132 Pattern | Leetcode 456 | Stack | Day-7

[@mailatkaushal]

very good explanation for better approach to arrive at optimal solution

[@AyushiSharmaDSA]

thank you :) glad u liked it :)

[@anubhavsonics-3699]

for brute force it will be int n = nums.size(); if(n < 3) return false; for(int i = 0; i < n-2 ; i++){ for(int j = i+1 ; j < n-1 ;j++){ for(int k = j+1 ; k < n ; k++){ if(nums[i] < nums[k] && nums[k] < nums[j]) { return true; } } } } return false;

[@sourjadhar9332]

Doesn't matter u put n-2 or n..it will run in both cases

[@VivekKumar-hb9dv]

@@sourjadhar9332 It matter. It will get give runtime error due to indexOutOfBound

[@souravjoshi2293]

This is what you said exactly for why we traverse from n-1 to 0 --> ""Because we need to find nums[i] and along with that we need to store mums[j] and nums[k], so that nums[j] & nums[k] we have to store and obviously that will be i < j < k, so from here like some elements here because j & k are greater than i right so we have to start from right and go to left so that nums[j] & nums[k] we can find before nums[I]"" How on earth this makes sense ? Someone please explain me what does the line above mean. Thank you

[@gauravkungwani]

Best Explanation Found !! 😃

[@AyushiSharmaDSA]

Thank you Gaurav :)

[@alankritraj5816]

so well explained concept

[@AyushiSharmaDSA]

thank you :)

[@AJDJ108]

Thanks mam you made it so easy to understand

[@AyushiSharmaDSA]

thank you, glad it was helpful :)

[@shubhanshjain2384]

Lucid explanation Ayushi. 🙂

[@AyushiSharmaDSA]

Thank you Shubhansh 🙂

[@priyanshusuryawanshi3302]

love your way to teach didi

[@AyushiSharmaDSA]

thank you, glad it was helpful :)

[@shubamgoswami]

you make it so easy for us

[@AyushiSharmaDSA]

Thanks Shubam :)

[@afzalghani5499]

Well Explained. Thank you

[@AyushiSharmaDSA]

Welcome, glad it was helpful :)

[@sanatangyansagar]

Thank you sister it's great one 😊I will watch each video hope it will help me to build right logic because I'm spending all day but easy problem is also taking 1-2 hour from me

[@AyushiSharmaDSA]

Don't worry Shubham, initially this happens😊

[@manishbolbanda9872]

very well explained.

[@AyushiSharmaDSA]

Thanks Manish :)

[@ninadkheratkar6716]

Very well explained!

[@AyushiSharmaDSA]

Thanks Ninad 🤗

[@Akash-dd6ev]

Thanks a lot

[@AyushiSharmaDSA]

welcome :)

[@amitrajput7019]

your explanation was awesome sis

[@AyushiSharmaDSA]

thank you, glad it was helpful :)

[@dayashankarlakhotia4943]

class Solution { public boolean Find132pattern (int []nums){ Stackst=new Stack(); int third = Integer.MIN_VALUE; for(int i=nums.length-1;i>=0;i--){ if(nums[i]

[@SuyashTalks]

Thanks !!

[@AyushiSharmaDSA]

welcome :)

[@yashikagupta3963]

Your Consistency and Explanation 👏

[@AyushiSharmaDSA]

Thank you Yashika 🙂

[@abhayj8706]

Please make video for flatten list iterator

[@samrat8413]

Mistake spotted at 1:00 ... --> nums[i] nums[j] nums[k]

[@shwetaj8189]

Why we are taking minimum element in O(n^2) approach?

[@adityaturumella6649]

Subscribed

[@AyushiSharmaDSA]

thank you😊

[@amitsinha806]

👍👍👍👍👍👍👍👍👍

[@karandeepsingh6201]

🙏🙏

[@experimentboysam1549]

Please reply ! Is software engineering job are permanent and many people are saying that after 5 ya 10 years software will be not there please reply .🙏

[@LordSarcasticVlogger]

YES IT IS THE HARSH TRUTH

[@7billon680]

How can i learn each topic with example...?

[@dreadnought8543]

I want to learn DSA from and I know java so please will you suggest me ?

[@LittleWheelsBigDream]

is this playlist completed ??

[@kalyanipharkandekar4314]

can you explain the test case-[3,5,0,3,4]??? I am not getting the dry run for this test case ...

[@hunt5624]

3,5,4

[@RajGupta-fv7xb]

Didi how to identify if we can solve this question by stack?

[@AyushiSharmaDSA]

This was similar to next greater element on right problem (in that we use stack) :)

[@RajGupta-fv7xb]

@@AyushiSharmaDSA ya thanks

[@amitgupta2890]

Dependency of one loop to other can be optimised using stack. This is the way to identify

[@RajGupta-fv7xb]

@@amitgupta2890 Thanks mate

[@amitgupta2890]

Ayushi what you were downloading from mega

[@AyushiSharmaDSA]

Not sure, some courses ig :)

[@souravjoshi2293]

14:53 (not a very intuitive statement)

[@jatingrover8621]

hi ayushi, why are you comparing mine with arr[j]and not with arr[k] in approach 2.

[@harshitgupta6635]

so that i

[@akshitmangotra5370]

I too had some doubt..it's just that we want to maintain the series for index like i

[@sakshiawadhiya7267]

Mam logic kaise build karti hai app

[@pmcollection2.071]

Mind use kar ke ,

[@nvm3172]

this question made me puke ngl. thankyou for the explanation

[@AyushiSharmaDSA]

Welcome :) Glad it was helpful

[@priyanshusingh100]

Di check linkdein msg

[@hunt5624]

9:55 can anyone explain me please

[@hunt5624]

the mini should be present in outerloop

[@tusharnain6652]

System is lagging, Android Studio, MS Office tools, browser, and screen recorder. That's too much!

[@AyushiSharmaDSA]

Yes😅

[@7billon680]

Hey....!

[@ak47_ff68]

Plz speak Hindi yrr

[@ujjwalabhishek]

Can you tell in which line I have done a logical mistake? It is failing on testcase: [1,3,2,4,5,6,7,8,9,10] class Solution { public boolean find132pattern(int[] nums) { int n = nums.length; if (n < 3) return false; Stack stack = new Stack(); int third = Integer.MIN_VALUE; for (int i = n - 1; i >= 0; i--) { if (stack.isEmpty()) { stack.push(nums[i]); } else { int top = stack.peek(); if (nums[i] > top) { third = top; stack.pop(); stack.push(nums[i]); } else { if (nums[i] < third) return true; } } System.out.println(i+" "+stack+" "+third); } return false; } }

[@AyushiSharmaDSA]

You need to keep popping elements from stack till nums[i] > top element So there will be while loop (check code in desc of video) :)

[@ujjwalabhishek]

@@AyushiSharmaDSA see 10 is the biggest element, i pushed that into the stack then there is no any smaller element so how can I pop that one.

[@ujjwalabhishek]

I got the logic from ur video only but I wanna implement this in my own way, that's y I have not focussed much more on your code.

[@AyushiSharmaDSA]

@@ujjwalabhishek that's great, you are doing it correct way only. First write code urself and then refer :)

[@ujjwalabhishek]

@@AyushiSharmaDSA thankyou so much now I got where I have done the mistake. Got it!!