Problem Description

You are given a survey with n questions, where each question can be answered with either 0 (no) or 1 (yes).

The survey was completed by m students (numbered from 0 to m-1) and m mentors (also numbered from 0 to m-1). Their answers are stored in two 2D arrays:

• students[i] contains the answers of the i-th student
• mentors[j] contains the answers of the j-th mentor

Your task is to pair each student with exactly one mentor (and each mentor with exactly one student) to maximize the total compatibility score.

The compatibility score between a student and a mentor is calculated as the number of questions they answered the same way. For instance, if a student answered [1, 0, 1] and a mentor answered [0, 0, 1], their compatibility score would be 2 (they both answered 0 for question 2 and 1 for question 3).

The goal is to find the optimal pairing arrangement that produces the maximum sum of all compatibility scores.

Example:

• Student answers: [1, 0, 1]
• Mentor answers: [0, 0, 1]
• Compatibility score: 2 (matching answers at positions 1 and 2)

Return the maximum possible sum of compatibility scores when optimally pairing all students with mentors.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: While we have students and mentors that need to be paired, this isn't a traditional graph problem with nodes and edges. It's an assignment/matching problem.

Need to solve for kth smallest/largest?

• No: We're looking for the maximum sum of compatibility scores, not the kth smallest or largest element.

Involves Linked Lists?

• No: The problem uses 2D arrays to store survey answers, not linked lists.

Does the problem have small constraints?

• Yes: The problem involves pairing m students with m mentors in a one-to-one mapping. Since we need to try all possible pairings to find the optimal solution, and the number of possible pairings is m! (factorial), the constraints must be small for this to be feasible. Typically, m would be at most 8-10 for such problems.

Brute force / Backtracking?

• Yes: We need to explore all possible ways to assign students to mentors. This is a perfect fit for backtracking - we assign a mentor to each student one by one, keep track of which mentors are already assigned, and backtrack when we need to try different assignments.

Conclusion: The flowchart correctly leads us to use a Backtracking approach. We need to systematically try all possible student-mentor pairings, keeping track of the maximum compatibility score sum found. The backtracking algorithm assigns mentors to students one by one, marking used mentors, and backtracks to explore alternative assignments until all possibilities are exhausted.

Intuition

The key insight is recognizing this as an assignment problem where we need to find the optimal one-to-one matching between two equal-sized groups. Since each student must be paired with exactly one mentor and vice versa, this is essentially a permutation problem - we're looking for the best permutation of mentors to assign to students.

Why backtracking? Consider what happens if we try to solve this greedily - always pairing each student with their most compatible available mentor. This might lead to suboptimal results because a greedy choice early on could prevent better overall pairings later. For example, if student 0 takes the best mentor for them, that mentor might have been the only good match for student 1, forcing student 1 into a poor pairing.

Since we need to explore all possible pairing combinations to guarantee finding the maximum sum, and the constraints are small (typically m ≤ 8), we can afford to try all m! possible assignments. Backtracking is perfect for this because:

1. We can systematically build up a complete assignment by assigning mentors to students one by one
2. We can track which mentors are already taken using a visited array
3. When we've assigned all students, we can check if this assignment gives us a better total score
4. We can easily backtrack by unmarking a mentor as visited and trying the next option

To optimize the backtracking process, we can precompute all compatibility scores between every student-mentor pair in a 2D grid g[i][j]. This way, during backtracking, we can quickly look up the compatibility score without recalculating it each time. The compatibility score is simply the count of matching answers: sum(a == b for a, b in zip(student_answers, mentor_answers)).

The recursive structure naturally emerges: for each student i, try pairing them with each available mentor j, add their compatibility score to the running sum, mark mentor j as used, and recursively solve for the next student. When all students are assigned, we've found one complete pairing and can update our maximum score.

Learn more about Dynamic Programming, Backtracking and Bitmask patterns.

Solution Approach

The solution implements a preprocessing step followed by backtracking to find the optimal pairing.

Step 1: Preprocessing Compatibility Scores

First, we create a 2D grid g[i][j] to store the compatibility score between each student i and mentor j. This avoids recalculating scores during backtracking:

g = [[0] * m for _ in range(m)]
for i, x in enumerate(students):
    for j, y in enumerate(mentors):
        g[i][j] = sum(a == b for a, b in zip(x, y))

For each student-mentor pair, we zip their answer arrays together and count how many answers match. This gives us an m × m matrix where g[i][j] represents the compatibility score if student i is paired with mentor j.

Step 2: Backtracking with DFS

We use a depth-first search function dfs(i, s) where:

• i represents the current student index we're trying to assign
• s represents the cumulative compatibility score so far

The backtracking algorithm works as follows:

1. Base Case: When i >= m, all students have been assigned. We update the global answer with the maximum of the current answer and the accumulated score s.

2. Recursive Case: For the current student i, we try pairing them with each mentor j from 0 to m-1:

   • Check if mentor j is available using the vis[j] array
   • If available, mark vis[j] = True to indicate this mentor is now assigned
   • Recursively call dfs(i + 1, s + g[i][j]) to assign the next student, adding the compatibility score g[i][j] to our running sum
   • After the recursive call returns, backtrack by setting vis[j] = False to try other pairings

Step 3: Tracking State

• vis: A boolean array of size m that tracks which mentors have been assigned. vis[j] = True means mentor j is already paired with a student.
• ans: A global variable (using nonlocal in Python) that maintains the maximum compatibility score sum found so far.

Step 4: Initialization and Execution

We initialize:

• ans = 0 to track the maximum score
• vis = [False] * m to mark all mentors as initially unassigned
• Call dfs(0, 0) to start assigning from student 0 with an initial score of 0

The algorithm explores all m! possible pairings through backtracking, guaranteeing we find the optimal solution. The time complexity is O(m! × m) for trying all permutations and calculating scores, while the space complexity is O(m²) for the compatibility grid plus O(m) for the recursion stack and visited array.

Example Walkthrough

Let's walk through a small example with 3 students and 3 mentors, each answering 4 questions:

Input:

• Students: [[1,1,0,0], [1,0,1,1], [0,0,1,0]]
• Mentors: [[1,0,0,0], [0,0,1,0], [1,1,0,1]]

Step 1: Build Compatibility Grid

First, we calculate the compatibility score between each student-mentor pair:

g[0][0]: Student 0 [1,1,0,0] vs Mentor 0 [1,0,0,0] → matches at positions 0,2,3 → score = 3
g[0][1]: Student 0 [1,1,0,0] vs Mentor 1 [0,0,1,0] → matches at position 3 → score = 1  
g[0][2]: Student 0 [1,1,0,0] vs Mentor 2 [1,1,0,1] → matches at positions 0,1,2 → score = 3

g[1][0]: Student 1 [1,0,1,1] vs Mentor 0 [1,0,0,0] → matches at positions 0,1 → score = 2
g[1][1]: Student 1 [1,0,1,1] vs Mentor 1 [0,0,1,0] → matches at positions 1,2 → score = 2
g[1][2]: Student 1 [1,0,1,1] vs Mentor 2 [1,1,0,1] → matches at positions 0,3 → score = 2

g[2][0]: Student 2 [0,0,1,0] vs Mentor 0 [1,0,0,0] → matches at positions 1,3 → score = 2
g[2][1]: Student 2 [0,0,1,0] vs Mentor 1 [0,0,1,0] → matches at all positions → score = 4
g[2][2]: Student 2 [0,0,1,0] vs Mentor 2 [1,1,0,1] → matches at position 2 → score = 1

Compatibility Grid:

    M0  M1  M2
S0  3   1   3
S1  2   2   2  
S2  2   4   1

Step 2: Backtracking Process

Starting with dfs(0, 0) and vis = [False, False, False]:

1. Assign Student 0:

   • Try Mentor 0: vis[0] = True, call dfs(1, 0+3=3)

     • Assign Student 1:
       • Try Mentor 1: vis[1] = True, call dfs(2, 3+2=5)
         • Assign Student 2:
           • Try Mentor 2: vis[2] = True, call dfs(3, 5+1=6)
             • Base case: All assigned, ans = max(0, 6) = 6
           • Backtrack: vis[2] = False
         • Backtrack: vis[1] = False
       • Try Mentor 2: vis[2] = True, call dfs(2, 3+2=5)
         • Assign Student 2:
           • Try Mentor 1: vis[1] = True, call dfs(3, 5+4=9)
             • Base case: All assigned, ans = max(6, 9) = 9
           • Backtrack: vis[1] = False
         • Backtrack: vis[2] = False
     • Backtrack: vis[0] = False

   • Try Mentor 1: vis[1] = True, call dfs(1, 0+1=1)

     • Assign Student 1:
       • Try Mentor 0: vis[0] = True, call dfs(2, 1+2=3)
         • Assign Student 2:
           • Try Mentor 2: vis[2] = True, call dfs(3, 3+1=4)
             • Base case: All assigned, ans = max(9, 4) = 9
           • Backtrack: vis[2] = False
         • Backtrack: vis[0] = False
       • Try Mentor 2: vis[2] = True, call dfs(2, 1+2=3)
         • Assign Student 2:
           • Try Mentor 0: vis[0] = True, call dfs(3, 3+2=5)
             • Base case: All assigned, ans = max(9, 5) = 9
           • Backtrack: vis[0] = False
         • Backtrack: vis[2] = False
     • Backtrack: vis[1] = False

   • Try Mentor 2: vis[2] = True, call dfs(1, 0+3=3)

     • (Similar process continues...)

Step 3: Optimal Solution Found

The maximum score of 9 is achieved with the pairing:

• Student 0 → Mentor 0 (score: 3)
• Student 1 → Mentor 2 (score: 2)
• Student 2 → Mentor 1 (score: 4)
• Total: 3 + 2 + 4 = 9

The backtracking ensures we explore all possible pairings and find the optimal one, avoiding the pitfall of greedy choices that might seem good initially but lead to suboptimal total scores.

Solution Implementation

Time and Space Complexity

Time Complexity: O(m! × m)

The algorithm uses depth-first search (DFS) to explore all possible assignments of students to mentors. At each level of recursion:

• Level 0: We have m choices for the first student
• Level 1: We have m-1 choices for the second student
• Level 2: We have m-2 choices for the third student
• And so on...

This gives us m × (m-1) × (m-2) × ... × 1 = m! different permutations to explore. Additionally, at each recursive call, we iterate through all m mentors to check availability and calculate scores, adding a factor of m. However, since we also spend O(m) time preprocessing to build the compatibility matrix g, and within each DFS path we perform O(1) operations per level, the dominant factor is O(m!) for exploring all permutations.

Note: The preprocessing step to build matrix g takes O(m² × n) where n is the length of each answer array, but this is done once outside the DFS.

Space Complexity: O(m²)

The space complexity consists of:

• The compatibility matrix g: O(m²) space to store compatibility scores between all student-mentor pairs
• The visited array vis: O(m) space to track which mentors have been assigned
• The recursion call stack: O(m) depth in the worst case as we make at most m recursive calls

The dominant space usage is the compatibility matrix, giving us O(m²) overall space complexity.

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

Common Pitfalls

1. Not Using Preprocessing - Recalculating Compatibility Scores

A common mistake is calculating compatibility scores repeatedly during the backtracking process instead of precomputing them once.

Pitfall Example:

def dfs(student_idx, current_sum):
    if student_idx >= m:
        # update answer
        return
  
    for mentor_idx in range(m):
        if not visited[mentor_idx]:
            # BAD: Recalculating compatibility every time
            compatibility = sum(students[student_idx][q] == mentors[mentor_idx][q] 
                              for q in range(n))
            visited[mentor_idx] = True
            dfs(student_idx + 1, current_sum + compatibility)
            visited[mentor_idx] = False

Solution: Always precompute the compatibility matrix before starting the backtracking. This reduces time complexity from O(m! × m × n) to O(m! × m + m² × n).

2. Incorrect Base Case Handling

Another pitfall is forgetting to properly handle the base case or using the wrong condition.

Pitfall Example:

def dfs(student_idx, current_sum):
    # BAD: Using > instead of >=, missing the last student
    if student_idx > m:  
        max_sum = max(max_sum, current_sum)
        return

Solution: Use student_idx >= m to ensure all m students (indexed 0 to m-1) have been assigned before updating the answer.

3. Forgetting to Backtrack (Restore State)

A critical error is not properly restoring the state after exploring a branch in the recursion tree.

Pitfall Example:

for mentor_idx in range(m):
    if not visited[mentor_idx]:
        visited[mentor_idx] = True
        dfs(student_idx + 1, current_sum + compatibility[student_idx][mentor_idx])
        # BAD: Forgot to reset visited[mentor_idx] = False

Solution: Always restore the state after the recursive call returns:

visited[mentor_idx] = True
dfs(student_idx + 1, current_sum + compatibility[student_idx][mentor_idx])
visited[mentor_idx] = False  # Critical: Reset for other branches

4. Using Global Variables Incorrectly in Python

Python requires the nonlocal keyword to modify variables from an enclosing scope.

Pitfall Example:

def maxCompatibilitySum(self, students, mentors):
    max_sum = 0
  
    def dfs(i, s):
        if i >= m:
            # BAD: This creates a local variable instead of modifying the outer one
            max_sum = max(max_sum, s)  # UnboundLocalError

Solution: Declare the variable as nonlocal:

def dfs(i, s):
    if i >= m:
        nonlocal max_sum
        max_sum = max(max_sum, s)

5. Attempting Greedy Approach

A tempting but incorrect approach is trying to solve this greedily by always choosing the best available mentor for each student.

Pitfall Example:

# BAD: Greedy approach - doesn't guarantee optimal solution
total = 0
used = [False] * m
for student_idx in range(m):
    best_score = -1
    best_mentor = -1
    for mentor_idx in range(m):
        if not used[mentor_idx]:
            score = compatibility[student_idx][mentor_idx]
            if score > best_score:
                best_score = score
                best_mentor = mentor_idx
    used[best_mentor] = True
    total += best_score

Solution: This is an assignment problem that requires exploring all possibilities. The greedy approach can miss the globally optimal solution because choosing the best mentor for one student might prevent a better overall pairing. Always use the complete backtracking approach to guarantee finding the maximum sum.