Polyglot Contest - Amazon Top Interview Questions


Problem Statement :


You are given a two-dimensional list of strings languages, where languages[i] is a list of programming languages person i is fluent in.

Consider any list of programming languages such that everyone knows at least one language in it. Return the minimum size of such list.

Constraints

1 ≤ n, m ≤ 16 where n and m are the number of rows and columns in languages.
1 ≤ l ≤ 32 where l is the total number of distinct strings in languages.

Example 1

Input

languages = [
    ["Java", "Perl"],
    ["C++", "Python"],
    ["Haskell"]
]

Output

3

Explanation

There is no overlap between the languages. Therefore any combination that uses one language from each participant is valid.

Example 2

Input

languages = [
    ["Java", "C++", "Python"],
    ["Python", "Cobol", "Java"],
    ["C++", "Haskell"],
    ["Ruby", "C++"]
]

Output

2

Explanation

Valid combinations are ["Cobol", "C++"], ["Java", "C++"] and ["Python", "C++"].

Example 3

Input

languages = [
    ["C", "Python", "Haskell", "Kotlin"],
    ["Java", "JavaScript", "C++", "Rust"],
    ["JavaScript", "Python", "C++"],
    ["Ruby", "C++"],
    ["Rust", "Python", "Java"]
]

Output

2

Explanation

The only minimal combination is ["Python", "C++"].



Solution :



title-img




                        Solution in C++ :

int solve(vector<vector<string>>& languages) {
    int persons_count = languages.size();

    vector<int> dp(1 << persons_count, INT32_MAX);

    unordered_map<string, int> language_masks;
    for (int i = 0; i < languages.size(); ++i) {
        for (const auto& language : languages[i]) {
            language_masks[language] |= (1 << i);
        }
    }

    dp[0] = 0;
    for (int mask = 1; mask < (1 << persons_count); ++mask) {
        int bit = __builtin_ctz(mask);

        for (auto& l : languages[bit]) {
            int new_mask = mask ^ (language_masks[l] & mask);
            dp[mask] = min(dp[mask], 1 + dp[new_mask]);
        }
    }

    return dp[(1 << persons_count) - 1];
}
                    




                        Solution in Python : 
                            
class Solution:
    def solve(self, languages):
        language_set = functools.reduce(set.union, map(set, languages))
        language_index = {language: index for index, language in enumerate(language_set)}
        language_masks = [
            sum((1 << language_index[language]) for language in participant_languages)
            for participant_languages in languages
        ]
        n = len(languages)
        l = len(language_set)

        @functools.lru_cache(None)
        def dp(language_index, participant_mask):
            if not participant_mask:
                return 0
            if language_index == l:
                return math.inf

            new_participant_mask = participant_mask
            for participant_index in range(n):
                if new_participant_mask & (1 << participant_index) and language_masks[
                    participant_index
                ] & (1 << language_index):
                    new_participant_mask ^= 1 << participant_index

            return min(
                1 + dp(language_index + 1, new_participant_mask),
                dp(language_index + 1, participant_mask),
            )

        return dp(0, (1 << n) - 1)
                    


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