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 :
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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