Problem Description

The problem presents a scenario where we have n survey questions answered with either 0 or 1, by m students and m mentors. Each student and each mentor have provided their answers, and these answers are represented by two 2D integer arrays: students and mentors. Every student is to be paired with exactly one mentor, and vice versa.

A compatibility score is defined for each student-mentor pair, which is just the count of the number of identical answers they have given. The goal is to find the pairing between students and mentors that maximizes the total sum of all compatibility scores.

Flowchart Walkthrough

Let's analyze the problem using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: The problem does not involve graph theory where connections or paths between nodes are crucial.

Need to solve for kth smallest/largest?

• No: The issue is not about identifying a specific order or rank-related element in a dataset.

Involves Linked Lists?

• No: The problem does not involve manipulating or navigating through linked lists.

Does the problem have small constraints?

• Yes: The problem constraints indicate that the maximum number of mentors or students is small enough that it's computationally feasible to consider many combinations or permutations.

Brute force / Backtracking?

• Yes: Given the small constraints and the need to explore various ways to pair students and mentors to maximize the compatibility score, brute force or backtracking is suitable to try all possibilities and select the maximum.

Conclusion: The flowchart suggests using a backtracking approach for enumerating and evaluating all possible pairings of students and mentors to find the combination with the maximum compatibility score sum.

Intuition

To solve this problem, we can think in terms of permutations and combinations because we want to find the maximally compatible pairing. The approach is to try, for each student, every possible mentor that has not been paired yet and calculate the compatibility score. By exploring all possible pairings, we can ensure that no compatibility score is left unchecked.

An efficient way to explore all the possibilities without repetition is to use a depth-first search (DFS) approach. We traverse the possible pairings in a recursive manner, effectively creating a search tree where each level corresponds to a student and each branch from a node represents trying out a mentor for that student.

Here are the steps we follow in the solution:

1. Create a matrix g where g[i][j] stores the compatibility score between student i and mentor j, which we calculate by comparing their answers.

2. Keep a visited array vis to track which mentors have been paired already.

3. Perform DFS starting from the first student (index 0), and for each student, try pairing them with each unvisited mentor. Update the running total of the score and mark the mentor as visited.

4. After trying out a mentor for a student, we backtrack by marking this mentor as unvisited and trying another unvisited mentor for the same student.

5. Whenever we reach a pairing where all students are paired (base case of the DFS where i == m), we compare the total score with the current maximum score ans and update ans if the new score is higher.

6. After all possibilities have been explored, ans holds the maximum compatibility score sum, which is the final answer we return.

Learn more about Dynamic Programming, Backtracking and Bitmask patterns.

Solution Approach

Algorithm and Patterns

The solution uses a Depth-First Search (DFS) recursive algorithm to explore all potential student-mentor pairings.

Data Structures

• A 2D array g that holds the compatibility scores between all students and mentors.
• A vis array acting as a boolean visited tracker, indicating whether a mentor has already been paired.

Pattern Used

The pattern used in this problem is backtracking, where we try different options, and as soon as we determine that a current path won't lead us to a solution, or we find a solution, we go back and try other options.

Solution Implementation

Let's break down the given Python function, maxCompatibilitySum, into steps:

1. Initialization:

   • The function calculates the length m of the students array, which will be used for iterations.
   • It initializes a 2D list g with dimensions m x m where g[i][j] will store the compatibility score of the ith student with the jth mentor.
   • It calculates these scores using a nested for loop and a generator expression with sum(a == b for a, b in zip(students[i], mentors[j])), which sums the number of answers that are identical for student i and mentor j.
   • A list vis is created with m boolean values set to False, representing that initially no mentor has been paired with a student.
   • A variable ans is initialized to 0, which will eventually store the maximum compatibility score sum.

2. Defining DFS function: The dfs function takes two parameters:

   • Index i that represents the current student being considered.
   • A temporary score t which is the running total of the compatibility scores.

   In this function:

   • The base case checks if i == m, meaning all students have been assigned mentors. It then updates ans if the total compatibility score t is greater than the current ans.
   • The recursive case iterates over all mentors using a loop. For each mentor j that has not yet been visited (vis[j] is False), the function:
     • Sets vis[j] to True to mark the mentor as paired.
     • Recursively calls dfs with the next student i + 1 and the updated score t + g[i][j].
     • Resets vis[j] to False to backtrack and allow the mentor to be paired with another student in subsequent recursion calls.

3. DFS Traversal:

   • After initializing all data structures and defining the dfs function, the function dfs(0, 0) is called to begin the DFS traversal from the first student and with an initial score of 0.

4. Return Result:

   • Once all possible pairings have been explored, the function returns ans, which now contains the maximum compatibility score sum that can be achieved.

The solution effectively pairs each student with each possible mentor, backtracks, and tries different pairings, keeping track of the maximum sum found along the way. By searching in this manner, it ensures that the optimal pairing is found, thus maximizing the compatibility score sum.

Example Walkthrough

Let's consider a case where we have n = 3 survey questions answered by m = 2 students and m = 2 mentors:

The students array is [[1, 0, 1], [0, 1, 0]], and the mentors array is [[0, 0, 1], [1, 1, 0]].

Here is the compatibility matrix g:

• The compatibility score between student 1 and mentor 1 is 1 (they have one answer in common, the last question).
• The compatibility score between student 1 and mentor 2 is 2 (they have two answers in common, the first and second question).
• The compatibility score between student 2 and mentor 1 is 2 (they have two answers in common, the first and second question).
• The compatibility score between student 2 and mentor 2 is 1 (they have one answer in common, the last question).

Thus, g will look like this: [[1, 2], [2, 1]]

Now, we initialize our visited array vis to [False, False] and set ans to 0.

We apply DFS to calculate the maximum compatibility score:

1. We start with student 1 (index 0) and try to pair them with each unassigned mentor:
   • If we pair them with mentor 1, we get a score of 1, vis is updated to [True, False].
   • We then move to student 2 (index 1) and since mentor 1 is already paired, we pair student 2 with mentor 2 and get a score of 1. Total score is now 2. We then backtrack since all students are paired.
   • We reset vis to [False, False] and try to pair student 1 with the next mentor.
2. Now, we pair student 1 with mentor 2, so vis is updated to [False, True], and score is 2.
   • For student 2, the only available mentor is mentor 1, yielding a score of 2. Total score is now 4 which is greater than the previous total. We update ans to 4.

Since we've explored all possible pairings, the final ans is 4, which is the maximum compatibility score sum we can achieve with these pairings.

Thus, the best pairing is student 1 with mentor 2 and student 2 with mentor 1, with each pair having a compatibility score of 2.

The solution methodology ensured every possible pairing was considered, and the best matching was found by exploring all options recursively and backtracking as necessary to try different configurations.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the provided code can be analyzed by examining the following parts:

1. The compatibility matrix computation (g): For each pair of student and mentor, we calculate the compatibility score which takes O(n) time where n is the number of questions, since we are comparing their answers question-by-question. As there are m students and m mentors, we perform this operation m^2 times. Thus, the time complexity for creating this matrix is O(m^2 * n).

2. The depth-first search (dfs) with backtracking: The dfs function explores all possible pairings between students and mentors. In the worst case, it has to explore m! (factorial of m) different configurations (each student can be matched with any of the m mentors, the next student with any of the m-1 remaining mentors, and so forth). For each recursive call, we also calculate the total compatibility score t + g[i][j], which is done in O(1) time. Therefore, the total time complexity for this part is O(m!).

Taking both parts into account, the overall time complexity is O(m^2 * n + m!).

Space Complexity

The space complexity of the code is dominated by the following components:

1. The compatibility matrix (g): This matrix is of size m x m, resulting in O(m^2) space.

2. The visited array (vis): An array keeping track of which mentors have been visited. This requires O(m) space.

3. The recursion call stack for dfs: In the worst case, the depth of the recursion is m (matching each student with a mentor), therefore, the recursive call stack will require O(m) space.

Since the space for the compatibility matrix is the largest component, the overall space complexity is O(m^2).

Learn more about how to find time and space complexity quickly using problem constraints.