algoadvance

Given an m x n matrix M initialized with all 0’s and several update operations.

Operations are represented by a 2D array ops, where each ops[i] = [ai, bi] signifies all cells M[x][y] where 0 <= x < ai and 0 <= y < bi should be incremented by 1.

Return the number of maximum integers in the matrix after performing all the operations.

Clarifying Questions

1. What should happen if ops is empty?
   • The entire matrix remains 0, so the number of maximum integers would be m * n.
2. Can ai or bi exceed m or n?
   • No, it’s expected that all operations are within the bounds of the matrix.
3. What are the constraints of m, n, and the number of operations?
   • This would help determine the expected performance and optimize accordingly.

Strategy

1. Initial Observations:
   • Each operation imposes an increment on a subrectangle from the top-left corner (0, 0) to (ai-1, bi-1).
   • Therefore, the most incremented area will be the smallest overlapping subrectangle defined by the minimum of all ai values and the minimum of all bi values from the operations.
2. Implementation Steps:
   • If there are no operations, return m * n.
   • Find the minimum ai and bi from the ops list.
   • The number of maximum integers will be the area of the rectangle defined by these minimum values, i.e., min_ai * min_bi.

Code

Here’s the Python code implementing this strategy:

def maxCount(m: int, n: int, ops: List[List[int]]) -> int:
    if not ops:
        return m * n
    
    min_ai = m
    min_bi = n
    
    for ai, bi in ops:
        if ai < min_ai:
            min_ai = ai
        if bi < min_bi:
            min_bi = bi
    
    return min_ai * min_bi

# Example usage:
print(maxCount(3, 3, [[2, 2], [3, 3]]))  # Output should be 4

Time Complexity

• Time Complexity: O(k) where k is the number of operations in ops.
  • We iterate through each operation only once to find the minimum ai and bi.
• Space Complexity: O(1)
  • Only a constant amount of additional space is used for maintaining the minimum values.

This ensures that the solution is efficient and scales well with the number of operations while keeping the implementation straightforward.

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