LeetCode 1659: Maximize Grid Happiness Solution

Master LeetCode problem 1659 (Maximize Grid Happiness), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1659. Maximize Grid Happiness

Hard

Dynamic Programming Bit Manipulation Memoization Bitmask

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming with bitmasking to keep track of the state of the grid and the types of people assigned to each cell. We can iterate through all possible combinations of assigning introverts and extroverts to each cell and calculate the happiness accordingly.

The key idea is to define a state for each cell indicating whether it is empty, assigned to an introvert, or assigned to an extrovert. We can then iterate through all possible states of the grid and for each state, calculate the happiness based on the neighbors and the type of person assigned to each cell.

By memoizing the results for each state, we can avoid recalculating the same states multiple times, which helps in reducing the time complexity of the solution.

Time Complexity:

The time complexity of this solution is O(3^(mn) * m * n) where 3^(mn) represents the number of possible states for each cell in the grid.

Space Complexity:

The space complexity of this solution is O(3^(m*n)) to store the memoization table.

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Solution Code

class Solution {
    public int getMaxGridHappiness(int m, int n, int introvertsCount, int extrovertsCount) {
        int[][][] dp = new int[m * n + 1][introvertsCount + 1][extrovertsCount + 1];
        return dfs(0, m, n, introvertsCount, extrovertsCount, 0, dp);
    }
    
    private int dfs(int idx, int m, int n, int intro, int extro, int state, int[][][] dp) {
        if (idx == m * n) {
            return 0;
        }
        
        if (dp[idx][intro][extro] != 0) {
            return dp[idx][intro][extro];
        }
        
        int r = idx / n, c = idx % n;
        int res = 0;
        
        // Skip assigning anyone to this cell
        res = dfs(idx + 1, m, n, intro, extro, state / 3, dp);
        
        // Assign introvert
        if (intro > 0) {
            int cur = 120;
            if (r > 0 && (state % 3) == 1) cur -= 30;
            if (c > 0 && (state % 3) == 1) cur -= 30;
            
            res = Math.max(res, cur + dfs(idx + 1, m, n, intro - 1, extro, (state * 3) % 9 + 1, dp));
        }
        
        // Assign extrovert
        if (extro > 0) {
            int cur = 40;
            if (r > 0 && (state % 3) == 2) cur -= 30;
            if (c > 0 && (state % 3) == 2) cur -= 30;
            
            res = Math.max(res, cur + dfs(idx + 1, m, n, intro, extro - 1, (state * 3) % 9 + 2, dp));
        }
        
        dp[idx][intro][extro] = res;
        return res;
    }
}

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