BinarySearch 0329 Task Schedule

Problem statement

https://binarysearch.com/problems/Task-Schedule/

Solution

Equal to Leetcode 0621. Task Scheduler.

Complexity

It is O(n) for time and space.

Code

class Solution:
    def solve(self, tasks, k):
        freq = Counter(tasks)
        Most_freq = freq.most_common()[0][1]
        Found_most = sum([freq[key] == Most_freq for key in freq])
        return max(len(tasks), (Most_freq - 1) * (k + 1) + Found_most)