Minimum Number of Operations to Move All Balls to Each Box - Leetcode 1769 - Python

[@FifaHades]

O(n) is the biggest enemy of the interview, great video!

[@aadil4236]

I solved it in O(n) by myself without looking at the solution or watching this video. Without you, I wouldn't have gotten here. I used to struggle with even easy problems in the beginning, but you improved me. Thank you so much for your kindness!

[@staywithmeforever]

88% acceptance rate

[@alexgabriel5877]

@@staywithmeforever because bruteforce passes

[@staywithmeforever]

@@alexgabriel5877 oh really, i didn't knew that, O(n) is kinda obvious though. Some problems don't even give me the thought of the solution.

[@anantakesharipanda4085]

How long did it take for you to overcome your struggle phase?

[@staywithmeforever]

@@anantakesharipanda4085 just remember the patterns like prefix sum, reversing the question, sorting, monotonic stack etc that will help solving questions quickly.

[@soupayt]

Nice video! I didn't think of it as product of array except self, but you're correct that this is basically the same problem. I calculated the left side and right side prefix & suffix on the fly instead when building the solution array, but still the same! class Solution: def minOperations(self, boxes: str) -> List[int]: # You can simplify this problem to keeping track of the distance between the current box and the boxes on the left and right with a non-zero # amount of balls. We can use a running prefix sum to pull this off. We keep track of the number of 1's on the left and right sides and can adjust the left and right sums accordingly with that after each iteration. boxes = [int(b) for b in boxes] left = 0 right = 0 left_ones = 0 right_ones = 0 res = [] # precalculate right for i, box in enumerate(boxes): if box == 1: right_ones += 1 right += i for i, box in enumerate(boxes): res.append(left + right) if box == 1: left_ones += 1 right_ones -= 1 left += left_ones right -= right_ones return res

[@Kaviarasu_NS]

I rewrote NeetCode's approch and I personally I find it more easy to understand def minOperations(self, boxes: str) -> List[int]: res = [0] * len(boxes) balls, moves = 0, 0 for i in range(len(boxes)): moves += balls res[i] += moves balls += int(boxes[i]) balls, moves = 0, 0 for i in range(len(boxes) - 1, -1, -1): moves += balls res[i] += moves balls += int(boxes[i]) return res

[@indian_geocacher]

I have one question @Neetcode Can you please explain how did you know that the problem will be this one as you have published the solution at 00:00 as soon as the problem appeared, is it prerecorded for all the problems and you just post it?

[@NeetCodeIO]

Yes I've actually recorded a video for every leetcode problem just to be safe

[@vinayakbajpeyi931]

@@NeetCodeIO 😂

[@joeAnon796]

Also he said it's NYE... it's actually the 5th. Something fishy going on here! I think he gets the questions in advance 🤔

[@freecourseplatformenglish2829]

@@joeAnon796 Actually you can see the upcoming question in advance. Navdeep mentioned it in one video. I couldn't found the source but it is possible.

[@leo881010able]

If we follow these dailies everyday we would know, he explained this awhile ago. It was by some Editorial updating logic, editorial often updates roughly a week before a question becomes a daily problem

[@staywithmeforever]

2 balls still left

[@eznx616]

lmao

[@AjayKumar-gs9sg]

Navdeep can you please explain solutions for yesterday's contest problems ? @neetcode

[@rode_atharva]

yesterday solved by myself using brute force. after listening that "you might be genius if you haven't seen the pattern and come up with the solution by yourself !". but i actually know the pattern of dynamic programming (so can't say am i genious or not 😅).. you are genoius. beacause of "you" and "codestory-with-mik" i am able to come at this level ! .... here is my solution class Solution: def minOperations(self, boxes: str) -> List[int]: count = 0 ans_left = [0] * len(boxes) ans_right = [0] * len(boxes) for i in range(len(boxes)): if i != 0: ans_left[i] = ans_left[i-1] + count if boxes[i] == '1': count += 1 count = 0 for i in range(len(boxes)-1, -1, -1): if i != len(boxes)-1: ans_right[i] = ans_right[i+1] + count if boxes[i] == '1': count += 1 for i in range(len(ans_left)): ans_left[i] += ans_right[i] return ans_left

[@STACYEMMA-y2e]

Did it in O(n^2) , came here for O(n) solution

[@ramez3038]

Solved it in o(n^2) after that optimized it to o(n) and came here to see that i thought about it the same way neet did seems hardwork and consistency pays off thanks alot neet 🔥🔥

[@dixittilaji6147]

came here after solving a problem with a n^2 approach, I knew it there has to be optimized way haha, love your videos man!

[@luciell4782]

Thanks! Amazing solution!

[@sri_harsha_dv]

Explanation felt little confusing while building the right part of the array and in solution where moves should be updated before balls. I was able to solve this problem surprisingly. Moves variable is not necessary atleast in the first array of your solution. My solution is one pass by building these left and right arrays. The intuition is that the right array is built using complete opposite steps of the left array. Only tricky part is understanding the indexing for the right array. Solution: def minOperations(self, boxes: str) -> List[int]: length = len(boxes) left, right = [0]*length, [0]*length left_count = right_count = 0 for i in range(length): if i: left[i] = left[i-1]+left_count right[length-i-1] = right[length-i]+right_count if boxes[i] == "1": left_count += 1 if boxes[length-i-1] == "1": right_count += 1 return [l+r for l,r in zip(left,right)]

[@eznx616]

damn playing with balls is hard

[@TheSmashten]

Hey Neetcode can you please do LC #37 (Sudoku Solver)? I can't even submit the editorial solutions for Python

[@sauravsingh4497]

Did somebody count how many times he said balls during the video

[@MP-ny3ep]

Thank you for the daily

[@fdasfsadfsdfa]

Does it count as O(n) if, in this video, the problem is actually solved in O(n + n)?

[@fancypants6062]

Yes, it is standard to reduce to O(n) for situations like this.

[@user-my6yf1st8z]

thanks boss, but that's a fuckin magic you did, most of the vid i thought what is he talking about, but in the end it kinda all made sense

[@Kaviarasu_NS]

I think this problems solution approch is similar to Leetcode 135. Candy

[@shivamvats2108]

Thanks!

[@yhbarve]

Goated!

[@andybnhquang]

How many days until that leetcode tshirt lol

[@studymotivation-c8t]

2:08 and i solved it myself lol

[@business_central]

wait it's New year's eve ??

[@eznx616]

real

[@karthi7102]

Goat

[@adilansari-hq9ge]

Recording on the eve of christmas, Big hug to you Bro, Thank you for all hthe ard work , effort , dedication and motivation.

[@freecourseplatformenglish2829]

After solving yeserday's question this was a cakewalk.

[@user-my6yf1st8z]

you posted a solution to a different problem, bruv

[@freecourseplatformenglish2829]

@@user-my6yf1st8z Thanks man, Solved the diffrent problem 😅😅🤣

[@ritikpatra4757]

damn

[@leechaolan6227]

like if you are here after finishing it with Brute force

[@ThaoRatke]

hi