# Solving Leetcode Interviews in Seconds with AI: Range Addition II


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "598" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given an m x n matrix M initialized with all 0's and an array of operations ops, where ops[i] = [ai, bi] means M[x][y] should be incremented by one for all 0 <= x < ai and 0 <= y < bi. Count and return the number of maximum integers in the matrix after performing all the operations.   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  

	# Explanation
	Here's the breakdown of the solution:

*   **Core Idea:** The maximum integer in the matrix will be determined by the smallest `a` and `b` values across all operations. This is because the increment operations affect rectangular regions from (0, 0) to (a-1, b-1). The overlapping region defined by the minimum `a` and `b` will be incremented the most.

*   **Optimization:** We don't need to perform the actual matrix increment operations. We only need to track the minimum `a` and `b` values from the `ops` array.

*   **Result:** The count of maximum integers is simply the product of the minimum `a` and minimum `b` found in the operations array. If no operations are present, the entire matrix remains filled with zeros, and the number of maximum integers is m * n.

*   **Complexity:** O(N) runtime, O(1) storage, where N is the length of `ops`.

	
	# Code
	```python
	def maxCount(m: int, n: int, ops: list[list[int]]) -> int:
    """
    Calculates the number of maximum integers in a matrix after applying operations.

    Args:
        m: The number of rows in the matrix.
        n: The number of columns in the matrix.
        ops: A list of operations, where each operation is a list [ai, bi].

    Returns:
        The number of maximum integers in the matrix.
    """

    min_a = m
    min_b = n

    for op in ops:
        min_a = min(min_a, op[0])
        min_b = min(min_b, op[1])

    return min_a * min_b
	```