Solution Explanation for Maximize Grid Happiness

This problem involves finding the optimal arrangement of introverts and extroverts in a grid to maximize their total happiness. The happiness of each individual depends on their neighbors. The solution uses dynamic programming with memoization to efficiently explore the state space.

Key Ideas:

1. State Representation: The core idea is representing the grid state efficiently. Since each cell can be empty, have an introvert, or have an extrovert, a ternary number system is used. Each row is represented as a base-3 number, where 0 means empty, 1 means introvert, and 2 means extrovert. This allows for concise representation and traversal of the search space.

2. Memoization: To avoid redundant computations, dynamic programming with memoization is used. The dfs function stores previously computed results to avoid recalculating them.

3. Transition Calculation: The dfs function calculates the happiness for each possible state transition. It considers the current row's configuration and its effect on the happiness of the current individuals and their interaction with the neighbors (above and to the left).

4. Base Cases: The base cases for the dfs function are when all rows are filled or when all people are placed.

Time Complexity Analysis:

• Solution 1 (Ternary State Compression): The time complexity is O(32n * (m * ic * ec + n)). The 3n represents the possible states for a single row, and 3n again for the previous row. The (m * ic * ec + n) factor comes from iterating through rows, people, and columns respectively during the recursive calls.

• Solution 2 (Contour Line Memorized Search): The time complexity is O(3n+1 * m * n * ic * ec). This solution explores states cell by cell. The 3n+1 comes from the possible states of the current cell and its neighbors. The rest (m * n * ic * ec) factor represents iterations during the recursive calls.

Space Complexity Analysis:

• Solution 1: O(32n + 3n * m * ic * ec). The first part is from the memoization table, and the second from the space used by the other variables during the recursion.

• Solution 2: O(3n * m * n * ic * ec). This is dominated by the size of the memoization table.

Code Explanation (Python3 - Solution 1):

class Solution:
    def getMaxGridHappiness(self, m, n, introvertsCount, extrovertsCount):
        @cache  # This is a Python decorator for memoization
        def dfs(i, pre, ic, ec):
            # Base Cases
            if i == m or (ic == 0 and ec == 0):
                return 0
            ans = 0
            for cur in range(mx): # Iterate through possible row states
                if ix[cur] <= ic and ex[cur] <= ec: # Check constraints (enough people left)
                    ans = max(ans, f[cur] + g[pre][cur] + dfs(i + 1, cur, ic - ix[cur], ec - ex[cur]))
            return ans
 
        # Preprocessing: Calculate happiness contributions and counts
        mx = 3**n
        f = [0] * mx
        g = [[0] * mx for _ in range(mx)]
        # ... (rest of the preprocessing which calculates f and g, as shown in the original response)...
 
        return dfs(0, 0, introvertsCount, extrovertsCount)
 

The code first preprocesses to compute f (initial happiness of each row state) and g (interaction happiness between adjacent rows). Then the recursive dfs function efficiently explores all possible states, using memoization to speed up the process. The base cases handle situations where all rows are filled or all people are placed.

The Java, C++, and TypeScript solutions follow a similar structure, utilizing memoization and optimized state representations to solve the problem efficiently. They differ slightly in syntax and data structure handling but share the core dynamic programming logic.