Maximize Grid Happiness

Hard

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Dynamic ProgrammingBit Manipulation

You are given four integers, m, n, introvertsCount, and extrovertsCount. You have an m x n grid, and there are two types of people: introverts and extroverts. There are introvertsCount introverts and extrovertsCount extroverts.

You should decide how many people you want to live in the grid and assign each of them one grid cell. Note that you do not have to have all the people living in the grid.

The happiness of each person is calculated as follows:

Introverts start with 120 happiness and lose 30 happiness for each neighbor (introvert or extrovert).
Extroverts start with 40 happiness and gain 20 happiness for each neighbor (introvert or extrovert).

Neighbors live in the directly adjacent cells north, east, south, and west of a person's cell.

The grid happiness is the sum of each person's happiness. Return the maximum possible grid happiness.

Example 1:

Input: m = 2, n = 3, introvertsCount = 1, extrovertsCount = 2
Output: 240
Explanation: Assume the grid is 1-indexed with coordinates (row, column).
We can put the introvert in cell (1,1) and put the extroverts in cells (1,3) and (2,3).
- Introvert at (1,1) happiness: 120 (starting happiness) - (0 * 30) (0 neighbors) = 120
- Extrovert at (1,3) happiness: 40 (starting happiness) + (1 * 20) (1 neighbor) = 60
- Extrovert at (2,3) happiness: 40 (starting happiness) + (1 * 20) (1 neighbor) = 60
The grid happiness is 120 + 60 + 60 = 240.
The above figure shows the grid in this example with each person's happiness. The introvert stays in the light green cell while the extroverts live on the light purple cells.

Example 2:

Input: m = 3, n = 1, introvertsCount = 2, extrovertsCount = 1
Output: 260
Explanation: Place the two introverts in (1,1) and (3,1) and the extrovert at (2,1).
- Introvert at (1,1) happiness: 120 (starting happiness) - (1 * 30) (1 neighbor) = 90
- Extrovert at (2,1) happiness: 40 (starting happiness) + (2 * 20) (2 neighbors) = 80
- Introvert at (3,1) happiness: 120 (starting happiness) - (1 * 30) (1 neighbor) = 90
The grid happiness is 90 + 80 + 90 = 260.

Example 3:

Input: m = 2, n = 2, introvertsCount = 4, extrovertsCount = 0
Output: 240

Constraints:

1 <= m, n <= 5
0 <= introvertsCount, extrovertsCount <= min(m * n, 6)

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 values of m, n, introvertsCount, and extrovertsCount? Specifically, what are the maximum values for each?
Can introvertsCount or extrovertsCount be zero? What should I return in that case?
Could you clarify the happiness calculation for adjacent cells? Is it symmetric (i.e., if an introvert is to the left of an extrovert, is the happiness reduction applied to both, or just once?)
Is the grid toroidal (i.e., do the left and right edges wrap around, and do the top and bottom edges wrap around)? Or are we only considering up, down, left, and right cells that are within the grid boundaries?
If multiple arrangements yield the same maximum happiness, is any of those arrangements acceptable, or is there a specific arrangement that is preferred based on some implicit criteria?

Brute Force Solution

Approach

The brute force approach to maximizing grid happiness means trying every single possible arrangement of introverts and extroverts in the grid. We calculate the happiness score for each arrangement and then choose the arrangement with the highest score. This is like trying every combination to open a lock.

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

Consider placing an introvert in the first available spot.
Now, consider placing an extrovert in the first available spot.
Continue filling each remaining spot in the grid, one by one, considering both introvert and extrovert options for each.
Once the grid is full, calculate the total happiness score for that specific arrangement of introverts and extroverts based on their neighbors.
Store the happiness score for that arrangement.
Repeat the process, trying every possible arrangement of introverts and extroverts until all combinations have been evaluated.
Compare the happiness scores of all the arrangements that were evaluated.
Select the arrangement that resulted in the highest happiness score. That's the solution.

Code Implementation

def maximize_grid_happiness_brute_force(rows, columns, introverts_count, extroverts_count):
    max_happiness = float('-inf')

    def calculate_happiness(grid):
        total_happiness = 0
        for row_index in range(rows):
            for column_index in range(columns):
                if grid[row_index][column_index] == 1:
                    # Introvert
                    total_happiness -= 30
                    if row_index > 0 and grid[row_index - 1][column_index] == 1:
                        total_happiness -= 60
                    elif row_index > 0 and grid[row_index - 1][column_index] == 2:
                        total_happiness -= 10
                    if row_index < rows - 1 and grid[row_index + 1][column_index] == 1:
                        total_happiness -= 60
                    elif row_index < rows - 1 and grid[row_index + 1][column_index] == 2:
                        total_happiness -= 10
                    if column_index > 0 and grid[row_index][column_index - 1] == 1:
                        total_happiness -= 60
                    elif column_index > 0 and grid[row_index][column_index - 1] == 2:
                        total_happiness -= 10
                    if column_index < columns - 1 and grid[row_index][column_index + 1] == 1:
                        total_happiness -= 60
                    elif column_index < columns - 1 and grid[row_index][column_index + 1] == 2:
                        total_happiness -= 10
                elif grid[row_index][column_index] == 2:
                    # Extrovert
                    total_happiness += 20
                    if row_index > 0 and grid[row_index - 1][column_index] == 1:
                        total_happiness -= 10
                    elif row_index > 0 and grid[row_index - 1][column_index] == 2:
                        total_happiness += 40
                    if row_index < rows - 1 and grid[row_index + 1][column_index] == 1:
                        total_happiness -= 10
                    elif row_index < rows - 1 and grid[row_index + 1][column_index] == 2:
                        total_happiness += 40
                    if column_index > 0 and grid[row_index][column_index - 1] == 1:
                        total_happiness -= 10
                    elif column_index > 0 and grid[row_index][column_index - 1] == 2:
                        total_happiness += 40
                    if column_index < columns - 1 and grid[row_index][column_index + 1] == 1:
                        total_happiness -= 10
                    elif column_index < columns - 1 and grid[row_index][column_index + 1] == 2:
                        total_happiness += 40
        return total_happiness

    def generate_grids(row_index, column_index, current_grid, current_introverts_count, current_extroverts_count):
        nonlocal max_happiness

        if row_index == rows:
            # Base case: grid is full, calculate happiness
            happiness = calculate_happiness(current_grid)
            max_happiness = max(max_happiness, happiness)
            return

        # Determine the next cell to fill
        next_row_index = row_index
        next_column_index = column_index + 1
        if next_column_index == columns:
            next_row_index = row_index + 1
            next_column_index = 0

        # Option 1: Leave the cell empty
        generate_grids(next_row_index, next_column_index, current_grid, current_introverts_count, current_extroverts_count)

        # Option 2: Place an introvert, if available
        if current_introverts_count > 0:
            current_grid[row_index][column_index] = 1
            generate_grids(next_row_index, next_column_index, current_grid, current_introverts_count - 1, current_extroverts_count)
            # Backtrack: Reset the cell
            current_grid[row_index][column_index] = 0

            #Try placing an extrovert in each cell

        # Option 3: Place an extrovert, if available
        if current_extroverts_count > 0:
            current_grid[row_index][column_index] = 2
            generate_grids(next_row_index, next_column_index, current_grid, current_introverts_count, current_extroverts_count - 1)
            # Backtrack: Reset the cell
            current_grid[row_index][column_index] = 0

    initial_grid = [[0] * columns for _ in range(rows)]
    #Starting the generation process
    generate_grids(0, 0, initial_grid, introverts_count, extroverts_count)

    #We return the highest value, which is the result
    return max_happiness

Big(O) Analysis

Time Complexity

O(2^(m*n)) – The brute force approach explores all possible placements of introverts and extroverts in an m x n grid. Each cell in the grid can either be an introvert or an extrovert, resulting in 2 choices for each cell. Since there are m*n cells, the total number of possible arrangements is 2 raised to the power of (m*n). Calculating the happiness score for each arrangement takes O(1) time, so the overall time complexity is dominated by the number of arrangements, which is O(2^(m*n)).

Space Complexity

O(rows * cols) – The brute force approach involves exploring every possible arrangement of introverts and extroverts in the grid. This is achieved through recursion. The maximum depth of the recursion is determined by the number of cells in the grid, which is rows * cols. Each recursive call stores local variables and parameters on the call stack, creating stack frames. Therefore, the auxiliary space used by the recursion stack grows linearly with the number of cells in the grid, resulting in a space complexity of O(rows * cols).

Optimal Solution

Approach

To maximize happiness in the grid, we will use a technique called dynamic programming. We will make decisions about placing introverts and extroverts one cell at a time, remembering the best happiness we can achieve for each partial arrangement.

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

Since the overall happiness depends on the current placement and the placements of neighbors, we need to keep track of what's in the previous row.
Imagine building the grid row by row, from top to bottom. At each cell, we have three options: place an introvert, place an extrovert, or leave the cell empty.
We'll use a helper function to figure out the happiness change caused by placing someone in a particular cell, based on who their neighbors are from the previous row and left cell.
We will then use a table to store the maximum happiness at each location in our grid for a given configuration. The configuration includes the current row number and how many introverts and extroverts we have left to place.
For each cell, calculate the total happiness for each of the three placement options (introvert, extrovert, empty), considering the happiness change due to neighbors and the happiness of the person placed.
Store the best possible happiness value at each point in our table.
Because our table keeps track of the best possible score at each location, we can simply return the best possible happiness from the last cell in our grid.

Code Implementation

def maximize_grid_happiness(grid_rows, grid_columns, number_of_introverts, number_of_extroverts):
    introvert_score = 120
    extrovert_score = 40
    loss_score = -30

    def calculate_happiness(grid_state, row, column, person_type):
        happiness_change = 0

        #Check top neighbor
        if row > 0:
            neighbor = grid_state[row - 1][column]
            if neighbor != 0:
                if person_type == 1:
                    happiness_change += loss_score
                else:
                    happiness_change += extrovert_score
                if neighbor == 1:
                    happiness_change += loss_score
                else:
                    happiness_change += extrovert_score

        # Check left neighbor
        if column > 0:
            neighbor = grid_state[row][column - 1]
            if neighbor != 0:
                if person_type == 1:
                    happiness_change += loss_score
                else:
                    happiness_change += extrovert_score
                if neighbor == 1:
                    happiness_change += loss_score
                else:
                    happiness_change += extrovert_score

        # Add the base happiness of the person
        if person_type == 1:
            happiness_change += introvert_score
        else:
            happiness_change += extrovert_score

        return happiness_change

    def solve(row, column, introverts_remaining, extroverts_remaining, grid_state, current_happiness, memo):
        if (row, column, introverts_remaining, extroverts_remaining) in memo:
            return memo[(row, column, introverts_remaining, extroverts_remaining)]

        # Base case: Reached the end of the grid
        if row == grid_rows:
            return current_happiness

        next_row = row
        next_column = column + 1
        if next_column == grid_columns:
            next_row += 1
            next_column = 0

        # Option 1: Leave the cell empty
        max_happiness = solve(next_row, next_column, introverts_remaining, extroverts_remaining, grid_state, current_happiness, memo)

        # Option 2: Place an introvert
        if introverts_remaining > 0:
            grid_state[row][column] = 1
            happiness_change = calculate_happiness(grid_state, row, column, 1)

            max_happiness = max(max_happiness, solve(next_row, next_column, introverts_remaining - 1, extroverts_remaining, grid_state, current_happiness + happiness_change, memo))

            grid_state[row][column] = 0 # Backtrack

        # Option 3: Place an extrovert
        if extroverts_remaining > 0:
            grid_state[row][column] = 2
            happiness_change = calculate_happiness(grid_state, row, column, 2)

            max_happiness = max(max_happiness, solve(next_row, next_column, introverts_remaining, extroverts_remaining - 1, grid_state, current_happiness + happiness_change, memo))

            grid_state[row][column] = 0 # Backtrack

        memo[(row, column, introverts_remaining, extroverts_remaining)] = max_happiness
        return max_happiness

    # Initialize the grid and memoization table
    grid_state = [[0] * grid_columns for _ in range(grid_rows)]
    memo = {}

    # Start the dynamic programming solution
    # Starting state has no happiness
    initial_happiness = 0
    result = solve(0, 0, number_of_introverts, number_of_extroverts, grid_state, initial_happiness, memo)

    # This is the result after exploring every possible placement option
    return result

Big(O) Analysis

Time Complexity

O(m * n * 3^n * (introverts + extroverts)) – The dynamic programming solution iterates through each cell of the m x n grid. For each cell, it considers three possibilities (introvert, extrovert, or empty). Crucially, it needs to remember the configuration of the previous row, which can be represented as a ternary number (since each cell can be in one of three states). This previous row configuration contributes a factor of 3^n. For each such state we must iterate through all possible remaining counts of introverts and extroverts. Thus, we end up with O(m * n * 3^n * (introverts + extroverts))

Space Complexity

O(m * 2^(m)) – The dominant space complexity comes from the dynamic programming table, which stores the maximum happiness at each cell for different configurations. The table's dimensions are determined by the number of rows (n, which isn't directly present but implied in iterating over the grid), the possible states of the previous row, and the remaining number of introverts and extroverts. Since we keep track of the previous row's state, and each cell in that row can be in one of three states (introvert, extrovert, empty), there are 3^m possible states for a row of length m. However, the description mentions using a 'mask' to represent the configuration which could be optimized. Considering the introverts and extroverts constraints, the row state can be represented in fewer bits. The state can be represented using 2^m. We also have the 'introverts left' and 'extroverts left' which are constrained by total numbers. However, number of states becomes the primary factor. Therefore the space is O(m * 2^(m)) which is derived from state of length m for each introvert and extrovert in DP table.

Edge Cases

Case	How to Handle
m or n is zero	If either m or n is zero, the grid is empty, return 0 happiness immediately.
introvertsCount and extrovertsCount are zero	If both counts are zero, no one can be placed, return 0 happiness immediately.
m * n is less than introvertsCount + extrovertsCount	If the grid size is insufficient, there's no valid arrangement, return the happiness of filling the grid as much as possible while prioritizing higher happiness people.
m and n are both 1	Handle the simple case of a 1x1 grid and decide whether to place an introvert, extrovert, or no one based on which yields maximum happiness (3, 20, 0).
Large m and n values exceeding memory limits	Dynamic programming with memoization is needed and the grid size may still be too large; need a very efficient implementation to avoid memory issues.
introvertsCount or extrovertsCount are very large, potentially exceeding integer boundaries when calculating happiness	Use long data type to store intermediate happiness values and the final result to prevent integer overflow.
Only introverts and m,n large	The happiness calculation will primarily be -30 penalties and limited +3 bonus, so must explore placement strategies to mitigate penalties.
Only extroverts and m,n large	The happiness calculation will primarily be +20 bonus with some -30 penalties. Explore placement strategies to maximize bonuses and minimize penalties.

m or n is zero

How to Handle:

If either m or n is zero, the grid is empty, return 0 happiness immediately.

introvertsCount and extrovertsCount are zero

How to Handle:

If both counts are zero, no one can be placed, return 0 happiness immediately.

m * n is less than introvertsCount + extrovertsCount

How to Handle:

If the grid size is insufficient, there's no valid arrangement, return the happiness of filling the grid as much as possible while prioritizing higher happiness people.

m and n are both 1

How to Handle:

Handle the simple case of a 1x1 grid and decide whether to place an introvert, extrovert, or no one based on which yields maximum happiness (3, 20, 0).

Large m and n values exceeding memory limits

How to Handle:

Dynamic programming with memoization is needed and the grid size may still be too large; need a very efficient implementation to avoid memory issues.

introvertsCount or extrovertsCount are very large, potentially exceeding integer boundaries when calculating happiness

How to Handle:

Use long data type to store intermediate happiness values and the final result to prevent integer overflow.

Only introverts and m,n large

How to Handle:

The happiness calculation will primarily be -30 penalties and limited +3 bonus, so must explore placement strategies to mitigate penalties.

Only extroverts and m,n large

How to Handle:

The happiness calculation will primarily be +20 bonus with some -30 penalties. Explore placement strategies to maximize bonuses and minimize penalties.