Is Maximum Value Sum by Placing Three Rooks I easy or hard?

Maximum Value Sum by Placing Three Rooks I is classified as a Hard problem because it requires reducing a large combinational search space. The challenge is recognizing that only a few high-value cells per row need to be considered, allowing enumeration of row triplets with efficient pruning.

Maximum Value Sum by Placing Three Rooks I Python/Java solution

Python and Java implementations typically follow the same logic: enumerate row triplets, collect the top column values for each row, and check combinations that use distinct columns. Python solutions often rely on sorting or heap operations, while Java implementations use arrays and priority queues for candidate selection.

How to solve Maximum Value Sum by Placing Three Rooks I in O(n)?

An O(n) solution is not feasible because the algorithm must evaluate combinations of rows and columns to enforce the rook constraints. The best practical strategy is O(m^3) by enumerating row triplets and testing only the highest-value columns per row. This drastically prunes the search while guaranteeing correctness.

What is the best approach for Maximum Value Sum by Placing Three Rooks I?

The most practical approach enumerates all triplets of rows and then greedily selects the best column candidates for each row while ensuring columns remain distinct. This reduces the search space significantly compared to brute force. The resulting complexity is around O(m^3) with small constant checks for column combinations.

Is Maximum Value Sum by Placing Three Rooks I asked at Google/Amazon/Meta?

Grid optimization problems with row and column constraints appear frequently in interviews at companies like Google, Amazon, and Meta. Variants involving rook placement, matrix selection, or constrained maximum sums are common because they test enumeration strategies and pruning techniques.

What data structure is used in Maximum Value Sum by Placing Three Rooks I?

The solution primarily uses arrays to store the matrix and track candidate column values per row. Some implementations maintain small lists or heaps of top elements in each row to quickly evaluate valid combinations while avoiding column conflicts.

What is the time complexity of Maximum Value Sum by Placing Three Rooks I?

A naive backtracking solution can reach O((m*n)^3) because it explores combinations of three cells across the grid. The optimized approach enumerates three rows and tests a limited set of top column candidates, reducing the complexity to roughly O(m^3). Space usage remains small, typically O(m) or constant depending on implementation.

Description Hints 2 Solution Video 22:04

3256. Maximum Value Sum by Placing Three Rooks I

Hard17.0% acceptance

You are given a m x n 2D array board representing a chessboard, where board[i][j] represents the value of the cell (i, j).

Rooks in the same row or column attack each other. You need to place three rooks on the chessboard such that the rooks do not attack each other.

Return the maximum sum of the cell values on which the rooks are placed.

Example 1:

Input: board = [[-3,1,1,1],[-3,1,-3,1],[-3,2,1,1]]

Output: 4

Explanation:

We can place the rooks in the cells (0, 2), (1, 3), and (2, 1) for a sum of 1 + 1 + 2 = 4.

Example 2:

Input: board = [[1,2,3],[4,5,6],[7,8,9]]

Output: 15

Explanation:

We can place the rooks in the cells (0, 0), (1, 1), and (2, 2) for a sum of 1 + 5 + 9 = 15.

Example 3:

Input: board = [[1,1,1],[1,1,1],[1,1,1]]

Output: 3

Explanation:

We can place the rooks in the cells (0, 2), (1, 1), and (2, 0) for a sum of 1 + 1 + 1 = 3.

Constraints:

3 <= m == board.length <= 100
3 <= n == board[i].length <= 100
-10⁹ <= board[i][j] <= 10⁹

Topics

Hints open one at a time

Hint 1

Store the largest 3 values for each row.

Hint 2

Select any 3 rows and brute force all combinations.

22:04

Still stuck?Read 2 approaches, or watch a walkthrough.

Open solution

3256. Maximum Value Sum by Placing Three Rooks I

Topics

Hints open one at a time

Similar problems