Ancient Astronaut Theory- Amazon Top Interview Questions


Problem Statement :


You are given a string dictionary, representing a partial lexicographic ordering of ancient astronauts' dictionary. Given a string s, return whether it's a lexicographically sorted string according to this ancient astronaut dictionary.

Example 1

Input

dictionary = "acb"

s = "aaaa h ccc i bbb"

Output

True

Explanation

The only constraint is that a comes before c which comes before b .

Example 2

Input

dictionary = "acb"

s = "aaaacccbc"

Output

False

Explanation

This is false because of the last c, which comes after b.


Solution :



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                        Solution in C++ :

bool isValid(char x, vector<int>& dict) {
    return (x - 'a') >= 0 && (x - 'a') <= 25 && (dict[x - 'a'] != -1);
}

bool solve(string dictionary, string s) {
    vector<int> dict(26, -1);
    for (int i = 0; i < dictionary.size(); ++i) dict[dictionary[i] - 'a'] = i;
    int n = s.size();
    int last = -1;
    for (int i = 0; i < n; ++i) {
        if (isValid(s[i], dict)) {
            if (dict[s[i] - 'a'] < last) return false;
            last = dict[s[i] - 'a'];
        }
    }
    return true;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public boolean solve(String dictionary, String s) {
        Map<Character, Integer> map = new HashMap();
        int dLen = dictionary.length();
        for (int i = 0; i < dLen; i++) {
            if (map.containsKey(dictionary.charAt(i))) {
                continue;
            }
            map.put(dictionary.charAt(i), i);
        }
        int index = 0, sLen = s.length(), minVal = Integer.MIN_VALUE;
        for (int i = 0; i < sLen; i++) {
            if (!map.containsKey(s.charAt(i))) {
                continue;
            }
            if (map.containsKey(s.charAt(i))) {
                index = map.get(s.charAt(i));
            }
            if (index < minVal) {
                return false;
            }
            minVal = Math.max(index, minVal);
        }
        return true;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, dictionary, s):
        dic = {}
        for i in range(len(dictionary)):
            dic[dictionary[i]] = i
        prev = 0
        for i in s:
            if i in dictionary:
                ind = dic[i]
                if ind < prev:
                    return False
                prev = ind
        return True


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