How do you solve Maximize Grid Happiness on LeetCode?

Maximize Grid Happiness (LeetCode #1659) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n * 3^(n)) time complexity and O(m * introvertsCount * extrovertsCount * 2^n) space complexity. Key insight: Understanding the happiness calculation is crucial for optimizing placements.

What is the brute force solution for Maximize Grid Happiness?

The brute force approach for Maximize Grid Happiness is Brute Force, with O(3^(m*n)) time complexity and O(1) space complexity. The brute-force approach explores every possible configuration of introverts and extroverts in the grid. It calculates the total happiness for each configuration and keeps track of the maximum happiness found. The optimal Optimal Solution approach improves this to O(m * n * 3^(n)).

What algorithm or data structure is used to solve Maximize Grid Happiness?

Maximize Grid Happiness uses the following concepts: Dynamic Programming, Bit Manipulation, Memoization, Bitmask. The recommended patterns to study are: Dynamic Programming, Bitmasking.

What are the interview tips for Maximize Grid Happiness?

Clearly explain your thought process and how you arrived at your solution. Discuss the trade-offs of different approaches, especially between brute force and dynamic programming. Practice explaining your code as if you were teaching someone else.

What are common mistakes when solving Maximize Grid Happiness?

Failing to account for neighbors when calculating happiness. Not using memoization effectively, leading to redundant calculations.

#1659

Maximize Grid Happiness

Hard

Dynamic ProgrammingBit ManipulationMemoizationBitmaskDynamic ProgrammingBitmasking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(3^(m*n))	O(m * n * 3^(n))
Space	O(1)	O(m * introvertsCount * extrovertsCount * 2^n)

💡

Intuition

Time O(m * n * 3^(n))Space O(m * introvertsCount * extrovertsCount * 2^n)

The optimal solution uses dynamic programming with bitmasking to efficiently explore configurations of the grid. It keeps track of the maximum happiness by considering the state of each row and the remaining introverts and extroverts.

⚙️

Algorithm

3 steps

1Step 1: Define a DP function that takes the current row, remaining introverts, remaining extroverts, and the previous row's configuration.
2Step 2: For each cell in the current row, decide whether to place an introvert, an extrovert, or leave it empty.
3Step 3: Calculate the happiness based on the current configuration and recursively call the DP function for the next row.

solution.py19 lines

1def maxGridHappiness(m, n, introvertsCount, extrovertsCount):
2    dp = {}  # memoization
3    def dfs(row, introverts_left, extroverts_left, prev_mask):
4        if row == m:
5            return 0
6        if (row, introverts_left, extroverts_left, prev_mask) in dp:
7            return dp[(row, introverts_left, extroverts_left, prev_mask)]
8        max_happiness = 0
9        for mask in range(1 << n):
10            happiness = calculate_happiness(mask, prev_mask, introverts_left, extroverts_left)
11            if happiness != -1:
12                max_happiness = max(max_happiness, happiness + dfs(row + 1, introverts_left, extroverts_left, mask))
13        dp[(row, introverts_left, extroverts_left, prev_mask)] = max_happiness
14        return max_happiness
15
16    return dfs(0, introvertsCount, extrovertsCount, 0)
17
18# Helper function to calculate happiness based on mask
19...

ℹ

Complexity note: The complexity is reduced by using bitmasking to represent the state of each row, allowing us to efficiently calculate happiness without generating all configurations.

1Understanding the happiness calculation is crucial for optimizing placements.
2Dynamic programming can significantly reduce the number of configurations we need to evaluate.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.