Range Addition Ii - Leetcode Solution

Solution:

The problem is quite straightforward. We start with a matrix of size m * n, where all elements are initialized as 0. Then we are given a list of operations, each of which consists of two integers a and b. The operation adds 1 to all elements in the submatrix of size a * b located at the top-left corner of the matrix.

Our task is to perform all operations and return the number of cells in the matrix that contain the maximum value after all operations are performed.

Algorithm:

1. Initialize the matrix of size m * n with all elements as 0.
2. For each operation (a, b), add 1 to all elements in the submatrix of size a * b located at the top-left corner of the matrix.
3. Count the number of cells in the matrix that contain the maximum value. This can be done by finding the maximum value and then traversing through the matrix to count the number of cells that contain this value.

Code:

Here is the Python implementation of the above algorithm:

class Solution:
    def maxCount(self, m: int, n: int, ops: List[List[int]]) -> int:
        # Initialize the matrix with all elements as 0
        matrix = [[0]*n for _ in range(m)]
        
        # Perform all operations
        for op in ops:
            for i in range(op[0]):
                for j in range(op[1]):
                    matrix[i][j] += 1
        
        # Find the maximum value in the matrix
        max_val = max([max(row) for row in matrix])
        
        # Count the number of cells that contain the maximum value
        count = 0
        for row in matrix:
            count += row.count(max_val)
        
        return count

Time Complexity:

The time complexity of the above algorithm is O(m * n * k), where k is the number of operations. This is because we traverse through the matrix for each operation. However, since we are only traversing through the submatrix of size a * b for each operation, the actual time complexity may be much less than O(m * n * k) in practice.

Space Complexity:

The space complexity of the above algorithm is O(m * n), which is the size of the matrix.

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