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Product of array except self (LeetCode 238) | Neetcode 7 / 150 | Bharath Chandra (Telugu)

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Hello guys, cheers to another piece of learning. Today I talked about the "Product of array except self" problem which is the first in Neetcode 150 list. It is one of the most asked questions in interviews and has some interesting ways to solve it.
In the above video, I showed all the different logics with which the problem can be solved and also the codes to those logics in Python. Let me know if you got a hold of the logic?
Link to Prefix sum - • Prefix Sum Arrays| Pro...
Link to the problem - leetcode.com/p...
Link to Neetcode 150 - neetcode.io/pr...
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00:00 - Introduction and Problem Explanation
02:12 - Straight forward approach
05:47 - Brute Force
11:43 - Optimisation
18:59 - Pseudocode
21:16 - Bonus question
28:16 - Conclusion

Опубликовано:

17 авг 2024

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Комментарии : 15

@niramalakanukolanu3222 Месяц назад

Bro I Have Doubt How It Would Be suffix[0]= 1 it is 24 kadha

@bharathh_chandraa Месяц назад

Avunu Niramala, Thanks for pointing this out. It is 24, my mistake. Luckily, it did not change any answers

@electron-u8p Месяц назад

7/150 ✅️

@lohithaadapala6989 Месяц назад

Ah! Found the perfect channel! Thankyou🙏

@tradeiteasyy 20 дней назад

Idea : iterate Evey element and get the left product and right product and multiply them and store the value in result array .

@vaneetha_here_navaneetha 29 дней назад

class Solution: def productExceptSelf(self, nums): n=len(nums) suffix=[1]*n suffix[n-1]=nums[n-1] for i in range(n-2,-1,-1): suffix[i]=suffix[i+1]*nums[i] for i in range(1,n): nums[i] = nums[i-1] *nums[i] for i in range(n): if i==0: suffix[i]=suffix[i+1] elif i==(n-1): suffix[i]=nums[-2] else: suffix[i]=(suffix[i+1] * nums[i-1]) return suffix

@vaneetha_here_navaneetha 29 дней назад

class Solution: def productExceptSelf(self, nums): n=len(nums) prefix=[1]*n prefix[0]=nums[0] for i in range(1,len(nums)): prefix[i]=prefix[i-1]*nums[i] suffix=[1]*n suffix[n-1]=nums[n-1] for i in range(n-2,-1,-1): suffix[i]=suffix[i+1]*nums[i] l=[] for i in range(len(nums)): if i==0: l.append(suffix[i+1]) elif i==(n-1): l.append(prefix[n-2]) else: l.append(suffix[i+1] * prefix[i-1]) return l after watching video

@Manipinnaka Месяц назад

Baga chepav bro❤

@SaiBharath_Kanasani12 Месяц назад

Hi bro, I have got the O(1) approach you told me in the last part of the video, but I am having a hard time coding that approach. can you tell me the code for that approach it would be helpful. Thank You!

@lohithaadapala6989 Месяц назад

class Solution { public int[] productExceptSelf(int[] nums) { int n = nums.length; int[] output = new int[n]; output[n-1] = nums[n-1]; for(int i=n-2;i>=0;i--){ output[i] = output[i+1] * nums[i]; } for(int i=1;i

@vaneetha_here_navaneetha Месяц назад

class Solution: def productExceptSelf(self, nums: List[int]) -> List[int]: c=nums l=[] for j in range(len(nums)): n = c[:j]+c[j+1:] p=1 for i in n: p=p*i l.append(p) return l before starting the video