"""

598. Range Addition II (Easy)

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

Operations are represented by a 2D array, each operation is represented by an array with two positive integers a and b,
which means M[i][j] should be added by one for all 0 <= i < a and 0 <= j < b.

You need to count and return the number of maximum integers in the matrix after performing all the operations.

Example 1:
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]]

So the maximum integer in M is 2, and there are four of it in M. So return 4.
Note:
The range of m and n is [1,40000].
The range of a is [1,m], and the range of b is [1,n].
The range of operations size won't exceed 10,000.


"""

class Solution:
    def maxCount(self, m, n, ops):
        """
        :type m: int
        :type n: int
        :type ops: List[List[int]]
        :rtype: int
        """

        r = [0 for i in range(m)]
        c = [0 for i in range(n)]
        max_val = 0

        for o in ops:
            for i in range(o[0]):
                r[i] += 1
                if r[i] > max_val:
                    max_val = r[i]

            for i in range(o[1]):
                c[i] += 1
                if c[i] > max_val:
                    max_val = c[i]

        rcount = ccount = 0
        for i in range(m):
            if r[i]== max_val:
                rcount += 1

        for i in range(n):
            if c[i] == max_val:
                ccount += 1

        return rcount*ccount

if __name__ == "__main__":

    sol = Solution()
    print(sol.maxCount(3,3,[[2,2],[3,3]]))
    print(sol.maxCount(40000, 40000, []))
