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Range Addition II

Number: 598

Difficulty: Easy

Paid? No

Companies: IXL

Problem Description

Given an m x n matrix initialized with all 0's, and an array of operations where each operation ops[i] = [ai, bi] increments all matrix cells in the top-left submatrix (0 <= x < ai and 0 <= y < bi) by 1, return the count of the maximum elements in the matrix after all operations.

Key Insights

  • The maximum value in the matrix is achieved at those cells that are incremented in every operation.
  • The overlapping area of all operations is defined by the minimum values of ai and bi across all operations.
  • If no operations are given, the entire matrix retains its initial value which is the maximum.

Space and Time Complexity

Time Complexity: O(P) where P is the number of operations. Space Complexity: O(1)

Solution

The solution hinges on recognizing that each operation increases every cell in a submatrix from (0,0) to (ai-1, bi-1). Therefore, after performing all operations, the cells that have been incremented the most (and thus are maximal) are the ones that lie in the intersection of all these submatrices. This intersection submatrix is defined by the smallest ai and the smallest bi from all operations.

Step-by-step approach:

  1. Initialize minRow = m and minCol = n.
  2. For each operation, update minRow = min(minRow, ai) and minCol = min(minCol, bi).
  3. If no operations exist, the entire matrix (m*n cells) have the same value.
  4. The result is simply the product of minRow and minCol which represents the area of the overlap.

Data structures involved include simple integer tracking for minimum values. This straightforward greedy approach avoids unnecessary iterations over the entire matrix.

Code Solutions

# Python implementation for Range Addition II
def maxCount(m: int, n: int, ops: list) -> int:
    # If no operations, the entire matrix remains unchanged with value 0
    if not ops:
        return m * n

    # Initialize the minimum row and column limits to the full matrix dimensions
    minRow, minCol = m, n

    # Iterate through each operation and update the minimum row and column boundaries
    for op in ops:
        a, b = op
        minRow = min(minRow, a)
        minCol = min(minCol, b)
    
    # The maximum value will be present in the top-left minRow x minCol sub-matrix
    return minRow * minCol

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