Count the maximum number of ways to form a rectangle by replacing one digit in matrix?

anyone could give me some hint how to solve this problem? .............................................................................................................................................................................................................. Hey, we're trying to form rectangles over a grid of numbers, where all four corners have the same number. The rectangles must have vertical and horizontal edges (no rotated rectangles). Here's an example:

In order to help maximize the total number of these rectangles that can be formed, we're allowed to change one number on the grid.

Given a matrix of integers, grid, return the indices of the element that we should change to yield the highest number of total rectangles. If there's a tie for the maximum number of rectangles, choose the answer with the smallest row index. If there's still a tie, choose the one with the smallest column index.

Example

For

grid = [[2,2,2,0,2], [0,1,0,0,1], [2,2,2,1,2], [0,2,0,0,2], [2,0,1,1,2]] the output should be rectangleReplacement(grid) = [4, 1].

There are 5 rectangles that can be formed by replacing the 0 at grid[4][1] with a 2:![]

It's also possible to form 5 new rectangles by replacing the 0 at grid[3][0] with a 2: but replacing that 0 would destroy 2 existing rectangles with a vertex at grid[3][0]: so effectively, this would only create 3 additional rectangles, which means that [4, 1] is a better choice.

For

grid = [[2, 3], [0, 7]] the output should be rectangleReplacement(grid) = [0, 0]. In this case, we can't form any rectangles by replacing any single element, so why is the answer [0, 0]? Well to put it formally, the maximum number of total rectangles we can form is 0, and since any of the indices will produce this value, we'll choose the one with the lowest row index and the lowest column index (which is [0, 0]).
could anyone give me a general way to solve this task?

#ComputerScience

Note by Kien Huu Nguyen
3 years ago

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