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Maximum Number of Accepted Invitations

Medium

Asked by:

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Topics:

ArraysGreedy Algorithms

There are m students and n invitations. You are given a 0-indexed array genders of length m where genders[i] denotes the gender of the i^th student. Similarly, you are given a 0-indexed array wishes of length n where wishes[j] is a list of students that the j^th invitation wants to invite.

A student can accept at most one invitation, and an invitation can be accepted by at most one student.

Return the maximum number of accepted invitations.

Example 1:

Input: genders = [0,1,1], wishes = [[0,1,2],[1,2,3],[0,2]]
Output: 3
Explanation: The following are the steps to arrive at the maximum number of accepted invitations:
- Invitation 0 is accepted by student 0.
- Invitation 1 is accepted by student 1.
- Invitation 2 is accepted by student 2.

Example 2:

Input: genders = [1,1,0], wishes = [[0,2], [1,2]]
Output: 2
Explanation: The following are the steps to arrive at the maximum number of accepted invitations:
- Invitation 0 is accepted by student 0.
- Invitation 1 is accepted by student 1.

Constraints:

m == genders.length
n == wishes.length
1 <= m, n <= 200
0 <= genders[i] <= 1
0 <= wishes[j].length <= m
0 <= wishes[j][k] < m

Try coding it in LeetCode

Maximum Number of Accepted Invitations

Medium

Asked by:

23 views

Topics:

ArraysGreedy Algorithms

A student can accept at most one invitation, and an invitation can be accepted by at most one student.

Return the maximum number of accepted invitations.

Example 1:

Input: genders = [0,1,1], wishes = [[0,1,2],[1,2,3],[0,2]]
Output: 3
Explanation: The following are the steps to arrive at the maximum number of accepted invitations:
- Invitation 0 is accepted by student 0.
- Invitation 1 is accepted by student 1.
- Invitation 2 is accepted by student 2.

Example 2:

Input: genders = [1,1,0], wishes = [[0,2], [1,2]]
Output: 2
Explanation: The following are the steps to arrive at the maximum number of accepted invitations:
- Invitation 0 is accepted by student 0.
- Invitation 1 is accepted by student 1.

Constraints:

m == genders.length
n == wishes.length
1 <= m, n <= 200
0 <= genders[i] <= 1
0 <= wishes[j].length <= m
0 <= wishes[j][k] < m

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 constraints on the number of students and available seats; specifically, what are the maximum possible values for `students.length` and `seats.length`?
Can a student prefer multiple seats, and similarly, can a seat be preferred by multiple students? Are there duplicates within `preferences[i]` for any student i?
If a student has no preferences (i.e., `preferences[i]` is empty), should they be considered as not being able to accept any invitation?
If it's impossible to maximize the number of accepted invitations, what should the return value be? Should I return 0, -1, or something else?
Is the input data guaranteed to be valid? For example, will `preferences[i]` always contain seat numbers within the valid range of 0 to `seats.length - 1`?

Brute Force Solution

Approach

The brute force approach involves exploring every single possible combination of accepting or rejecting invitations. We check each combination to see how many invitations can be accepted based on the given constraints. Ultimately we're trying every path to find the one that yields the maximum number of accepted invitations.

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

Imagine each person on one side can either accept or reject each invitation from the other side.
Start by considering the scenario where nobody accepts any invitation. Count how many acceptances there are (which is zero).
Then, try the scenario where just the first person accepts the first invitation they received, and everyone else rejects everything. Count the acceptances.
Next, consider the case where the first person accepts the second invitation they received, and everyone else rejects everything. Count the acceptances.
Keep going through all possible combinations of the first person accepting only one invitation at a time, while others reject.
Now, start experimenting where the first person accepts two invitations and others reject, and keep track of acceptances.
Continue this pattern where each person can accept different combinations of invitations they received.
For each combination of acceptances and rejections, check if any person on either side has accepted more invitations than they are allowed.
If a combination violates the acceptance limit for anyone, discard that entire possibility.
Out of all the valid combinations (those that don't violate any limits), find the one where the total number of accepted invitations is the highest.
The maximum number of accepted invitations found across all the valid combinations is the answer.

Code Implementation

def maximum_number_of_accepted_invitations_brute_force(
    invitations, row_limits, col_limits
):
    number_of_rows = len(invitations)
    number_of_cols = len(invitations[0])

    maximum_accepted_invitations = 0

    # Iterate through all possible combinations of accept/reject
    for i in range(2 ** (number_of_rows * number_of_cols)):
        current_combination = []
        temp = i
        for _ in range(number_of_rows * number_of_cols):
            current_combination.append(temp % 2)
            temp //= 2

        accepted_invitations_matrix = [
            current_combination[
                row * number_of_cols : (row + 1) * number_of_cols
            ]
            for row in range(number_of_rows)
        ]

        row_acceptances = [sum(row) for row in accepted_invitations_matrix]
        col_acceptances = [
            sum(accepted_invitations_matrix[row][col] for row in range(number_of_rows))
            for col in range(number_of_cols)
        ]

        valid_combination = True
        # Check if the limits for each row are exceeded
        for row in range(number_of_rows):
            if row_acceptances[row] > row_limits[row]:
                valid_combination = False
                break

        if not valid_combination:
            continue

        # Check if the limits for each column are exceeded
        for col in range(number_of_cols):
            if col_acceptances[col] > col_limits[col]:
                valid_combination = False
                break

        if not valid_combination:
            continue

        # Calculate the total number of accepted invitations
        total_accepted_invitations = sum(current_combination)

        # Update the maximum if the current combination is better
        maximum_accepted_invitations = max(
            maximum_accepted_invitations, total_accepted_invitations
        )

    return maximum_accepted_invitations

Big(O) Analysis

Time Complexity

O(2^(m*n)) – The algorithm explores all possible combinations of accepting or rejecting invitations. If there are m people on one side and n people on the other side, and we're examining all possible pairings, each possible pairing has two states (accepted or rejected). Therefore, we have 2 possible states for each invitation, and with m*n potential invitations, we end up with 2^(m*n) possible combinations. Hence, the time complexity is O(2^(m*n)), where m and n represent the number of people on each side, determining the potential pairings.

Space Complexity

O(2^(M*N)) – The brute force approach described explores every possible combination of accepting or rejecting invitations, where M is the number of people on one side and N is the number of people on the other side. Each person on one side can either accept or reject an invitation from the other side, leading to 2^(M*N) possible combinations. Although the description doesn't explicitly state the storage of all combinations, the exploration of each one inherently requires tracking the current acceptance state. This implicitly uses space proportional to the number of combinations to manage the recursive call stack and potentially store intermediate results or flags representing accepted/rejected invitations, leading to a space complexity of O(2^(M*N)).

Optimal Solution

Approach

This problem asks us to maximize the number of successful invitations between two groups. The clever approach involves matching individuals from one group to the other in a way that ensures the most pairings possible, avoiding any conflicts or overlaps. Think of it like an efficient dating service or job fair, where you want to connect as many people as you can.

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

Imagine each person in the first group is looking for a partner from the second group, and vice versa. Represent all possible pairings as a network or a web of connections.
Start with someone from the first group. See if they have any available partners in the second group.
If they do, provisionally pair them up. Now, consider the impact of this pairing on the remaining people and connections.
Look for the most constrained person: someone who has very few options available. Prioritize pairing them up.
Keep pairing people strategically, always focusing on maximizing the use of connections that would otherwise go unused.
If you reach a point where someone has no available partners, you know you've made a suboptimal choice somewhere along the way. Instead of backtracking through all previous choices, use a technique to efficiently try alternative pairings.
This optimization technique involves looking for alternate paths through the network where changing one pairing can open up new opportunities for the overall arrangement.
Continue until you've exhausted all possibilities and determined the maximum number of successful pairings you can achieve.

Code Implementation

def maximum_invitations(grid):
    number_of_rows = len(grid)
    number_of_columns = len(grid[0]) if number_of_rows > 0 else 0

    matching = [-1] * number_of_columns
    visited = [False] * number_of_rows
    max_accepted_invitations = 0

    def find_match(row_index):
        visited[row_index] = True

        for column_index in range(number_of_columns):
            if grid[row_index][column_index] == 1:
                # Try to match the current row with an available column.
                if matching[column_index] == -1 or (
                    not visited[matching[column_index]] and find_match(matching[column_index])
                ):
                    matching[column_index] = row_index
                    return True

        return False

    for row_index in range(number_of_rows):
        # Reset visited array for each person in the first group.
        visited = [False] * number_of_rows
        if find_match(row_index):
            # Increment counter if matching was found.
            max_accepted_invitations += 1

    return max_accepted_invitations

Big(O) Analysis

Time Complexity

O(m*n*k) – The algorithm iterates through each person in the first group (size m) and attempts to find a suitable partner in the second group (size n). For each person in the first group, it might need to explore multiple potential matches, and for each match, it may need to trace alternate paths or adjust existing pairings to optimize the overall matching. The 'k' factor represents the average number of potential partners considered or the depth of path exploration required before finding a valid matching or exhausting the possibilities for a given person from the first group. Therefore, the overall time complexity approximates O(m*n*k).

Space Complexity

O(N*M) – The provided plain English explanation suggests representing pairings as a network or web of connections. In the worst-case scenario, this could involve storing an adjacency matrix or adjacency list to track possible pairings between the first group of N people and the second group of M people. This network data structure, whether implemented as a matrix or a list of lists, would require space proportional to the product of N and M to store the connections. Therefore, the auxiliary space complexity is O(N*M), where N and M are the sizes of the two groups.

Edge Cases

Case	How to Handle
Empty guest list or empty chairs list.	Return 0, as no invitations can be accepted.
Guest list has length 1 or chairs list has length 1.	Iterate through the smaller list to find compatible pairs in the longer list, returning 1 if a match is found, otherwise 0.
Maximum number of guests and chairs (scalability concern).	Ensure the algorithm's time complexity is optimized (e.g., using a bipartite matching algorithm or efficient data structures) to avoid exceeding time limits for large inputs.
All guests can sit in all chairs.	If there are more guests than chairs return the number of chairs; if more chairs than guests return the number of guests, and if they're equal, return either.
No guest can sit in any chair (incompatible lists).	Return 0, as no invitations can be accepted.
Duplicate preferences; a guest prefers multiple chairs or a chair is preferred by multiple guests.	The algorithm should correctly account for multiple preferences and still produce a valid maximum matching.
Disjoint preference graph (guests only prefer one set of chairs and another set of guests only prefer a different set).	The algorithm should independently maximize matching in each disconnected component of the preference graph.
Integer overflow when representing the number of accepted invitations.	Use a data type that can accommodate the maximum possible number of invitations (e.g., long) if necessary.

Empty guest list or empty chairs list.

How to Handle:

Return 0, as no invitations can be accepted.

Guest list has length 1 or chairs list has length 1.

How to Handle:

Iterate through the smaller list to find compatible pairs in the longer list, returning 1 if a match is found, otherwise 0.

Maximum number of guests and chairs (scalability concern).

How to Handle:

Ensure the algorithm's time complexity is optimized (e.g., using a bipartite matching algorithm or efficient data structures) to avoid exceeding time limits for large inputs.

All guests can sit in all chairs.

How to Handle:

If there are more guests than chairs return the number of chairs; if more chairs than guests return the number of guests, and if they're equal, return either.

No guest can sit in any chair (incompatible lists).

How to Handle:

Return 0, as no invitations can be accepted.

Duplicate preferences; a guest prefers multiple chairs or a chair is preferred by multiple guests.

How to Handle:

The algorithm should correctly account for multiple preferences and still produce a valid maximum matching.

Disjoint preference graph (guests only prefer one set of chairs and another set of guests only prefer a different set).

How to Handle:

The algorithm should independently maximize matching in each disconnected component of the preference graph.

Integer overflow when representing the number of accepted invitations.

How to Handle:

Use a data type that can accommodate the maximum possible number of invitations (e.g., long) if necessary.

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 constraints on the number of students and available seats; specifically, what are the maximum possible values for `students.length` and `seats.length`?
Can a student prefer multiple seats, and similarly, can a seat be preferred by multiple students? Are there duplicates within `preferences[i]` for any student i?
If a student has no preferences (i.e., `preferences[i]` is empty), should they be considered as not being able to accept any invitation?
If it's impossible to maximize the number of accepted invitations, what should the return value be? Should I return 0, -1, or something else?
Is the input data guaranteed to be valid? For example, will `preferences[i]` always contain seat numbers within the valid range of 0 to `seats.length - 1`?

Brute Force Solution

Approach

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

Imagine each person on one side can either accept or reject each invitation from the other side.
Start by considering the scenario where nobody accepts any invitation. Count how many acceptances there are (which is zero).
Then, try the scenario where just the first person accepts the first invitation they received, and everyone else rejects everything. Count the acceptances.
Next, consider the case where the first person accepts the second invitation they received, and everyone else rejects everything. Count the acceptances.
Keep going through all possible combinations of the first person accepting only one invitation at a time, while others reject.
Now, start experimenting where the first person accepts two invitations and others reject, and keep track of acceptances.
Continue this pattern where each person can accept different combinations of invitations they received.
For each combination of acceptances and rejections, check if any person on either side has accepted more invitations than they are allowed.
If a combination violates the acceptance limit for anyone, discard that entire possibility.
Out of all the valid combinations (those that don't violate any limits), find the one where the total number of accepted invitations is the highest.
The maximum number of accepted invitations found across all the valid combinations is the answer.

Code Implementation

def maximum_number_of_accepted_invitations_brute_force(
    invitations, row_limits, col_limits
):
    number_of_rows = len(invitations)
    number_of_cols = len(invitations[0])

    maximum_accepted_invitations = 0

    # Iterate through all possible combinations of accept/reject
    for i in range(2 ** (number_of_rows * number_of_cols)):
        current_combination = []
        temp = i
        for _ in range(number_of_rows * number_of_cols):
            current_combination.append(temp % 2)
            temp //= 2

        accepted_invitations_matrix = [
            current_combination[
                row * number_of_cols : (row + 1) * number_of_cols
            ]
            for row in range(number_of_rows)
        ]

        row_acceptances = [sum(row) for row in accepted_invitations_matrix]
        col_acceptances = [
            sum(accepted_invitations_matrix[row][col] for row in range(number_of_rows))
            for col in range(number_of_cols)
        ]

        valid_combination = True
        # Check if the limits for each row are exceeded
        for row in range(number_of_rows):
            if row_acceptances[row] > row_limits[row]:
                valid_combination = False
                break

        if not valid_combination:
            continue

        # Check if the limits for each column are exceeded
        for col in range(number_of_cols):
            if col_acceptances[col] > col_limits[col]:
                valid_combination = False
                break

        if not valid_combination:
            continue

        # Calculate the total number of accepted invitations
        total_accepted_invitations = sum(current_combination)

        # Update the maximum if the current combination is better
        maximum_accepted_invitations = max(
            maximum_accepted_invitations, total_accepted_invitations
        )

    return maximum_accepted_invitations

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Imagine each person in the first group is looking for a partner from the second group, and vice versa. Represent all possible pairings as a network or a web of connections.
Start with someone from the first group. See if they have any available partners in the second group.
If they do, provisionally pair them up. Now, consider the impact of this pairing on the remaining people and connections.
Look for the most constrained person: someone who has very few options available. Prioritize pairing them up.
Keep pairing people strategically, always focusing on maximizing the use of connections that would otherwise go unused.
If you reach a point where someone has no available partners, you know you've made a suboptimal choice somewhere along the way. Instead of backtracking through all previous choices, use a technique to efficiently try alternative pairings.
This optimization technique involves looking for alternate paths through the network where changing one pairing can open up new opportunities for the overall arrangement.
Continue until you've exhausted all possibilities and determined the maximum number of successful pairings you can achieve.

Code Implementation

def maximum_invitations(grid):
    number_of_rows = len(grid)
    number_of_columns = len(grid[0]) if number_of_rows > 0 else 0

    matching = [-1] * number_of_columns
    visited = [False] * number_of_rows
    max_accepted_invitations = 0

    def find_match(row_index):
        visited[row_index] = True

        for column_index in range(number_of_columns):
            if grid[row_index][column_index] == 1:
                # Try to match the current row with an available column.
                if matching[column_index] == -1 or (
                    not visited[matching[column_index]] and find_match(matching[column_index])
                ):
                    matching[column_index] = row_index
                    return True

        return False

    for row_index in range(number_of_rows):
        # Reset visited array for each person in the first group.
        visited = [False] * number_of_rows
        if find_match(row_index):
            # Increment counter if matching was found.
            max_accepted_invitations += 1

    return max_accepted_invitations

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty guest list or empty chairs list.	Return 0, as no invitations can be accepted.
Guest list has length 1 or chairs list has length 1.	Iterate through the smaller list to find compatible pairs in the longer list, returning 1 if a match is found, otherwise 0.
Maximum number of guests and chairs (scalability concern).	Ensure the algorithm's time complexity is optimized (e.g., using a bipartite matching algorithm or efficient data structures) to avoid exceeding time limits for large inputs.
All guests can sit in all chairs.	If there are more guests than chairs return the number of chairs; if more chairs than guests return the number of guests, and if they're equal, return either.
No guest can sit in any chair (incompatible lists).	Return 0, as no invitations can be accepted.
Duplicate preferences; a guest prefers multiple chairs or a chair is preferred by multiple guests.	The algorithm should correctly account for multiple preferences and still produce a valid maximum matching.
Disjoint preference graph (guests only prefer one set of chairs and another set of guests only prefer a different set).	The algorithm should independently maximize matching in each disconnected component of the preference graph.
Integer overflow when representing the number of accepted invitations.	Use a data type that can accommodate the maximum possible number of invitations (e.g., long) if necessary.

Empty guest list or empty chairs list.

How to Handle:

Return 0, as no invitations can be accepted.

Guest list has length 1 or chairs list has length 1.

How to Handle:

Iterate through the smaller list to find compatible pairs in the longer list, returning 1 if a match is found, otherwise 0.

Maximum number of guests and chairs (scalability concern).

How to Handle:

Ensure the algorithm's time complexity is optimized (e.g., using a bipartite matching algorithm or efficient data structures) to avoid exceeding time limits for large inputs.

All guests can sit in all chairs.

How to Handle:

If there are more guests than chairs return the number of chairs; if more chairs than guests return the number of guests, and if they're equal, return either.

No guest can sit in any chair (incompatible lists).

How to Handle:

Return 0, as no invitations can be accepted.

Duplicate preferences; a guest prefers multiple chairs or a chair is preferred by multiple guests.

How to Handle:

The algorithm should correctly account for multiple preferences and still produce a valid maximum matching.

Disjoint preference graph (guests only prefer one set of chairs and another set of guests only prefer a different set).

How to Handle:

The algorithm should independently maximize matching in each disconnected component of the preference graph.

Integer overflow when representing the number of accepted invitations.

How to Handle:

Use a data type that can accommodate the maximum possible number of invitations (e.g., long) if necessary.