Problem Description

You start with an m x n matrix filled with all zeros. You're given a list of operations ops, where each operation ops[i] = [ai, bi] tells you to add 1 to all cells in the rectangle from the top-left corner (0, 0) to position (ai-1, bi-1) inclusive.

In other words, for each operation [ai, bi], you increment every cell M[x][y] by 1 where 0 <= x < ai and 0 <= y < bi.

After performing all the operations, some cells will have higher values than others. Your task is to find how many cells contain the maximum value in the final matrix.

For example, if you have a 3x3 matrix and perform operation [2, 2], you would add 1 to the top-left 2x2 submatrix. After multiple operations, certain cells that are included in all operations will have the highest values.

The key insight is that cells that get incremented by every single operation will have the maximum value. These cells form a rectangle starting from (0, 0) with dimensions equal to the minimum ai and minimum bi across all operations. The solution finds these minimum dimensions and returns their product, which gives the count of cells with the maximum value.

If no operations are performed (empty ops array), all cells remain at 0, so all m * n cells have the maximum value.

Intuition

Let's think about what happens when we perform these operations. Each operation [ai, bi] adds 1 to a rectangle starting from (0, 0) to (ai-1, bi-1).

The key observation is that all operations start from the same top-left corner (0, 0). This means the operations create overlapping rectangles, all anchored at the origin.

Which cells will have the maximum value after all operations? The cells that are included in every single operation. These are the cells that get incremented the most times - once for each operation.

Since all operations start from (0, 0), the intersection of all these rectangles is also a rectangle starting from (0, 0). The dimensions of this intersection rectangle are determined by the smallest ai and smallest bi values across all operations.

Think of it this way: if we have operations [3, 3], [2, 4], and [4, 2]:

• The first operation covers a 3×3 area
• The second operation covers a 2×4 area
• The third operation covers a 4×2 area

The overlap of all three is a 2×2 area (minimum of 3, 2, 4 for rows = 2; minimum of 3, 4, 2 for columns = 2). Only the cells in this 2×2 area get incremented by all three operations, so they'll have the maximum value of 3.

This is why the solution simply finds min(all ai values) and min(all bi values), then multiplies them together to get the count of cells with maximum value. The product m * n gives us the area of the intersection rectangle, which equals the number of cells with the maximum value.

Learn more about Math patterns.

Solution Approach

The implementation is remarkably simple once we understand the problem's nature. We need to find the intersection of all operation rectangles, which is determined by the minimum dimensions across all operations.

Here's the step-by-step approach:

1. Initialize with matrix dimensions: Start with m and n as the initial row and column bounds. These represent the maximum possible intersection area if no operations are performed.

2. Find minimum dimensions: Iterate through each operation [a, b] in the ops array:

   • Update m = min(m, a) to track the smallest row dimension
   • Update n = min(n, b) to track the smallest column dimension

3. Calculate the result: Return m * n, which gives the area of the intersection rectangle.

The algorithm works because:

• Each operation covers a rectangle from (0, 0) to (a-1, b-1)
• The intersection of all these rectangles is from (0, 0) to (min_a - 1, min_b - 1)
• The number of cells in this intersection is min_a * min_b

Edge Case: When the ops array is empty, no operations are performed, so all cells remain at value 0. The function correctly returns m * n since the initial values aren't modified.

Time Complexity: O(k) where k is the length of the ops array, as we iterate through it once.

Space Complexity: O(1) as we only use two variables to track the minimum dimensions.

The beauty of this solution is that we don't actually need to create or modify the matrix - we can determine the answer just by analyzing the operation boundaries.

Example Walkthrough

Let's walk through a concrete example with a 3×3 matrix and operations [[2,2], [3,3], [3,2]].

Initial Setup:

• Matrix dimensions: m = 3, n = 3
• All cells start at 0

Operation 1: [2,2]

• Adds 1 to rectangle from (0,0) to (1,1)
• Affects a 2×2 area
• Matrix after operation:

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

Operation 2: [3,3]

• Adds 1 to rectangle from (0,0) to (2,2)
• Affects a 3×3 area (entire matrix)
• Matrix after operation:

[2, 2, 1]
[2, 2, 1]
[1, 1, 1]

Operation 3: [3,2]

• Adds 1 to rectangle from (0,0) to (2,1)
• Affects a 3×2 area
• Final matrix:

[3, 3, 1]
[3, 3, 1]
[2, 2, 1]

Finding the Maximum:

• Maximum value in the matrix = 3
• Cells with value 3 form a 2×2 rectangle at the top-left

Using Our Algorithm: Instead of building the matrix, we find:

• min(2, 3, 3) = 2 (minimum row dimension)
• min(2, 3, 2) = 2 (minimum column dimension)
• Result = 2 × 2 = 4 cells with maximum value

The cells at positions (0,0), (0,1), (1,0), and (1,1) are included in all three operations, so they have the maximum value of 3. Our algorithm correctly identifies that there are 4 such cells.

Solution Implementation

Time and Space Complexity

The time complexity is O(k), where k is the length of the operations array ops. This is because the algorithm iterates through the ops list exactly once, performing constant-time operations (minimum comparisons and assignments) for each operation pair [a, b].

The space complexity is O(1). The algorithm only uses two variables m and n to track the minimum dimensions, which are modified in place. No additional data structures are created that scale with the input size, resulting in constant space usage.

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

Common Pitfalls

1. Misunderstanding the Operation Indices

A common mistake is confusing the operation parameters [ai, bi] with the actual cell indices. The operation [ai, bi] affects cells from (0, 0) to (ai-1, bi-1) inclusive, meaning it covers ai rows and bi columns, not the cell at position (ai, bi).

Incorrect interpretation:

# Wrong: Thinking [2, 2] only affects cell at position (2, 2)
# or affects cells from (0, 0) to (2, 2) which would be 3x3

Correct understanding:

# Right: [2, 2] affects cells from (0, 0) to (1, 1), which is a 2x2 area

2. Handling Empty Operations Array Incorrectly

Some implementations might not properly handle the case when ops is empty, potentially returning 0 or causing an error.

Problematic approach:

def maxCount(self, m: int, n: int, ops: List[List[int]]) -> int:
    if not ops:
        return 0  # Wrong! Should return m * n
  
    min_row = float('inf')
    min_col = float('inf')
    for a, b in ops:
        min_row = min(min_row, a)
        min_col = min(min_col, b)
    return min_row * min_col  # This would return inf * inf if ops is empty

Correct approach:

def maxCount(self, m: int, n: int, ops: List[List[int]]) -> int:
    min_row = m  # Initialize with matrix dimensions
    min_col = n
    for a, b in ops:
        min_row = min(min_row, a)
        min_col = min(min_col, b)
    return min_row * min_col  # Returns m * n when ops is empty

3. Actually Creating the Matrix

A significant inefficiency is actually creating and updating the matrix when you only need to find the intersection area.

Inefficient approach:

def maxCount(self, m: int, n: int, ops: List[List[int]]) -> int:
    # Unnecessarily creates and modifies the entire matrix
    matrix = [[0] * n for _ in range(m)]
    for a, b in ops:
        for i in range(a):
            for j in range(b):
                matrix[i][j] += 1
  
    # Then counts maximum values
    max_val = max(max(row) for row in matrix)
    return sum(1 for row in matrix for val in row if val == max_val)

This approach has O(mnk) time complexity and O(m*n) space complexity, where k is the number of operations, making it inefficient for large matrices.

4. Off-by-One Errors with Boundaries

Confusion about whether the operation boundaries are inclusive or exclusive can lead to errors.

Potential mistake:

# Thinking you need to subtract 1 from the result
return (min_row - 1) * (min_col - 1)  # Wrong!

The operation [ai, bi] already represents the count of rows and columns (ai rows from 0 to ai-1), so no adjustment is needed in the final calculation.