Leetcode:2141. The longest time to run n computers at the same time [the maximum value is two points]

analysis

n <= batteries.length There will be no more computers
There should be greedy thinking
We know if it's fixed n In the case of computers , We can judge the operation x Whether minutes is feasible
The specific method is to take each battery min(x, battery[i]) This is its effective use often
Put it all together and see n * x The size is ok
One is x The maximum power provided in minutes One is x Minutes of power needed
then check return BOOl, Finally, use the skill of 1.2 points to solve

ac code

class Solution:
    def maxRunTime(self, n: int, batteries: List[int]) -> int:
        #  Consider two points for the best value problem 
        #  Given minutes , You can get the most running computer books 
        #  The longer the minutes , The fewer computers running at the same time 
        m = len(batteries)
        #  Directly see if it can run mins minute 
        def check(mins):
            valid_mins = [min(mins, batteries[i]) for i in range(m)]
            return sum(valid_mins) >= n * mins


        l, r = 1, 10 ** 14 # 1 should be starter
        while l + 1 < r:
            mid = (l + r) // 2
            if check(mid):
                l = mid
            else:
                r = mid - 1
        
        for ans in range(r, l - 1, -1):
            if check(ans):
                return ans

summary

Attention to detail , Attention simulation , Pay attention to the simplification of thinking
Consider two points for the best value problem