Problem Description

You have a matrix M with dimensions width × height. Every cell in this matrix can only contain either 0 or 1. There's a special constraint: any square sub-matrix of size sideLength × sideLength within M can have at most maxOnes ones.

Your task is to find the maximum possible number of ones that can be placed in the entire matrix M while respecting this constraint.

For example, if you have a 3×3 matrix with sideLength = 2 and maxOnes = 1, then every 2×2 square region in the matrix can contain at most 1 one. You need to strategically place ones to maximize the total count while ensuring no 2×2 square violates this rule.

The key insight is that when you place a 1 at position (i, j), it affects all square sub-matrices that include this position. The solution leverages the fact that positions at coordinates (i mod sideLength, j mod sideLength) are "equivalent" - they appear in the same relative position within their respective square regions. By counting how many times each equivalent position appears across the entire matrix and selecting the maxOnes most frequent positions, we can maximize the total number of ones.

Intuition

Think about overlaying a grid of sideLength × sideLength squares across the entire matrix. Each square can have at most maxOnes ones. The key observation is that these squares repeat in a pattern - they tile the matrix.

Imagine we have a single sideLength × sideLength template square. When we tile this template across the matrix, position (0, 0) in the template appears at positions (0, 0), (0, sideLength), (0, 2×sideLength), ... and also at (sideLength, 0), (2×sideLength, 0), etc. in the actual matrix.

Each position (i, j) in the matrix belongs to exactly one position in this template square, specifically position (i % sideLength, j % sideLength). This means that if we decide to place a 1 at position (r, c) in the template (where r < sideLength and c < sideLength), then ALL positions (i, j) in the matrix where i % sideLength = r and j % sideLength = c must also be 1.

Since we can only choose maxOnes positions in the template to set as 1, we want to choose the positions that appear most frequently when the template is tiled across the entire matrix. A position (r, c) in the template appears exactly ⌈width/sideLength⌉ × ⌈height/sideLength⌉ times (accounting for partial tiles at the edges).

By counting how many times each template position appears in the full matrix and selecting the maxOnes positions with the highest counts, we maximize the total number of ones in the matrix while respecting the constraint that each sideLength × sideLength square has at most maxOnes ones.

Learn more about Greedy, Math, Sorting and Heap (Priority Queue) patterns.

Solution Approach

The implementation follows the pattern-based counting strategy identified in our intuition. Let's denote x = sideLength for simplicity.

We create an array cnt of size x * x to represent all possible positions in our template square. Each index in this array corresponds to a unique position (i % x, j % x) in the template.

For each cell (i, j) in the matrix:

1. Calculate its corresponding position in the template: (i % x, j % x)
2. Convert this 2D position to a 1D index: k = (i % x) * x + (j % x)
3. Increment the counter at cnt[k] by 1

This counting process tells us how many times each template position appears when tiled across the entire width × height matrix.

After counting all positions:

1. Sort the cnt array in descending order to prioritize positions that appear most frequently
2. Select the top maxOnes positions (these are the positions we'll set to 1 in our template)
3. Return the sum of counts for these selected positions

The key insight is that by selecting the maxOnes most frequent positions, we maximize the total number of ones. Since each selected template position forces all its corresponding positions in the matrix to be 1, choosing the positions with highest frequency gives us the maximum possible ones.

Time complexity: O(width * height + x²log(x²)) where the first term is for counting and the second is for sorting. Space complexity: O(x²) for the counting array.

Example Walkthrough

Let's work through a concrete example with a 5×4 matrix where sideLength = 2 and maxOnes = 1.

Step 1: Understanding the Template We have a 2×2 template square with positions:

Template positions:
(0,0) (0,1)
(1,0) (1,1)

Step 2: Mapping Matrix Positions to Template Each position in our 5×4 matrix maps to a template position based on (i % 2, j % 2):

Matrix (5×4):         Template Position:
[0,0] [0,1] [0,2] [0,3]    (0,0) (0,1) (0,0) (0,1)
[1,0] [1,1] [1,2] [1,3]    (1,0) (1,1) (1,0) (1,1)
[2,0] [2,1] [2,2] [2,3]    (0,0) (0,1) (0,0) (0,1)
[3,0] [3,1] [3,2] [3,3]    (1,0) (1,1) (1,0) (1,1)
[4,0] [4,1] [4,2] [4,3]    (0,0) (0,1) (0,0) (0,1)

Step 3: Count Frequencies Now we count how many times each template position appears:

• Position (0,0): appears at matrix positions [0,0], [0,2], [2,0], [2,2], [4,0], [4,2] → 6 times
• Position (0,1): appears at matrix positions [0,1], [0,3], [2,1], [2,3], [4,1], [4,3] → 6 times
• Position (1,0): appears at matrix positions [1,0], [1,2], [3,0], [3,2] → 4 times
• Position (1,1): appears at matrix positions [1,1], [1,3], [3,1], [3,3] → 4 times

Converting to 1D array indices (using index = row * 2 + col):

• cnt[0] = 6 (for template position (0,0))
• cnt[1] = 6 (for template position (0,1))
• cnt[2] = 4 (for template position (1,0))
• cnt[3] = 4 (for template position (1,1))

Step 4: Select Top maxOnes Positions Sort counts in descending order: [6, 6, 4, 4]

Since maxOnes = 1, we select only 1 position. We can choose either position with count 6 (template positions (0,0) or (0,1)).

Step 5: Calculate Result Maximum ones possible = 6

Verification: If we choose template position (0,0), our matrix would look like:

[1] [0] [1] [0]
[0] [0] [0] [0]
[1] [0] [1] [0]
[0] [0] [0] [0]
[1] [0] [1] [0]

Every 2×2 square contains exactly 1 one:

• Square at [0,0]: contains [0,0]=1, others=0 ✓
• Square at [0,1]: contains [0,2]=1, others=0 ✓
• Square at [1,0]: contains [2,0]=1, others=0 ✓
• And so on...

The constraint is satisfied, and we've placed the maximum possible 6 ones.

Solution Implementation

Time and Space Complexity

Time Complexity: O(width × height)

The algorithm iterates through every position in the matrix exactly once using two nested loops - the outer loop runs width times and the inner loop runs height times. For each position (i, j), it performs constant time operations: calculating the modulo, computing index k, and incrementing a counter. The sorting operation takes O(sideLength² × log(sideLength²)) time, but since sideLength ≤ min(width, height), this is bounded by O(width × height). Therefore, the overall time complexity is O(width × height), which matches the reference answer's O(m × n) where m and n represent the dimensions of the matrix.

Space Complexity: O(sideLength²)

The algorithm uses an array cnt of size sideLength × sideLength to store the frequency counts for each unique position within a sideLength × sideLength subgrid. This is the dominant space usage in the algorithm, as all other variables use constant space. Therefore, the space complexity is O(sideLength²).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Confusion Between Matrix Dimensions and Iteration Order

A critical pitfall in this solution is mixing up the iteration variables with matrix dimensions. The problem states the matrix has dimensions width × height, which typically means width columns and height rows. However, the code iterates using:

for row in range(width):
    for col in range(height):

This treats width as the number of rows and height as the number of columns, which is counterintuitive and could lead to incorrect results if the matrix is not square.

Solution: Use clearer variable naming and iterate consistently with the matrix dimensions:

# Iterate through all cells in the height × width matrix
for row in range(height):
    for col in range(width):
        pattern_position = (row % sideLength) * sideLength + (col % sideLength)
        position_frequencies[pattern_position] += 1

2. Incorrect Pattern Position Calculation

Another potential pitfall is incorrectly mapping 2D coordinates to 1D array indices. The formula (row % sideLength) * sideLength + (col % sideLength) assumes row-major ordering, but developers might accidentally use column-major ordering or mix up the formula.

Solution: Add validation or use a helper function to make the mapping explicit:

def get_pattern_index(row, col, sideLength):
    """Convert 2D pattern position to 1D array index."""
    pattern_row = row % sideLength
    pattern_col = col % sideLength
    return pattern_row * sideLength + pattern_col

3. Edge Case: maxOnes Greater Than sideLength²

If maxOnes > sideLength * sideLength, the code would work correctly but it's worth noting that this means we can fill the entire matrix with ones since the constraint becomes meaningless.

Solution: Add a optimization check at the beginning:

if maxOnes >= sideLength * sideLength:
    return width * height  # Can place 1s everywhere

4. Integer Overflow in Large Inputs

For very large matrices, the frequency counts could potentially overflow in languages with fixed-size integers (though Python handles this automatically).

Solution: In Python this isn't an issue, but in other languages, ensure you're using appropriate data types for large values or add bounds checking.