Range Addition II

Problem Statement

You are provided with a matrix M of dimensions m x n, initially filled with zeros. Accompanying the matrix is an array of operations, ops, where each element of ops is a pair [ai, bi]. Each operation signifies that every element M[x][y] should be incremented by one in the submatrix bounded by rows 0 to ai - 1 and columns 0 to bi - 1. Your task is to determine the number of cells in the matrix M that have the maximum integer value after all the operations in ops have been applied.

Examples

Example 1

Input:

m = 3, n = 3, ops = [[2,2],[3,3]]

Output:

4

Explanation:

The maximum integer in M is 2, and there are four of it in M. So return 4.

Example 2

Input:

m = 3, n = 3, ops = [[2,2],[3,3],[3,3],[3,3],[2,2],[3,3],[3,3],[3,3],[2,2],[3,3],[3,3],[3,3]]

Output:

4

Example 3

Input:

m = 3, n = 3, ops = []

Output:

9

Constraints

• 1 <= m, n <= 4 * 104
• 0 <= ops.length <= 104
• ops[i].length == 2
• 1 <= ai <= m
• 1 <= bi <= n

Approach and Intuition

Understanding the effect of each operation and the constraints is crucial. From the given examples, we can derive a notable intuition:

1. Each operation [ai, bi] affects a rectangular area from the top left corner (0,0) to (ai-1, bi-1) in the matrix.

2. For each operation, the number of cells affected depends directly on the values of ai and bi. Therefore, the smallest values of ai and bi across all operations will determine the dimensions of the most frequently incremented submatrix because:

   • Every operation will contribute to overlapping increments in this submatrix of dimensions determined by the smallest ai and bi.

   • Particularly, if the smallest values of ai from any operation is min_ai and for bi is min_bi, then the top-left submatrix of dimension min_ai x min_bi will be incremented the most number of times.

3. If no operations are provided (ops is empty), the maximum value in the matrix remains 0, thus each cell still retains the original value. Therefore, the count of maximum integers should be the total number of cells m * n.

From these observations, the steps to find the number of maximum integers in the matrix after applying all operations are straightforward:

1. Initialize min_ai and min_bi with m and n respectively to capture the minimum extents of increments from all operations.
2. For each operation in ops, update min_ai and min_bi with the minimum of current and provided ai and bi values.
3. Calculate the product min_ai * min_bi, which gives the area of the submatrix that achieves the maximum incremented value, and is hence the answer.

This approach is efficient even for the maximum input sizes due to its linear traversal of the operations list and constant time complexity for decision-making and calculation.

Solutions

public class Solution {
    public int maximumArea(int x, int y, int[][] operations) {
        for (int[] operation : operations) {
            x = Math.min(x, operation[0]);
            y = Math.min(y, operation[1]);
        }
        return x * y;
    }
}