Minimize Maximum of Array - Leetcode 2439 - Python 

I have almost 400 questions on leetcode now and still got stuck at this , it is in my opinion a one off problem (it is what it is). Dont be discouraged if not able to solve this , keep grinding . We always have neetcode to help us out

he came when we needed the most

with every new problem, I'm getting intuition if I will be able to solve it on my own or not. For this one, I came straight to you without writing a single thing on paper.

I got upset when I could not come up with a Binary search approach as mentioned in the Hint section of the problem but after watching the first 3-4 mins of your video I was able to solve it. You are really great at breaking down complicated problem statements in their lucidest form 🙏🏻🙏🏻Thank u NeetcodeIO.

I don't even understand the question to begin with, thanks so much for the video!!!

Thank you so much for the daily leetcode problems ! Helps a lot !!! Keep it up

You have explained the solution so simply. Keep up the good work

You are amazing. You explain this problem as if explaining to 7 years old student. Great Video

Thank you so much, I actually thought of this idea but couldn't verify it for some reason myself. I watched your video and realized it was the same idea, so I implemented it and it worked. New sub!

After listening 2 min of your video, suddenly the solution occurred in my mind and I continued. It was precisely the same as yours. Thank you for clarifying as best as every time

Only video that helped me understand the problem fully thank you !!❤❤

Right on time! Thank you!

how did you even come with this type of intuition, its amazing

class Solution: def minimizeArrayValue(self, nums: List[int]) -> int: sum_val = nums[0] count = 1 max_val = nums[0] for r in range(1,len(nums)): sum_val+=nums[r] count+=1 avg_val = math.ceil(sum_val/count) max_val = max(max_val,avg_val) return max_val

thank you so much for the vedio it helps me a lot........

why a similar logic in c++ fails a test case? Code: class Solution { public: int minimizeArrayValue(vector& nums) { int total = nums[0]; int res = nums[0]; int n = nums.size(); for(int i = 1;i < n;i++){ total += nums[i]; int t = ceil(total/(i+1)); res = max(res,t); } return res; } }; Test case: nums = [13,13,20,0,8,9,9] Output 15 Expected 16

Very nice! Thanks !

good work , thank you

Really good video

please also solve this using binary search ,else this is also a fantastic way

thanks

why is average working here? why isn't median being used? i thought of something along these lines then went ahead with median and got WA.

you are genious

I one question about this in the explanation that they have provided what if we do this operation one more time and select i=3 then the array will become [5,5,3,4] So wouldn't this 4 be the minimum value ? I think I am missing something here

I had compilation error @line 7 because there is to +i add in total ( total + i ) / ( i + 1 )) not sure how it running right there. Please check

dayummmm love u

I came up with a solution in 35 minutes

Tricky one, yeah

Don't understand what you mean by "we can't move values to the left" / right. 🤔

man i tried heap sol, but couldnt get it to pass. Not sure where the error is: class Node: def __init__(self,val,left=None,right=None): self.val=val self.left=left self.right=right def __lt__(self, other): return self.val int: heap=[] curr=h=Node(float('inf')) for j in range(len(nums)): node=Node(-nums[j]) curr.right,node.left=node,curr heappush(heap,node) curr=curr.right out=heap[0].val while True: if heap[0].left==h or heap[0].val+1==0: #cant make smaller return -heap[0].val node=heappop(heap) node.val+=1 node.left.val-=1 heappush(heap,node) if heap[0].val>=out: out=heap[0].val else: break return -out Testcase: [13,13,20,0,8,9,9] output: 15 Expected: 16