#1659

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Maximize Grid Happiness

Maximize Grid Happiness is a dynamic programming problem focusing on state transitions with bitmasking to maximize happiness for introverts and extroverts.

dynamic programming bit manipulation memoization bitmask

Problem Statement

You are given a grid with dimensions m x n, where m is the number of rows and n is the number of columns. There are two types of people: introverts and extroverts. You also have the count of introverts (introvertsCount) and extroverts (extrovertsCount). You are tasked with placing these people in the grid in such a way that maximizes the total happiness of all people.

The happiness of each person depends on the type of their neighbors. An introvert starts with 120 happiness and loses 30 for each neighbor. An extrovert starts with 40 happiness and gains 20 for each neighbor. The task is to compute the maximum happiness achievable with the given m, n, introvertsCount, and extrovertsCount.

Examples

Example 1

Input: m = 2, n = 3, introvertsCount = 1, extrovertsCount = 2

Output: 240

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

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

Example details omitted.

Constraints

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

Solution Approach

Dynamic Programming with State Transitions

The solution revolves around dynamic programming and bitmasking to track different placements of introverts and extroverts across the grid. For each possible configuration, calculate the total happiness by considering the starting happiness and the effect of neighbors. Using state transitions, store the results of subproblems to avoid redundant calculations.

Bitmasking for Grid Configuration

Bitmasking is used to represent the state of each grid cell—whether it is empty, holds an introvert, or holds an extrovert. This allows for efficient storage and evaluation of all possible placements of people in the grid.

Memoization to Optimize Repeated States

Memoization is employed to store previously computed results for grid configurations, reducing the time complexity. This ensures that redundant state evaluations are avoided, significantly improving the performance of the algorithm.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is primarily influenced by the number of possible grid configurations, which is exponential in nature due to the use of bitmasking for each cell's state. Space complexity is similarly impacted by the storage of the dynamic programming table, which tracks the happiness values for each configuration.

What Interviewers Usually Probe

Look for understanding of dynamic programming and state transition problems.
Ensure the candidate is comfortable with bitmasking as a technique to represent grid states.
Check if the candidate can explain how memoization is used to optimize repeated states and why it is necessary for this problem.

Common Pitfalls or Variants

Common pitfalls

Overlooking the need for efficient state transitions, which may lead to an exponential time complexity.
Not properly handling the grid boundaries when calculating neighbors.
Failing to optimize the solution using memoization, which could result in redundant calculations.

Follow-up variants

Consider grids of different sizes beyond the given constraints to test the scalability of the approach.
Introduce a higher number of introverts and extroverts to examine the performance with larger inputs.
Modify the happiness formula or neighbor effect to create variations in difficulty and complexity.

FAQ

What is the primary technique to solve the Maximize Grid Happiness problem?

The problem is primarily solved using dynamic programming with state transitions and bitmasking to represent different grid configurations.

How does memoization help in solving the Maximize Grid Happiness problem?

Memoization optimizes the solution by storing previously computed grid configurations, which prevents redundant recalculations and speeds up the process.

Can the Maximize Grid Happiness problem be solved without dynamic programming?

Solving this problem without dynamic programming would likely be inefficient due to the large number of possible configurations. DP ensures an optimal solution.

How do you handle grid boundaries when calculating happiness in Maximize Grid Happiness?

You need to carefully check if the neighbors of each grid cell exist, especially when the cell is at the grid's edge, to avoid out-of-bounds errors.

What is the main challenge when solving the Maximize Grid Happiness problem?

The main challenge is efficiently managing the large number of possible grid configurations and maximizing happiness, which requires dynamic programming and memoization.

terminal

Solution

Solution 1: Ternary State Compression + Memoization

We notice that in the problem, $1 \leq m, n \leq 5$, and each grid cell has three states: no personnel assigned, introverted personnel assigned, and extroverted personnel assigned. Therefore, we can use $0$, $1$, $2$ to represent these three states, and each row in the grid can be represented by a ternary number of length $n$.

class Solution:
    def getMaxGridHappiness(
        self, m: int, n: int, introvertsCount: int, extrovertsCount: int
    ) -> int:
        @cache
        def dfs(i: int, pre: int, ic: int, ec: int) -> int:
            if i == m or (ic == 0 and ec == 0):
                return 0
            ans = 0
            for cur in range(mx):
                if ix[cur] <= ic and ex[cur] <= ec:
                    a = f[cur] + g[pre][cur]
                    b = dfs(i + 1, cur, ic - ix[cur], ec - ex[cur])
                    ans = max(ans, a + b)
            return ans

        mx = pow(3, n)
        f = [0] * mx
        g = [[0] * mx for _ in range(mx)]
        h = [[0, 0, 0], [0, -60, -10], [0, -10, 40]]
        bits = [[0] * n for _ in range(mx)]
        ix = [0] * mx
        ex = [0] * mx
        for i in range(mx):
            mask = i
            for j in range(n):
                mask, x = divmod(mask, 3)
                bits[i][j] = x
                if x == 1:
                    ix[i] += 1
                    f[i] += 120
                elif x == 2:
                    ex[i] += 1
                    f[i] += 40
                if j:
                    f[i] += h[x][bits[i][j - 1]]
        for i in range(mx):
            for j in range(mx):
                for k in range(n):
                    g[i][j] += h[bits[i][k]][bits[j][k]]
        return dfs(0, 0, introvertsCount, extrovertsCount)

Solution 2: Contour Line Memorized Search

We can consider searching each grid cell, each time searching a position $(i, j)$, we denote $pos = i \times n + j$. Then its left and upper adjacent grids will affect their happiness contribution.

class Solution:
    def getMaxGridHappiness(
        self, m: int, n: int, introvertsCount: int, extrovertsCount: int
    ) -> int:
        @cache
        def dfs(i: int, pre: int, ic: int, ec: int) -> int:
            if i == m or (ic == 0 and ec == 0):
                return 0
            ans = 0
            for cur in range(mx):
                if ix[cur] <= ic and ex[cur] <= ec:
                    a = f[cur] + g[pre][cur]
                    b = dfs(i + 1, cur, ic - ix[cur], ec - ex[cur])
                    ans = max(ans, a + b)
            return ans

        mx = pow(3, n)
        f = [0] * mx
        g = [[0] * mx for _ in range(mx)]
        h = [[0, 0, 0], [0, -60, -10], [0, -10, 40]]
        bits = [[0] * n for _ in range(mx)]
        ix = [0] * mx
        ex = [0] * mx
        for i in range(mx):
            mask = i
            for j in range(n):
                mask, x = divmod(mask, 3)
                bits[i][j] = x
                if x == 1:
                    ix[i] += 1
                    f[i] += 120
                elif x == 2:
                    ex[i] += 1
                    f[i] += 40
                if j:
                    f[i] += h[x][bits[i][j - 1]]
        for i in range(mx):
            for j in range(mx):
                for k in range(n):
                    g[i][j] += h[bits[i][k]][bits[j][k]]
        return dfs(0, 0, introvertsCount, extrovertsCount)

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