Minimum Size Subarray Sum | Leetcode

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This video explains the minimum size subarray sum problem which is a variable size sliding window technique-based frequent interview problem. In this problem, I have explained the problem statement using easy-to-understand examples. I have explained the intuition for the algorithm using graph diagrams and then followed it up by a dry run and code explanation.
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Опубликовано:

8 сен 2024

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Комментарии : 36

@CostaKazistov 2 года назад

10^5 size for N^2 problem becomes 10^10 - very interesting observation you noted. I never realised that about computation limits, such as 10^8 you mentioned. Learned something new today. Great walkthrough of Leetcode 209 btw. Clear and on point.

@leetcoder1159 2 года назад

Are you working as a SDE?

@techdose4u 2 года назад

😀

@CostaKazistov 2 года назад

@@leetcoder1159 Currently, no - but preparing for future interviews. Doing 4-5 Leetcode medium every day (434 so far).

@abhijitroy1958 2 года назад

@@CostaKazistov great sir

@oqant0424 Год назад

@@CostaKazistov now?

@KingBobXVI 2 года назад

Excellent explanation of the problem overall, I kind of feel like the explanation of the +2 and inclusion of a second loop make it a little more confusing though, since a lot of time is spent explaining why it's +2 but that's a design choice rather than a necessary part of the overall algorithm. Time complexity is the same, but I feel like it's a little more straightforward (if more lengthy) like this: sum = nums[0]; while (true) { int sub_array_size = right - left + 1; if (sum >= target) { if (sub_array_size == 1) return 1; // In the case where nums[i] >= target, you know the answer is the minimum. Early out prevents the left index going to the right of the right index. shortest = min(sub_array_size, shortest); sum -= nums[left]; ++left; if (left == n) break; } else { ++right; if (right == n) break; sum += nums[right]; } }

@vivekkumaragrawal3963 Год назад

class Solution { public: int minSubArrayLen(int target, vector& nums) { int len = nums.size(), ans = INT_MAX; int sum = 0; int left = 0,right = 0; while(right=target && left

@nirajgujarathi6796 2 года назад

found one modification, Correct me if I am wrong ! when , sum==target then no need to move left pointer because any ways it will lower the sum, In short if sum> target move left pointer towards right else if sum == taget update window size if it is smaller else if sum< target move right pointer

@KingBobXVI 2 года назад

With the way his logic is written, it just keeps the code consistent - since the "left" pointer is always over-shooting by one, so that once the sub-array is too small, he'll include the sub-array plus the previous element, so if the sub array is currently [1, 2, 4], he wants to iterate again to 1, [2, 4] because now he'll count the length of the sub array (2) and add 1 extra to account for the over-shoot. I think the way he did that actually complicates the verbal explanation significantly, for the sake of the video. Doing it with two loops is unnecessary, as you could just do a single loop along the lines of: while (not finished) { int sub_array_size = right - left + 1; if (total >= target) { minimum_sub_array_size = min(sub_array_size, minimum_sub_array_size); ++left; } else { ++right; } }

@oqant0424 Год назад

thanksssss a lot ............awesome explanation

@oladipotimothy6007 2 года назад

The best explanation Sensei

@techdose4u 2 года назад

Thanks 😊

@syedaqib2912 2 года назад

Well explained 🔥

@techdose4u 2 года назад

Thanks 😊

@sarthakgupta9858 2 года назад

very nice explanation

@techdose4u 2 года назад

Thanks :)

@mtex448 2 года назад

I am not very good in time complexity theory, 5:36 how its become 10^10 and why should we go under

@nehasomani3446 Год назад

Order is N^2 & max input size will be 10^5, so 2*5 = 10, therefore 10^10. 10^8, here 8/2 = 4, all inputs till 10^4 can be solved with it. Which means, if we go under 10^8, we can solve for further higher input size. I hope, this helps.

@Pegasus02Kr 2 года назад

Happy new year for you sir

@techdose4u 2 года назад

Same to you 😀

@VinothOfficialClub 2 года назад

For input target =7 and the array = {4,2,2,2,1,3}. how the solution find the correct answer based on your fix? the right pointer will reach end of an array and will tell the solution as 4.. not 2..

@PraveshGupta1992 2 года назад

Just curious to know , would it work for [7, 7, 8, 8] and target is 2 ?

@techdose4u 2 года назад

Yes. r-l+2 = 0-1+2 = 1 :)

@saivardhanpallerla9 2 года назад

Hi sir which app do u use for explaining

@programming3043 2 года назад

One note

@devbhattacharya153 2 года назад

Sir one suggestion this video could be much shorter about of 15 mins although a great explanation thank you :)

@vinayghadigaonkar8215 7 месяцев назад

Are you Still Teaching the DSA CRASH COURSE?

@_benon 2 года назад

i didnt understand what ur algorithm is

@pijushbiswas4352 9 месяцев назад

Could you help me understand, why won't this code work ? int minSubArrayLen(int target, vector& nums) { long long int s=0; int n=nums.size(); for(int i=0;i