Solution Explanation: Maximum Number of Ones

This problem asks to find the maximum number of ones that can be placed in a matrix of size width x height such that any square submatrix of size sideLength x sideLength contains at most maxOnes ones. The solution uses a clever observation about equivalent positions within the matrix.

Core Idea:

The key insight is that positions within the matrix that are equivalent modulo sideLength contribute equally to the number of ones in any sideLength x sideLength square submatrix.

Let's illustrate with an example: width = 5, height = 4, sideLength = 2, maxOnes = 2.

Consider a 2 x 2 submatrix. If we place a '1' at (0, 0), it's equivalent to placing a '1' at (0, 2), (2, 0), (2, 2), (4, 0), (4, 2), and so on. Similarly, (0, 1) is equivalent to (0, 3), (2, 1), (2, 3), (4, 1), (4, 3), etc. (1, 0) is equivalent to (1, 2), (3, 0), (3, 2), and (1, 1) to (1, 3), (3, 1), (3, 3).

Therefore, we only need to count the number of occurrences of each of these equivalent positions and determine how many ones can be placed in these positions while satisfying the maxOnes constraint per sideLength x sideLength submatrix.

Algorithm:

1. Count Equivalent Positions: Create an array cnt of size sideLength * sideLength. Iterate through each cell (i, j) of the matrix. The index k = (i % sideLength) * sideLength + (j % sideLength) represents the equivalent position. Increment cnt[k] to count the number of occurrences of this equivalent position.

2. Sort and Sum: Sort cnt in descending order. The top maxOnes elements represent the equivalent positions that contribute the most ones to the matrix. Sum these top maxOnes elements to get the maximum number of ones.

Time and Space Complexity:

• Time Complexity: O(width * height + sideLength² * log(sideLength²)). The dominant factor is iterating through the matrix (O(width * height)) and sorting the cnt array (O(sideLength² * log(sideLength²))). Since sideLength is much smaller than width and height in most cases, the overall complexity is effectively O(width * height).

• Space Complexity: O(sideLength²). The cnt array is the primary space used, with size sideLength * sideLength.

Code Examples (Python, Java, C++, Go, Javascript):

The code examples provided in the original response accurately implement this algorithm in several programming languages. They follow the steps outlined above: counting equivalent positions, sorting the counts, and summing the top maxOnes counts to get the final result. The time and space complexities as described above apply to all of these implementations.