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

Maximum Value Sum by Placing Three Rooks II (LeetCode #3257) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Precomputing values reduces redundant calculations.

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

The brute force approach for Maximum Value Sum by Placing Three Rooks II is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try every possible combination of placing three rooks on the board. This means checking all sets of three rows and three columns to find the maximum sum of the values where the rooks are placed. The optimal Optimal Solution approach improves this to O(n).

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

Maximum Value Sum by Placing Three Rooks II 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 II?

Always clarify the constraints and edge cases. Discuss your thought process and approach before jumping into coding. Test your solution with multiple test cases.

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

Not considering the column restrictions when placing rooks. Overlooking negative values in the board.

#3257

Maximum Value Sum by Placing Three Rooks II

Hard

ArrayDynamic ProgrammingMatrixEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of checking all combinations, we can precompute the top three values in each row, allowing us to quickly evaluate potential rook placements without redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: For each row, find the top three values and their respective column indices.
2Step 2: Use a nested loop to iterate through the top three values of three different rows, ensuring no two rooks are in the same column.
3Step 3: Calculate the sum of the selected values and keep track of the maximum sum.

solution.py14 lines

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

ℹ

Complexity note: The time complexity is O(n) because we only need to find the top three values in each row, and the nested loops are limited to three iterations each, making it efficient.

1Precomputing values reduces redundant calculations.
2Using combinations wisely can optimize the search.

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