BS-9. Find Peak Element

[@takeUforward]

Sorry for the delay :( But I had a product launch, for which I had to spend most of my time in the office. I am back, will try to come with 1/2 videos daily now

[@charan123rams3]

Thank you so much sir 🙏 Sir first take rest whenever you have free time then post sir 🥲

[@shivanshumishra0560]

Thankyou so much bhaiya for all videos ...🙏🙏❤️❤️ Your content is >>>> whole KZbin

[@sudhanshushekhar4222]

Understood

[@sayakghosh5104]

Was trying this for the past 2 days, couldn't solve it. Took 1 day break came back at it again with a fresh mind, could solve it in one go by myself. Thanks, your series instills such confidence in me always.

[@OrderEmperor]

no one asked

[@sayakghosh5104]

@@OrderEmperor 😂 ok bro.

[@paraskashyap7362]

@@OrderEmperor no one asked nested(no one asked if no one asked or not)

[@paraskashyap7362]

@@arunm619 fair enough

[@Kj89865]

@@paraskashyap7362 calling in return statement (recursion)no one asked if no one asked or not that no one asked :----) *without breaking condition

[@anujrathore5726]

NO ONE CAN TEACH LIKE STRIVER, HE GIVES HIS 100% to his lectures. he want to teach , he compleltly go into the state of teaching , like with full FOCUS & DEDICATION . he enjoy to teach . and all of this FREE OF COST , LEGEND AKA RAJ VIKRAMDITYA BHAIYA (STRIVER BHAI OP ) ❤

[@saswatrath4646]

at 21:15 correction* if(arr[o] > arr[1]) return 0; the first index is the peak

[@ishanshah3309]

the best about this video was the thinking process of how to arrive from linear to binary search in problems like this. execellent stuff.

[@yashgarg3899]

i have solved these questions already so before but still i watch every video of you because i know there will be something different which ultimately enhance my code writing or intuition skills.

[@priyeshtandel2101]

Are you from DAIICT ?

[@shivanshugarg807]

Thanks striver I completed all your sheets , like twice , they boost my confidence , before watching your solution , i try every question for 30 mins , and regardless i am able to solve it or not , i watch your video for better understanding and side by side mark improtance of question and make notes for revision

[@Kaif_Ali_Khan]

ok

[@3lpme]

Then plz explain why should we consider binary search even though it's not

[@girivardhanvelpula]

00:04 Find the peak element in an array 02:16 Identifying peak elements using the concept of infinity assumptions. 06:42 Peak element finding in a sorted array using binary search 09:03 Binary search to find peak element 13:59 Identifying peak in decreasing curve 16:22 Dealing with the cases of the first and last elements as peak elements. 20:38 Finding peak element in an array using binary search 22:54 In a decreasing curve, the mid element is greater than the right element. 27:08 Binary search can efficiently find the peak element 29:08 Algorithm to find peak element in an array

[@3lpme]

first of all it's not sorted array 06:42

[@ritvikryadav9667]

Hey Striver, Your videos are pure gem. Certainly the best DSA course on the planet. Keep going. God Bless You!

[@kishoregowda8210]

Below is my code. You may find this more easier. Initially start with low -> 0 and high -> n-1. Whenever we calculate the "mid", we will check if this "mid" is part of uphill curve or downhill curve. If it is part of downhill, then i would reduce the search space to left of the array because for sure we know that we would find the larger element to it's left. And similarly if mid element is part of uphill, then move to low pointer to right side of mid. public int findPeakElement(int[] nums) { int low = 0, high = nums.length-1, n = nums.length; while (low < high){ int mid = (high - low)/2 + low; if(nums[mid] > nums[mid + 1]) high = mid; else low = mid + 1; } return low; }

[@ankitdhattarwal2286]

There are one mistake in the code when try solve problem on GFG then not pass all the cases the mistake is if(arr[n-1] >= arr[n-2]) return n-1; // use the equal sign also so that can pass all the cases and thank you so much to provide best content

[@lavanya_m01]

yes, that is because the definition of peak is different in gfg. It says the peak can be greater than or equal to its adjacent elements (a=b) but in leetcode, the peak should be strictly greater (ab)

[@rakshan3323]

@@lavanya_m01 And that peak definition seems to be correct because here, it doesn't cover the case when all the elements are equal.

[@GalaxyStart62]

it's a blessing that you are in youtube, but on the other side , people are enjoying DSA now, hence , competition is drastically increasing too.

[@mriit3025]

the extra afforts for the edgecases is appreciated! keep working on content like this...

[@oguluribrahmaiah1301]

Commenting Understood is just a simple thing for your wonderful explanation !!! Thank you So much for your lectures Striver

[@hitanshujena7470]

Accidently i just wrote the code to find the peak element if there are multiple peaks are available in the array but all these happens for you so thanks a lot

[@iWontFakeIt]

was waiting for these past two weeks and you showed up with another awesome explanation! Thanks dada ❤

[@ishangujarathi10]

GOD LEVEL INTUITION AND EXPLANATION STRIVERRR TYSM < 3!!!

[@3lpme]

Kaise develop hogi iss intuition from inside?😢

[@tanaywhocodes]

Thank you ! wrote the same code without watching the video. All because of you.

[@mahendrachourasiya7444]

SUPERBBBB Explanation! Thankyou so much for your great efforts.

[@AdityaGarg-hl7ri]

Thankyou striver, still maintaining that low=m+1 and high =m-1 thing which makes it super easy to understand!!

[@Dipanshutripathi2407]

Filled with full confidence after watching ur videos.

[@viveksinghrajput2725]

Best explanation on the KZbin thank you soo much....each and every second of this video is informative.

[@RaunitJaiswal-s9v]

for those who are facing problem with the if else part for the right side here is the easy way to think If the element to the right of mid is greater (arr[mid] < arr[mid + 1]), we move to the right half (since there must be a peak in that direction).

[@Bigg_boss_trolls]

NO ONE CAN TEACH LIKE STRIVER

[@DeepikaKondamadugu]

Didn't even look for completion time of the video!!!.....So that much interesting your videos are❤

[@pizzagamingyt9712]

Bro u are the most helpful person for all engineering cse student 🧡❤

[@Dreamer4lyf]

Amazing explanation Striver my man! Hats off to you for all your hard work for the community.

[@Manishgupta200]

THe last one infinite loop example was very helpful to understand the problems condition clearly.

[@Harsh-x5o9f]

The man who is making every CSE student's career bright

[@madhulikaghosh3876]

Your explanation is crazy. You are really doing a very good job.

[@theengineer8301]

Hi Striver, this was an epic explanation. Hats off! However I think we can only check the left element of mid. If greater then eliminate right. Else eliminate left. No need for the other if condition. class Solution { public: int findPeakElement(vector& arr) { int n = arr.size(); if(n == 1) return 0; if(arr[0] > arr[1]) return 0; if(arr[n-1] > arr[n-2]) return n-1; int low = 1; int high = n - 2; int mid; while(low arr[mid+1] ) return mid; //mid-1 greater than mid, peak on left,eliminate right if(arr[mid-1] > arr[mid]) high = mid - 1; //peak on right, eliminate left else low = mid + 1; } return mid; } };

[@expanse8846]

he explained it by the end. it may lead to an infinite loop if there are multiple peaks.

[@deepanshudahiya8966]

An amazing question learning new patterns of conditions in binary search

[@graviton001]

Was able to do it myself 😮 thnx for building up my logical thinking ❤

[@rrt19254]

Woah, solved this at once! Think I'm learning this concept!

[@saambpatil9912]

iNSTEAD OF ADDING THE LAST CONDITION WE CAN CHANGE THE ABOE CONDITIONS TO : else if (nums[mid] < nums[mid + 1]) { low = mid + 1; } else if (nums[mid] < nums[mid - 1]) { high = mid - 1; } COMPLETE CODE: class Solution { public: int findPeakElement(vector &nums) { int n = nums.size(); if (n == 1) { return 0; } if (nums[0] > nums[1]) { return 0; } if (nums[n - 1] > nums[n - 2]) { return n - 1; } int low = 1; int high = n - 2; while (high >= low) { int mid = low + (high - low) / 2; if (nums[mid] > nums[mid + 1] && nums[mid] > nums[mid - 1]) { return mid; } else if (nums[mid] < nums[mid + 1]) { low = mid + 1; } else if (nums[mid] < nums[mid - 1]) { high = mid - 1; } } return 0; } };

[@gokulkrishnanj]

Nice explanation on that hidden test case

[@Josuke217]

This was pretty easy, finally solved it on my own.

[@Raj10185]

code for the same :- int findPeakElement(vector& arr) { int n = arr.size(); //checking done for initial cases where finding and is very easy if(n==1) //edgecase1 return 0; if(arr[0] > arr[1]) //edgecase 2 return 0; if(arr[n-1] > arr[n-2]) //edgecase 3 return n-1; // now apply bS on remaining part int l = 1; int r = n-2; while(larr[mid+1] && arr[mid]>arr[mid-1]) { return mid ; } else if (arr[mid]>arr[mid-1]) //mid on the increasing path that means our ans will be in right side l = mid+1; else // mid is on the decreasing path peak will be on left side r = mid -1; } return -1; }

[@AppaniMadhavi]

not working for all testcases in coding ninjas

[@noone1000_xyz]

at the end of video i am just surprised.......understood very clearly.....thankssss a lottttt striver......

[@ParkersFact_Finder]

how you think like that striver i mean your thought process is totally diff how can i achieve this type of thought process honestly Best DSA playlist

[@ashishrajoria2251]

public int findPeakElement(int[] nums) { int left = 0, right = nums.length - 1; while (left < right) { int mid = left + (right - left) / 2; if (nums[mid] > nums[mid + 1]) { right = mid; } else { left = mid + 1; } } return left; }

[@PiyushKumar-xe1ng]

int start = 0; int end = n - 1; while (start < end) { int mid = start + (end - start) / 2; if (arr[mid] < arr[mid + 1]) { start = mid + 1; } else { end = mid; } } return start; (how about this ,easy and covers every edge case)

[@RaunitJaiswal-s9v]

done with easy part of binary search from monday will start with medium part lots of love raj bhaiya

[@padmaja577]

Firstly the brute force psudeo code will only for: 1) array which has one peak element. 2) array is in ascending order 3) array is in descending order.

[@dank7044]

absolute beauty of a question

[@hashcodez757]

Understood very well sir!! No words for your appeciation bhaiya

[@socify4410]

solved by my own thanks a lot Striver Bro keep spreading the knowledge

[@piyushwandile4152]

great content no one teach like this atleast for free

[@Raj10185]

Tysm first time i got clear cut intution of finding the peak element :) tysm again UNDERSTOOD EVERYTHING:)

[@abhijittalukdar6995]

Bhaiya I solved this without seeing the video. Because i have learnt the concepts from your earlier videos. Thanks and LOTS of LOVE.

[@Codro09]

This is the brute linear solution : class Solution { public: int findPeakElement(vector& nums) { //Brute force: int n = nums.size(); int ans = INT_MIN; int peak_index = -1; if(n==1) return 0; for(int i = 0; i < n; i++){ if (nums[i] > ans) { ans = nums[i]; peak_index = i; } } return peak_index; } };

[@AtulKumar-c4x7l]

understood Thank you striver for such an amazing explanation...

[@manavsingh5919]

Thank You Striver..Understood everything🙂

[@hareshnayak7302]

Understood,thanks striver for this amazing video.

[@knownforunknowns]

Understand !! Thank u bhaiya

[@arunm619]

Reversal of a peak is called trough. 30:29

[@Mousam_Sharma]

Understood .best Explanation 👍

[@cinime]

Understood! Super awesome explanation as always, thank you very much for your effort, and congrats for your product launch!! Is this idea used to find the derivative local maximum / minimum?

[@siddunihary]

Strivers Sir You Are The Best

[@casafurix]

wow, the reversal peak, beautiful

[@HitanshuRathi-zz5vn]

understood thanks brother

[@TrinmoyDutta]

while(low < high-1){ mid = (low+high)/2; if(arr[mid]>arr[mid+1] && arr[mid]>arr[mid-1]) return arr[mid]; if(arr[mid+1]>arr[mid]) low = mid+1; else high = mid-1; } if(arr[low]>arr[high]) return arr[low]; else return arr[high]; initially came up with this solution, it was working and was accepted in leetcode

[@boomerofficial7177]

Kya question samjhaya hai bhai 👍👍👍

[@kenkaneki669]

Damn, was eagerly waiting for this!!!

[@rashikhandelwal4121]

Bestest explanation ❤

[@Ald0Abdiel]

Amazing explanation! thanks teacher :)

[@PrioritizingPeace]

Bs-9 Done ✅ Weekend 27 July 2024 7:43 pm

[@kingbadshah452]

thanks striver understood everything

[@Vivek-t9o8g]

Excellent explanation

[@agamsrivastava8037]

In the last example peak element will be 3 instead of 5 @3:34

[@muntasirayan6006]

if we give last condition (else high = mid -1;) that will work for both 1 peak element and also multi peak element.

[@snehashisratna9074]

luv the way u teach

[@rohitsai806]

Fantastic teaching...!

[@cursed_nerd]

welcome back, waited for you every single day

[@ammanchhetri6716]

there is also a brute force approach using stl ....that is we know whatever the case is.... the maximum element of the vector is always a peak element...so we just return the index of the max element.... int ind = max_element( nums.begin(), nums.end() ) - nums.begin() ; ig

[@kayathrivenkatesan7712]

Great video sir..thank u for saving us

[@LemesaElias]

Someone wrote and I quote "Striver can transfer his confidence to his students" .

[@dipingrover1970]

very good explanation .

[@ashokyadavfqplmgxjia4668]

far better than other vedio

[@charan123rams3]

Sir please complete as soon as possible please sir 🙏🙏😢

[@riteshbisht94]

God level content 🥵 🔥🔥

[@Aman_Panchal27]

understood, you are BEST !

[@impalash_ag]

Hi Raj, could you help me understand how did you approach this problem which made you think of writing a solution considering there's only a single peak element in the array?

[@t3ch_r4id]

Tried to code it by myself and it's working fine 🙂 but the only problem is it's returning 7 7 and only 7 no matter what changes you make it will give you seven 😢😐

[@draxxgamingf.f6369]

Thala for a reason

[@sukritinarang5016]

Great Playlist . 👍

[@hariomtiwari7605]

UNDERSTOOD 👌👌👌👌👏👏👏👏

[@akashddeepchitransh4537]

Mind blowing video..

[@LokeshKumar-lv3zq]

Compiling your Code... > Success! Running Test Cases... > TestCase - Easy Failed Wrong Answer Your program's output doesn't match the expected output. You can try testing your code with custom input and try putting debug statements in your code. Your submission failed for the following input Arg 1: An Integer Array, For e.g [1,2,3] [1,1000000000,1000000000] Test As Custom Input The expected return value: 1000000000 Your function returned the following: -1 Final Verdict > Wrong Answer Failing for input - [1,1000000000,1000000000] can someone help me Identify the error(Note: the code is same as mentioned in this video)

[@dayashankarlakhotia4943]

Very well explained. Understood

[@SatyaMantha]

Striver you are amazing

[@mehrabhossain8225]

you are really great one ..

[@harshitsingh3510]

can we simply find the largest and return that index , it will also give the peak😜

[@mohitt_amv]

it will take o(n) if im not wrong!?

[@KR_Technical-hj3bf]

@@mohitt_amvif i am not wrong we have given a condition is that it is should be greater then their next element

[@harshithapogala8845]

The condition you siad is right but we are assuming that ithe first elements left and last elements right would always be-infinity so automatically it will be the greatest element in the edge cases and@@KR_Technical-hj3bf

[@ganeshjaggineni4097]

NICE SUPER EXCELLENT MOTIVATED

[@AnmolGupta-oj4lm]

Understood Very Well!

[@BrownMonster17]

love you struggle bro

[@anshpalekar152]

But what if question is [1,2,1,3,5,6,4] , it will eliminate left half and go to right where peak is not there

[@NewtonVarma]

If we are to return an array of indices where peaks are found(all peak) can we still apply binary search by making some modifications and applying if else checkups. 🤔

[@RajNamdev_19]

Understood.

[@aayushshrestha1655]

TO AVOID THAT INFINITE LOOP CONDITION WE CAN JUST CHANGE THE LAST IF CONDITION TO : if(a[mid] < a[mid-1]) { .....} this will prevent the infinite loop

[@NazeerBashaShaik]

Understood, thank you.