Range Addition II

#leetcode #dsa #theabbie

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

SOLUTION:

import bisect

class Solution:
    def maxCount(self, m: int, n: int, ops: List[List[int]]) -> int:
        rowctr = [0 for i in range(m)]
        colctr = [0 for i in range(n)]
        for a, b in ops:
            for i in range(a):
                rowctr[m - i - 1] += 1
            for i in range(b):
                colctr[n - i - 1] += 1
        return (m - bisect.bisect_left(rowctr, rowctr[-1])) * (n - bisect.bisect_left(colctr, colctr[-1]))