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)
View More Similar Problems
Print the Elements of a Linked List
This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode
View Solution →Insert a Node at the Tail of a Linked List
You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink
View Solution →Insert a Node at the head of a Linked List
Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below
View Solution →Insert a node at a specific position in a linked list
Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e
View Solution →Delete a Node
Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo
View Solution →Print in Reverse
Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing
View Solution →