Longest Turbulent Array - Leetcode 978 - Python

[@khyunwoo1]

as someone learning through neetcode, this problem didn't really belong under the kadane's algo section. It had less to do with kadane's algo itself and more to do with sliding window or 2 pointers.

[@NeetCodeIO]

Today's daily LC is already on the main channel (kzbin.info/www/bejne/Z6HOgJqOeZtmr7c), so uploading another one instead today.

[@ary_21]

I thought it's gonna be a holiday today , appreciate your dedication

[@krakenjoka]

Actually from this explanation, variables tracking max and cur length would be enough. On sliding window problems we usually need the left pointer to check for some conditions. But on this problem it's not that relevant. Thanks for these videos

[@alexl2512]

My implementation def maxTurbulenceSize(self, arr: List[int]) -> int: l = 0 maxLen = 0 preS = "=" for r in range(len(arr)): if r == 0: maxLen = 1 continue if arr[r - 1] < arr[r] and preS != "": preS = ">" maxLen = max(maxLen, r - l + 1) elif arr[r - 1] == arr[r]: l = r preS = "=" else: l = r - 1 return maxLen

[@rafael84]

I think this solution is a bit more easy to read: var maxTurbulenceSize = function (arr) { if (arr.length curr ? 1 : -1; if (newSign === sign) { currCount = 2; } else { currCount += 1; } sign = newSign; } maxCount = Math.max(maxCount, currCount); } return maxCount; };

[@trueknowledge1981]

Holy I am the first this time Hello Neetcode

[@iamnoob7593]

Thanks neetcode

[@junjason4114]

My solution to this problem. class Solution: def maxTurbulenceSize(self, arr: List[int]) -> int: if len(arr) == 1: return 1 pre_diff, subarray_len = -(arr[0] - arr[1]), 1 longest = subarray_len for i in range(1, len(arr)): cur_diff = arr[i - 1] - arr[i] if cur_diff * pre_diff < 0: subarray_len, pre_diff = subarray_len + 1, cur_diff continue longest = max(longest, subarray_len) subarray_len = 1 if cur_diff == 0 else 2 pre_diff = cur_diff return max(longest, subarray_len)

[@EzraSchroeder]

@5:07 math died!!!! XD hahahahahaha!!!!!

[@sumitsharma6738]

also for famous problem "OPTIMAL STRATEGY FOR A GAME"

[@vivekmit06]

Thank you very much!!!

[@sumitsharma6738]

please make a video for "Super Egg Drop" problem

[@Mohit-uq7hq]

he sounds like john smilga from free code camp

[@menagamal7239]

Can't we Just Use Stacks for this one ?

[@BCS_AAMIRASHRAF]

❤❤❤❤❤

[@trueknowledge1981]

bdw what is use of left pointer exacltly?

[@MeanSoybean]

to keep track of the beginning of the currently examined candidate for turbulent subarray

[@krakenjoka]

I dont think it's necessary, 2 variables, one counting current length and another for max length would be enough

[@MeanSoybean]

@@krakenjoka and how would you compute what the current length is

[@krakenjoka]

@@MeanSoybean current length initial value can be 0, then during the for loop, add cur+=1 if it satisfy our condition. Otherwise reset it to 0 or 1 depending on stones[i-1] and stones[i] being equal or not. At every iteration compare global max so far to the current length.

[@nehatyagi5218]

hey can you debug my code class Solution { public: int maxTurbulenceSize(vector& arr) { int l =0; int r = 1; int ans =1; int n = arr.size(); string prev = ""; while(r arr[r] && prev != ">"){ ans = max(ans , r-l+1); r += 1; prev = ">"; } else if(arr[r-1] < arr[r] && prev != "

[@isseihyodou7111]

The issue with your code lies in how it's handling the case where the array elements are equal (arr[r] == arr[r-1]). The logic doesn't correctly reset the pointers l and r, resulting in incorrect computations for the turbulent subarray length. Here's a modified version of your code that addresses this issue: public: int maxTurbulenceSize(vector& arr) { int l = 0; int r = 1; int ans = 1; int n = arr.size(); string prev = ""; while (r < n) { if (arr[r - 1] > arr[r] && prev != ">") { ans = max(ans, r - l + 1); prev = ">"; } else if (arr[r - 1] < arr[r] && prev != "