Contiguous Array - Leetcode 525 - Python 

I would never think of this without seeing it before.

What sets you apart from the other youtubers is the fact that you don't just explain the solution. You also explain why other methods don't work? why only this method works? and how exactly do we come up with this solution? This video was an amazing explanation . Thank you for that.

What does everyone think of the new leetcode UI? Kinda annoying that they don't let you switch back to the old one anymore

I signed my Apple summer internship offer letter yesterday, thanks a lot for all your help. I hope you continue doing what you do the best

Very interesting approach. This is the TDLR for the comment section: 1. Maintain a count variable as you iterate through array. For each 0 subtract count by 1 and add by 1 for each 1. 2. Maintain a hash map in which you store the leftmost index for each count value and initialize it with 0: -1 3. As you iterate through the array if the current count value already exists in the hash map, update the size with curr_idx - map[curr_count] if that value is greater than the greatest size so far

Hi Navdeep, thank you for your solution! It will be extremely helpful if you include the brainstorming notes too in the future videos. Thanks!

Congrats on 100k 🎊

I was struggling with visualising the solutions, Thank for such easy explanation.

sick, thanks for continuing to make these

Congrats on 100k!

Finally an explanation I understood perfectly. Thanks!

Great solution as usual. 🔥 But, hope the following might make it less complex: In simple terms, in the array, if you see 1 -> preSum - 1 0 -> preSum + 1 Then, you just have to check two cases - 1. If the preSum is 0 (equal number of 1s and 0s till that index) -> update result index 2. If the preSum already exists -> check the difference - NOTE: Should not update the hashmap (we only want the first seen preSum to stay to maximise the result) If not both, we are seeing this preSum for the first time -> Add it to hashmap

The comments are a great addition, Navdeep Paji

Amazing explanation today, thanks

very smart. the solution is so clean

These problems are beautiful and set apart top-level coders from the rest. Hope one day to be able to solve this kind of problems myself.

Yes, please show me another solution where you use -1 and 1. I would really like to see him, in general, this is a very difficult task for me. Thank you for the video!

It's a Prefix sum problem

Just find the prefixSum arr, add 1 if 1 or -1 if 0 or vice versa, use hashMap to compare prev values but don't store when the sum repeats otherwise it will overwrite, check the sum=0 case also

please make complete playlist with problem sovling using python in from any online patfrom

hey @NeetCodeIO at 0:46 wont it be !n (factorial of n) rather than n^2 (n squared) ??

it was great explanation, I do understand the solution after watching your video. But how coming up with such solution is not intuitive for me, could you please help with that in any way?

How have you not won awards for your work. Been using your solutions for years now, I love you neetcode🌸

leetcode- Replace Question Marks in String to Minimize Its Value. Solve this question in python in easiest way.

Amazing :D

What if we count all zeroes and ones first time and will use sliding window by decreasing it from right and left. For example, we have 3 zeros and 5 ones. We will decrease the side where we have ones or zeros if we don't have ones from each side and each time update(decrease) zeros and ones values. Will it work?

Hey @NeetCodeIO, won't it be a better solution to use double pointers one at index 0 (s) and another at index n -1 (e), and increment or decrement s and e depending on the number of 1es. if sum(array[s:e]) * 2 > (e - s + 1), s++ if array[s] == 1 or e-- if array[e] == 1 else if sum(array[s:e]) * 2 < (e - s + 1) , s++ if array[s] == 0 or e-- if array[e] == 0 else return array[s:e]. let me know what you think.

Honestly I didn't think that much but I was able to solve the question :p. My thought process was "how about I get the largest array with sum = 0 and how to do that? consider every 0 as a -1"

This is so intelligent and I still can’t figure out how can you come up with this solution XD

love the solution but I aint never coming up with this during an interview without prior knowledge.

@NeetCodeIO, you can shift the sliding window. Suppose we have array = [1, 1, 1, 0, 0, 0]. As the the maximum length of the array is 6. We can consider the length of the array as a limit. If the sliding window has 3 one's and 2 zero's, and the length of the array is 6 then we can definitely shift our right side by 1. But if we have 4 one' and 1 zero's, the shifting the right side by does not make sense. So we will shift the left side of the window by 1.

one day ill be like you

does it matter if the array starts with a 0?

yup

I am in the final grade in Russian Highschool, and I actually saw this problem preparing for my informatics exam (it was not with 0s and 1s, it was the problem to find the longest window that has the largest sum that meets all the requirements, but the idea was also about to keep all the prefixes), it was the last problem in the previous year's exam. I hope I will not get something like this, because I wont be able to solve it under the pressure, haha

please keep the comments NC !!

How are you supposed to know this without seeing it before :(

Include comments! Sometimes the hardest part is intuiting the solution, so seeing how you got from Sliding Window to this Prefix Sum is interesting. Glad to have your videos & roadmap as I prep for interviews. Thx a bunch.

censoring dislikes from leetcode is annoying AF

am i the only one who wasn't able to pass all the testcases with this? here's my code, just in case someone can help point out my mistake class Solution: def findMaxLength(self, nums: List[int]) -> int: zeros, ones = 0, 0 diff_index = {} res = 0 for i, n in enumerate(nums): if i == 0: zeros += 1 else: ones += 1 if ones - zeros not in diff_index: diff_index[ones - zeros] = i if ones == zeros: res = ones + zeros else: idx = diff_index[ones - zeros] res = max(res, i - idx) return res

People are still learning dsa? Whym