LeetCode Minimum Size Subarray Sum Solution Explained - Java

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#coding #programming #softwareengineering

[@pl5778]

great vid. I like how you explained the 2 pointers is not from start and end, but its both front the left. I think that concept is really helpful.

[@faizanusmani2200]

Amazing solution. I was just very close to solving it but just could not think about the inner while loop

[@harshavardhanreddy7996]

Very well explained...two pointers from front concept - sliding window...!!

[@deeproy2719]

many places I am seeing that sliding window is being treated as dp problem

[@CyberMew]

Good explanation, I had the intuition to do sliding window using 2 pointers, but could not think of doing a while loop inside the forloop, and was fixated on moving both the right and left in the same loop, and was left figuring out why it is not working the way I want. ugh.. I think it was because I didn't want it to iterate so much, because it feels like O(n^2).

[@PeachHeartBlog]

Both left and right points will only iterate list once. After the left items are popped up, they won't loop again, So still O(n)

[@sumeetsapla]

Thanks Nick for explaining in simple way

[@SHASHANKRUSTAGII]

I dont know why, but i also thought of this method, before watching the video, my only issue is to develop the code. please guide me, i can see the way problem should be solved, but not able to code.

[@deepakbisht7764]

You made things looks so easy. thanks for the great video

[@yalansun4382]

glad i found this video, well explained. thank you so much.

[@suryanshrajdubey]

Thanks for the video!!

[@edwardnewgate2198]

Amazing Solution!

[@jaybabaofficial]

Loved this solution and explanation

[@sanketkoli8641]

Very good solution. Thank a lot! :)

[@chaoschao9432]

good one, thx

[@alankuo2316]

nice explain!

[@jatinbhatoya8420]

crystal clear explanation

[@nandanimadhukar]

good explanation!

[@gregoryrobertson]

Thank you for creating this video 👍

[@lifeofme3172]

Beautiful 👏

[@OMFGallusernamesgone]

What happens if one of the input numbers was much larger than s, such as 1000, you would need to check if left !> i

[@artsysoul1050]

can any one tell what is the time complexity ??

[@abishekkachroo938]

u can never be rusty ... hahaha I have got the rust from past 25 years

[@pavithraradhakrishnan8229]

wow thank you

[@kunalkheeva]

Appreciate you!

[@adarsh9555]

great , i liked it

[@bhavyanayyer3391]

how one can think of this solution without looking at the soln

[@rajeevmohan9049]

If u this type of question previously..😊

[@cpwithsundar]

@@dashrathsinghkaviya it isn't related to the qn asked? if we visit his channel......will we get the ans??

[@aj9706]

That's an interesting question ⁉️

[@mnchester]

great vid

[@davegould4940]

Slidin' window

[@akshitgarg09]

love you man

[@sebastianbaptiste4905]

You shouldn't ignore your Apple updates

[@harini3191]

Video is awesome!. Please create a video for implementing binary search using two pointers method for this problem

[@siobhanahbois]

The time complexity for this one will be O(n log n), is that right?

[@siobhanahbois]

A J even with binary search? If yes, do you know how to get to O(n log n)?

[@aj9706]

@@siobhanahbois binary search requires array to be sorted. That's not applicable in this problem.

[@zlzlinkinpark5337]

awesome

[@arijitroy8390]

Its called sliding window

[@harinijeyaraman8789]

Isn't this slightly similar to sliding window concept ?

[@TheGianaJinx]

This is sliding window. The window can vary in size as it moves over the structure.

[@harinijeyaraman8789]

@@TheGianaJinx Basically a variable window-sized sliding window. Thanks for confirming !

[@tensorflaw]

"Remind me tomorrow"

[@jakubk7158]

int[] arr= new int[]{5,2,2,1,1,4}; target = 9 output = 3 , should be 2 [5,4] , anyway this solution passes leetcode tests :)

[@dcodernz]

But 5,4 isn't a sub array. It's a sub sequence.

[@madhavram7138]

HELP! How will the return statement look like in Python?

[@madhavram7138]

I got, it should be: if result == math.inf: return 0 return result

[@raushanraj_99]

[@anoopkumar8732]

the solution is a little buggy. on line 12, he should add left

[@kevinwang3981]

nice catch!

[@austinpereira9204]

Its more like sliding window

[@vk1618]

Other solution

[@siobhanahbois]

This is O(n) right? What about the follow up question: "If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n)."

[@ashwani4569]

This is O(nlogn). There is inner while loop.

[@arjunreddy3996]

@@ashwani4569 lol no, it's O(N)

[@siobhanahbois]

Did any of you guys figure out the O(n log n) solution or did you find a video solving it in O(n log n) time?

[@pavithraradhakrishnan8229]

@@siobhanahbois O(n log n) would be using binary search

[@grandstarchief3226]

O(n log n) is prefixSum + binary search

[@Karthik__S_554]

wow

[@joydeep_]

My O(1) solution in JS ``` function minSubArrayLen(arr, n) { if (!arr.length) return undefined let i=0 let j=0 let tmpSum = arr[i] let minLen = Infinity while(i < arr.length && j < arr.length){ if (i === j){ if (tmpSum >= n){ minLen = Math.min(minLen, (j-i)+1) tmpSum-= arr[i] i++ j++ tmpSum+= arr[j] continue } if (tmpSum < n){ j++ tmpSum+= arr[j] continue } } if (tmpSum >= n){ minLen = Math.min(minLen, (j-i)+1) tmpSum-= arr[i] i++ continue } if (tmpSum < n){ j++ tmpSum+= arr[j] continue } } return minLen === Infinity ? 0 : minLen; } ```