Maximum Compatibility Score Sum

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There is a survey that consists of n questions where each question's answer is either 0 (no) or 1 (yes).

The survey was given to m students numbered from 0 to m - 1 and m mentors numbered from 0 to m - 1. The answers of the students are represented by a 2D integer array students where students[i] is an integer array that contains the answers of the i^th student (0-indexed). The answers of the mentors are represented by a 2D integer array mentors where mentors[j] is an integer array that contains the answers of the j^th mentor (0-indexed).

Each student will be assigned to one mentor, and each mentor will have one student assigned to them. The compatibility score of a student-mentor pair is the number of answers that are the same for both the student and the mentor.

For example, if the student's answers were [1, 0, 1] and the mentor's answers were [0, 0, 1], then their compatibility score is 2 because only the second and the third answers are the same.

You are tasked with finding the optimal student-mentor pairings to maximize the sum of the compatibility scores.

Given students and mentors, return the maximum compatibility score sum that can be achieved.

Example 1:

Input: students = [[1,1,0],[1,0,1],[0,0,1]], mentors = [[1,0,0],[0,0,1],[1,1,0]]
Output: 8
Explanation: We assign students to mentors in the following way:
- student 0 to mentor 2 with a compatibility score of 3.
- student 1 to mentor 0 with a compatibility score of 2.
- student 2 to mentor 1 with a compatibility score of 3.
The compatibility score sum is 3 + 2 + 3 = 8.

Example 2:

Input: students = [[0,0],[0,0],[0,0]], mentors = [[1,1],[1,1],[1,1]]
Output: 0
Explanation: The compatibility score of any student-mentor pair is 0.

Constraints:

m == students.length == mentors.length
n == students[i].length == mentors[j].length
1 <= m, n <= 8
students[i][k] is either 0 or 1.
mentors[j][k] is either 0 or 1.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the size constraints for the number of students and mentors (i.e., the dimensions of the input arrays)?
Can the compatibility scores within the student and mentor preference matrices be negative, zero, or only positive?
Are the preference lists guaranteed to be complete (i.e., each student ranks all mentors, and vice-versa)?
If there are multiple assignments that yield the maximum compatibility score sum, is any one of them acceptable?
Is there a guarantee that the number of students will always be equal to the number of mentors?

Brute Force Solution

Approach

The brute force approach to this problem involves considering every single possible matching between the two groups of items. We calculate the 'score' for each of these matchings, and then select the one that gives us the highest total score. It's like trying every combination to find the best one.

Here's how the algorithm would work step-by-step:

Start by taking the first item from the first group.
Consider matching it with each item in the second group, one at a time.
For each of these matches, calculate the compatibility score.
Now, for each of these first-item matches, take the second item from the first group and repeat the process.
Keep matching and scoring, ensuring that no item from the second group is used more than once.
Continue this process until every item from the first group has been matched with an item from the second group.
Keep track of the total compatibility score for each complete matching.
Finally, compare all the total compatibility scores, and select the highest one. This is the maximum compatibility score sum.

Code Implementation

def max_compatibility_score_sum(students, mentors):
    max_total_compatibility = 0

    def calculate_compatibility(student_index, mentor_matches, current_compatibility):
        nonlocal max_total_compatibility

        if student_index == len(students):
            # All students have been matched.
            max_total_compatibility = max(max_total_compatibility, current_compatibility)
            return

        for mentor_index in range(len(mentors)):
            if mentor_index not in mentor_matches:

                # Try matching the current student with this mentor.
                new_mentor_matches = mentor_matches | {mentor_index}

                compatibility_score = sum(
                    students[student_index][i] == mentors[mentor_index][i]
                    for i in range(len(students[0]))
                )

                # Recurse to match the next student.

                calculate_compatibility(
                    student_index + 1,
                    new_mentor_matches,
                    current_compatibility + compatibility_score,
                )

    # Start the recursion with the first student and an empty set of matches.
    calculate_compatibility(0, set(), 0)

    return max_total_compatibility

Big(O) Analysis

Time Complexity

O(n!) – The algorithm explores all possible permutations of matching n students to n mentors. For the first student, there are n possible mentors to match with. For the second student, there are (n-1) remaining mentors, and so on. This leads to n * (n-1) * (n-2) * ... * 1, which is n! permutations. Calculating the score for each permutation takes O(n) time, but the dominating factor is the generation of permutations. Therefore, the time complexity is O(n!).

Space Complexity

O(N) – The brute force approach, as described, explores all possible matchings. This can be implemented using recursion. The maximum depth of the recursion is equal to the number of items in the first group, let's denote it as N. At each level of recursion, we are essentially storing the current matching state and function call context on the call stack. Therefore, the auxiliary space used by the recursion stack can grow up to N levels deep, resulting in a space complexity of O(N).

Optimal Solution

Approach

The key to maximizing the compatibility score is realizing that we need to explore all possible pairings between students and mentors. Since each student must be assigned to exactly one mentor and vice versa, we can think of this as finding the best possible matching. The optimal solution involves cleverly exploring the different ways we can match students to mentors, ensuring we don't miss the highest possible compatibility score.

Here's how the algorithm would work step-by-step:

Start by considering the first student. We need to try assigning them to each of the available mentors and calculating the compatibility score for each of those assignments.
For each possible assignment of the first student, temporarily 'remove' that mentor from the list of available mentors.
Now, move on to the second student and repeat the process: try assigning them to each of the *remaining* available mentors and calculate the compatibility score for each of those assignments.
Keep doing this for each student, one at a time, always considering only the mentors that haven't already been assigned.
Each time you reach the last student, you've created a complete matching of students to mentors. Calculate the total compatibility score for that matching.
Keep track of the highest total compatibility score you've found so far.
After you've explored all possible matchings (by trying all assignments for each student), the highest total compatibility score you kept track of is the answer.

Code Implementation

def max_compatibility_score_sum(students, mentors):
    number_of_students = len(students)
    number_of_mentors = len(mentors)
    maximum_score = 0

    def calculate_compatibility_score(student, mentor):
        score = 0
        for i in range(len(student)): 
            if student[i] == mentor[i]:
                score += 1
        return score

    def find_max_score(student_index, assigned_mentors, current_score):
        nonlocal maximum_score

        # Base case: all students have been assigned.
        if student_index == number_of_students:
            maximum_score = max(maximum_score, current_score)
            return

        # Iterate through all mentors.
        for mentor_index in range(number_of_mentors):
            # Only consider mentors that haven't been assigned yet.
            if mentor_index not in assigned_mentors:
                # Calculate compatibility score.
                compatibility_score = calculate_compatibility_score(students[student_index], mentors[mentor_index])

                # Recursively explore the next student, marking the mentor as assigned.
                find_max_score(
                    student_index + 1,
                    assigned_mentors | {mentor_index},
                    current_score + compatibility_score,
                )

    # Start the recursive process with the first student and no assigned mentors.
    # The bitwise or operation is used for constant time lookup
    find_max_score(0, set(), 0)

    # Return the maximum compatibility score found.
    return maximum_score

Big(O) Analysis

Time Complexity

O(n!) – The algorithm explores all possible matchings between n students and n mentors. For the first student, there are n choices of mentors. For the second student, there are n-1 choices, and so on. This leads to n * (n-1) * (n-2) * ... * 1, which is n! (n factorial) possible combinations. Calculating the compatibility score for each combination takes O(n*m) where m is the length of the scores array rows, but the dominant factor is the number of permutations to explore. Thus, the time complexity is O(n!).

Space Complexity

O(N) – The algorithm uses recursion to explore all possible pairings. Each recursive call corresponds to assigning a student to a mentor. In the worst-case scenario, where we explore all possible assignments, the maximum depth of the recursion will be equal to the number of students, which we can denote as N. Each level of the recursion stack stores the current assignment and available mentors. Therefore, the space complexity is determined by the maximum depth of the recursion stack, resulting in O(N) space complexity, where N is the number of students. No other significant auxiliary data structures are utilized.

Edge Cases

Case	How to Handle
Empty students or mentors array	Return 0 immediately as there are no pairings possible.
Students and mentors arrays of different lengths	The problem statement implicitly states they are of equal length; if not, throw an IllegalArgumentException or return -1 to indicate invalid input.
Single student and single mentor	Calculate and return the compatibility score directly.
Maximum array size (n = m = 10)	Ensure the chosen algorithm (e.g., backtracking) has acceptable time complexity for n=10, avoiding timeouts.
All answers are the same for all student/mentor pairs	The algorithm should still produce a valid permutation and corresponding score, even with uniform data.
One student is perfectly compatible with all mentors, and the rest are incompatible.	The algorithm should prioritize pairing the perfectly compatible student with the best-suited mentor from that group.
Very large compatibility scores could lead to integer overflow.	Use a data type with sufficient range (e.g., long) to store the compatibility sum, or check for potential overflow during calculations.
Multiple optimal pairings exist with the same maximum compatibility score.	The algorithm only needs to find one of the optimal pairings, and the existence of multiple solutions is irrelevant to the correctness.

Empty students or mentors array

How to Handle:

Return 0 immediately as there are no pairings possible.

Students and mentors arrays of different lengths

How to Handle:

The problem statement implicitly states they are of equal length; if not, throw an IllegalArgumentException or return -1 to indicate invalid input.

Single student and single mentor

How to Handle:

Calculate and return the compatibility score directly.

Maximum array size (n = m = 10)

How to Handle:

Ensure the chosen algorithm (e.g., backtracking) has acceptable time complexity for n=10, avoiding timeouts.

All answers are the same for all student/mentor pairs

How to Handle:

The algorithm should still produce a valid permutation and corresponding score, even with uniform data.

One student is perfectly compatible with all mentors, and the rest are incompatible.

How to Handle:

The algorithm should prioritize pairing the perfectly compatible student with the best-suited mentor from that group.

Very large compatibility scores could lead to integer overflow.

How to Handle:

Use a data type with sufficient range (e.g., long) to store the compatibility sum, or check for potential overflow during calculations.

Multiple optimal pairings exist with the same maximum compatibility score.

How to Handle:

The algorithm only needs to find one of the optimal pairings, and the existence of multiple solutions is irrelevant to the correctness.