# 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"],
]

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"],
["Ruby", "C++"]
]

Output

2

Explanation

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

Example 3

Input

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

Output

2

Explanation

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

### Solution :

```                        ```Solution in C++ :

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

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

for (int i = 0; i < languages.size(); ++i) {
for (const auto& language : languages[i]) {
}
}

dp[0] = 0;

for (auto& l : languages[bit]) {
}
}

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)}
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)
return 0
if language_index == l:
return math.inf

for participant_index in range(n):
participant_index
] & (1 << language_index):

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

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

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