How do you solve Maximum Value Sum by Placing Three Rooks I on LeetCode?

Maximum Value Sum by Placing Three Rooks I (LeetCode #3256) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m^3) time complexity and O(m) space complexity. Key insight: Rooks cannot attack each other if they are in different rows and columns.

What is the brute force solution for Maximum Value Sum by Placing Three Rooks I?

The brute force approach for Maximum Value Sum by Placing Three Rooks I is Brute Force, with O(m^3 * n^3) time complexity and O(1) space complexity. We can try every possible combination of placing three rooks on the board. Since rooks cannot attack each other, we need to ensure they are in different rows and columns. This approach is straightforward but inefficient for larger boards. The optimal Optimal Solution approach improves this to O(m^3).

What algorithm or data structure is used to solve Maximum Value Sum by Placing Three Rooks I?

Maximum Value Sum by Placing Three Rooks I uses the following concepts: Array, Dynamic Programming, Matrix, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Value Sum by Placing Three Rooks I?

Think about how to reduce the problem size before diving into code. Always check edge cases, like negative values or small boards. Explain your thought process clearly while coding.

What are common mistakes when solving Maximum Value Sum by Placing Three Rooks I?

Not considering the column indices when selecting values. Overlooking negative values in the board.

#3256

Maximum Value Sum by Placing Three Rooks I

Hard

ArrayDynamic ProgrammingMatrixEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(m^3)Space O(m)

Instead of checking all combinations, we can first find the top three values in each row and store them. Then, we can iterate through combinations of three rows and calculate the maximum sum using pre-stored values, significantly reducing the number of calculations.

⚙️

Algorithm

3 steps

1Step 1: For each row, find the top three values and their respective column indices.
2Step 2: Store these values in a list of tuples for easy access.
3Step 3: Iterate through all combinations of three rows and calculate the maximum sum using the stored values, ensuring no column indices overlap.

solution.py17 lines

1from itertools import combinations
2
3def maxRookSum(board):
4    m, n = len(board), len(board[0])
5    top_values = []
6    for row in board:
7        values = sorted([(val, col) for col, val in enumerate(row)], reverse=True)[:3]
8        top_values.append(values)
9    max_sum = float('-inf')
10    for rows in combinations(range(m), 3):
11        for i in range(3):
12            for j in range(3):
13                for k in range(3):
14                    if top_values[rows[0]][i][1] != top_values[rows[1]][j][1] and top_values[rows[0]][i][1] != top_values[rows[2]][k][1] and top_values[rows[1]][j][1] != top_values[rows[2]][k][1]:
15                        current_sum = top_values[rows[0]][i][0] + top_values[rows[1]][j][0] + top_values[rows[2]][k][0]
16                        max_sum = max(max_sum, current_sum)
17    return max_sum

ℹ

Complexity note: This complexity is more efficient because we only calculate the top three values per row once, and then we only iterate through combinations of rows, significantly reducing the number of calculations.

1Rooks cannot attack each other if they are in different rows and columns.
2Finding the top values in each row reduces the number of combinations we need to check.

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