How do you solve Maximum Compatibility Score Sum on LeetCode?

Maximum Compatibility Score Sum (LeetCode #1947) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m!) time complexity and O(m) space complexity. Key insight: Understanding bit manipulation can significantly optimize the solution.

What is the brute force solution for Maximum Compatibility Score Sum?

The brute force approach for Maximum Compatibility Score Sum is Brute Force, with O(n!) time complexity and O(n) space complexity. We can try every possible pairing of students and mentors to calculate the compatibility scores. This method is straightforward but inefficient as it checks all combinations. The optimal Optimal Solution approach improves this to O(n * m!).

What are the interview tips for Maximum Compatibility Score Sum?

Explain your thought process clearly while coding. Consider edge cases, such as when there are no students or mentors. Practice writing clean and efficient code.

What are common mistakes when solving Maximum Compatibility Score Sum?

Forgetting to check if a mentor has already been used. Not properly calculating the compatibility score.

#1947

Maximum Compatibility Score Sum

Medium

ArrayDynamic ProgrammingBacktrackingBit ManipulationBitmaskBacktrackingBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n * m!)
Space	O(n)	O(m)

💡

Intuition

Time O(n * m!)Space O(m)

We can use backtracking with bit manipulation to efficiently explore the assignment of students to mentors without generating all permutations. This reduces the time complexity significantly.

⚙️

Algorithm

3 steps

1Step 1: Create a function to calculate the compatibility score between a student and a mentor.
2Step 2: Use a backtracking approach to assign students to mentors, keeping track of which mentors have been used using a bitmask.
3Step 3: Recursively calculate the maximum score by trying to assign each student to each available mentor.

solution.py15 lines

1def maxCompatibilitySum(students, mentors):
2    def compatibility(student, mentor):
3        return sum(s == m for s, m in zip(student, mentor))
4
5    def backtrack(student_index, used_mask):
6        if student_index == len(students):
7            return 0
8        max_score = 0
9        for mentor_index in range(len(mentors)):
10            if not (used_mask & (1 << mentor_index)):
11                score = compatibility(students[student_index], mentors[mentor_index])
12                max_score = max(max_score, score + backtrack(student_index + 1, used_mask | (1 << mentor_index)))
13        return max_score
14
15    return backtrack(0, 0)

ℹ

Complexity note: The time complexity is O(n * m!) because we explore all possible assignments of students to mentors, where n is the number of students and m is the number of mentors. The space complexity is O(m) due to the bitmask used to track mentor assignments.

1Understanding bit manipulation can significantly optimize the solution.
2Backtracking helps explore all combinations without generating all permutations explicitly.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.