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Leetcode 598. Range Addition II Sure, let’s go through the problem step by step.

Problem Statement

The problem is as follows:

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

Each operation is represented by a tuple (a, b), which means all cells (i, j) for 0 <= i < a and 0 <= j < b should be incremented by 1.

You need to find the number of maximum integers in the matrix after performing all the operations.

Example:

Input:
m = 3, n = 3
operations = [[2,2],[3,3]]
Output:
4

Explanation:
Initially, M = 
[[0, 0, 0],
 [0, 0, 0],
 [0, 0, 0]]

After performing [2,2], M = 
[[1, 1, 0],
 [1, 1, 0],
 0, 0, 0]]

After performing [3,3], M = 
[[2, 2, 1],
 [2, 2, 1],
 [1, 1, 1]]

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

Clarifying Questions

1. Can the matrix dimensions m and n be zero? (Assumption: No, they are always positive integers)
2. Can the operations list be empty? (Assumption: Yes, if no operations are applied, the matrix remains with all elements as 0)
3. Are a and b always within the bounds of the matrix dimensions? (Assumption: Yes)

Strategy

To solve this problem efficiently, instead of simulating the entire matrix operations, which could be computationally expensive, note the following key insight:

• Since each operation (a, b) increments all elements in a submatrix of size a x b, the cells that have been incremented the most times will be those in the intersection of the smallest values of a and b for all operations.

Thus:

1. Track the minimum value of a (let’s call it minA).
2. Track the minimum value of b (let’s call it minB).
3. The number of maximum integers will then be the area of this intersection submatrix, which is minA * minB.

Code

Here is the Java code implementing the strategy:

public class Solution {
    public int maxCount(int m, int n, int[][] ops) {
        // If ops is empty, the maximum number doesn't change the entire grid, so it's m * n.
        if (ops.length == 0) return m * n;
        
        int minA = m;
        int minB = n;
        
        // Iterate over all the operations to find the smallest a and b
        for (int[] op : ops) {
            minA = Math.min(minA, op[0]);
            minB = Math.min(minB, op[1]);
        }
        
        // The area of the top-left sub-matrix affected by all operations 
        return minA * minB;
    }
}

Time Complexity

The time complexity of this solution is:

• (O(K)), where K is the number of operations since we only perform a single loop over the operations array.

The space complexity is:

• (O(1)) since we only use a few extra variables for tracking the minimum values.

This solution ensures we handle the problem efficiently both in terms of time and space.

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