of res = total number of charachters
            return res;
        return "";
    }
};