Leaders in an Array | Brute - Optimal | Strivers A2Z DSA Course

[@takeUforward]

You don't need to sort the answer, as it will store the answer in the sorted order only when taken from back. Let's march ahead, and create an unmatchable DSA course! ❤ Use the problem links in the description.

[@kshitijkumre9895]

When will this course possibly complete? Placement season gonna start from august!

[@vishal_s583]

Great explanation ❤️

[@chandreshgourd3273]

We can also complete this in O(1) space complexity by updating the current element with the maximum on the right.

[@akshatranakoti3224]

STRIVER BHAIYA RULAOGE KYA ITNI MEHNAT TO BACHHE BHI NAHI KR RAHE . PHLE GRAPH AUR TREE SERIES NE DIL JEETA PHIR DP SERIES NE ABHI YE BHI , EK HI DIL HAI BHAIYA KITNI BAR JEETOGE❤️

[@vardhamanbhandari5644]

@@akshatranakoti3224 he is striving to get out better placement,Hands off to his efforts🙌🙌

[@yhbarve]

I've been learning DSA for almost 1 month now. When I saw this video pop-up, I read the question, worked on it myself, coded my solution on GFG, it worked in the 1st attempt, came back to this video to see the optimal approach, and it was the same as mine! I know this is an easy question, but I'm still proud of myself for getting the optimal approach in the 1st attempt! Thanks Striver!

[@sathvikreddy4807]

same, its been 3 months since you commented this, how's your dsa prep going>?

[@yhbarve]

@@sathvikreddy4807 Going strong. Thanks for asking

[@sathvikreddy4807]

@@yhbarve how's your motivation tho? are you doing it daily? cause after 20 days, I am burned out tbh don't feel like doing it idk, became lazy ig

[@yhbarve]

@@sathvikreddy4807 yeah that indeed happens to me sometime. I just take a 7-10 days gap and then come back to it. Works always!

[@ThattukadaCooks-um4yn]

thats nice and a relief to hear..! : )@@yhbarve

[@ayushdhiman9378]

PLEASE PLEASE PLEASE !! Do not stop this "Brute Better Optimal Approach technique" till the end of the series this is helpting a lott !! Lots of love

[@harshalborle]

Few enhancements: 1. I think ans is already sorted we don't need to sort that. Because we are placing new value only if it is bigger than older. 2. We can directly assign the "maxi = a[I]" in "if (a[I] > maxi)".

[@vadlacharan5442]

yeah that's right! and even if they ask to return the result in prior to their occurence we need reverse the result

[@padmaja577]

Yep no need for the thing he did abt maxi......

[@Sankalp-sd6fm]

how do i return ans as array a

[@ANKAMJANAKIRAM]

Please refer chatgpt for these doubts bro... I know your pain​@@Sankalp-sd6fm

[@itzmartin20]

Understood! 12 minutes of concise and crisp clear explaination! Thanks a lot for your great work and dedication!

[@kimpatel1832]

Best best best youtube channel to learn DSA. Striver is a blessing!!!💎💖

[@riteshbisht94]

After watching your previous videos Now I am able to think about optimal solution for this problem Thanks a lot 🌟

[@mehulthuletiya497]

03:21 AM 😮 00:41- Problem Statement 00:46- What is leader and explain through example 02:24- Brute force solution 05:10- Optimal solution 08:59- Code 10:36- Time & Space Complexity

[@KazHachiOreki]

if he is not in India time could be different

[@takeUforward]

The video was recorded yesterday at 01:00 AM, and uploaded today at 11:00 PM, the time mentioned are in my time zone Could not upload in the morning as I had office, and did not get time to edit, so came back from the office, edited, rendered, and uploaded :)

[@takeUforward]

Thank you for the time stamp ❤️

[@faizanahmed9304]

​@@takeUforward thank you bhaiya! Immense hardwork from u

[@pranayreddy5824]

@takeUforward where do you live, bro?

[@sksabir8063]

Seriously bro i am speechless 😶..... really u r doing great 👍 for tier 3 college.....

[@senpaidawwwggg]

For every tier colleges

[@Manishgupta200]

That's amazing 🔥. I successfully able to solved optimal approach by myself before watching your tutorial.. it's my improvement that now i can able to think optimal approach. (I stuck on > and >= but now clear).

[@san_7123]

yep

[@akashajith3120]

i have been learning from you for the past 2-3 weeks and I solved this one on my own on a similar level of optimization that gives in the order of the array without reversing, thank you for helping me build my logical thinking . here is my solution : vector leaders(int a[], int n){ // Code here vector ans; int size=0,cur; for(int i=0;i0 && cur>ans[size-1]){ ans.pop_back(); size--; } ans.push_back(a[i]); size++; } return ans; }

[@GuruPrasadShukla]

probably easiest question in the series! thanks man!

[@charlesbabbage6786]

Understood every part of the video!! Beautifully explained!!

[@kushagramishra5638]

done by myself + understood!

[@DevilJim-p1s]

Strivers My G I can't believe I'm able to solve questions by optimal approach directly now thanks bro I'm broke rn but as soon as I'll get the job I'll buy your premium course 😅

[@shubhamagarwal1434]

#Free Education For All.. # Bhishma Pitamah of DSA...You could have earned in lacs by putting it as paid couses on udamey or any other elaerning portals, but you decided to make it free...it requires a greate sacrifice and a feeling of giving back to community, there might be very few peope in world who does this...."विद्या का दान ही सर्वोत्तम दान होता है" Hats Off to you man, Salute from 10+ yrs exp guy from BLR, India.

[@sane_aalam]

Allah Tumhe Humesha Salamat rakhe! Tum Bahut Nek Kam kr rahe ho Tumhre Channel Se Graphs + Tree + backtracking sikhne ke baad, leetcode platforms ko aache se use kr raha hu😊

[@sumitkanth5349]

Hello bhaiyaaa, at 10:20 we don't need to check maxi = max(maxi, a[i]), we can change maxi only if, if condition will true.

[@snehashisratna9074]

luv you bhai , thank you for making our lives so easy 😭😭😭😭😭😭😭 lots of love to you

[@soumyadeepdas6441]

Great Explanation understood in first try 🎉🎉

[@gopibhoyar3]

This question is basically, reverse of "Buy and sell Stocks". In Buy and Sell stocks, we use to keep minimum of the value on the left side of ith position. Here we keep maximum of the value on the right side of the ith position.

[@hritikbhatnagar9616]

Your optimization technique is too good❤❤

[@cinime]

Understood! Awesome explanation as always, thank you very much for your effort!!

[@sarangkumarsingh7901]

Another Awesome lecture Sir......................

[@himadriroy7980]

Crystal Clear Striver Da

[@many987]

hey Stiver bhaiya, first of all thanks for such content and I feel we don't need to sort the answer in the end because it will definitely store the answer in a sorted manner. thanks

[@takeUforward]

Yes, you are correct, good observation.

[@shwetachoudhary9003]

you are such a gem❤

[@himanshupatel1053]

vector leaders(vector& arr) { // Code here vectorans; int maxElement = INT_MIN; for(int i=arr.size()-1;i>=0;i--){ maxElement = max(maxElement, arr[i]); if(arr[i]>=maxElement) ans.emplace_back(arr[i]); } reverse(ans.begin(),ans.end()); return ans; }

[@RohitRaj-kr7ti]

Hey, There will be a slight update in the if statement, it should be a[i] >= maxi, and the ans has to be reversed. Some test case will break if `=` is not put in the if condition(Tried on GFG, hence saying). @takeUforward. This may need a correction in the leaders explanation page on your website too.

[@RohitRaj-kr7ti]

Btw, thanks for a nice explanation.

[@yash1525]

Yes you are right, thanks for telling the "=" correction.

[@ravindrav1450]

Clear explanation and Great energy 🙏

[@torishi82]

Awesome!! Great!! Excellent!!

[@Raj-yr3vz]

kitna pyaar se samjhate ho bhaiya🥺

[@jagratgupta8392]

understood striver sir ....really nice concepts and questions

[@culeforever5408]

understood 😇

[@samiranroyy1700]

best teacher😇😇😇

[@GhostVaibhav]

Understood🔥

[@hareshnayak7302]

understood, Thanks striver for this amazing video.

[@DevashishJose]

Understood. Thank you so much striver.

[@MCA_HimanshuSeth]

Thanku Bhaiya for your support , I am super happy to see your success as well as your efforts for students making free content. You are my idol , so many wishesh for u Bhaiya 😊😊

[@striverdaaadi]

awesome bro, love from harry

[@iazizhussain]

Fully understood

[@rajatrathore5455]

Understood , great explanation

[@ganeshjaggineni4097]

NICE SUPER EXCELLENT MOTIVATED

[@knoxtsgamer186]

Thank you for your Hardwork...🙏🙏🙏

[@nice_random_guy5347]

understood! thanks

[@RaviKumar-sn6tu]

understood very well...thanks sir😍

[@kingbadshah452]

understood everything thanks striver

[@ashishpradhan6250]

understood

[@gauravsrivastava17]

Great explanation 🔥

[@shaikimran5441]

Love You Brooooooooooooooooooo........

[@konankikeerthi]

Understood. Thank you bro

[@YourCodeVerse]

Understood✅🔥🔥

[@rajaryaman6495]

It's the Second question, I solved myself right (optimal). Thanks! Striver for your teachings. Here's my code :- vector superiorElements(vector&a) { // Write your code here. vector ans; int n = a.size(); int max = a[n - 1]; ans.push_back(a[n - 1]); for(int i = n - 1; i >= 0; i--) { if(a[i] > max) { max = a[i]; ans.push_back(a[i]); } } return ans; }

[@niteshkhanna690]

It's east

[@RaunitJaiswal-s9v]

Imagine you are at a marathon race, standing at the finish line, watching the runners as they approach. The runners arrive one by one, and your goal is to identify the "leaders" of the race. A leader in this context is someone who is faster than all the runners who arrive after them. Start with the last runner (125 minutes): Since they are the last to arrive, there's no one after them. Therefore, this runner is a leader. Move to the previous runner (130 minutes): Compare 130 minutes with 125 minutes. Since 130 > 125, this runner is also a leader. Move to the next runner (115 minutes): Compare 115 minutes with 130 minutes. Since 115 < 130, this runner is not a leader. Move to the next runner (110 minutes): Compare 110 minutes with 130 minutes. Since 110 < 130, this runner is not a leader. Move to the next runner (105 minutes): Compare 105 minutes with 130 minutes. Since 105 < 130, this runner is not a leader. Move to the first runner (120 minutes): Compare 120 minutes with 130 minutes. Since 120 < 130, this runner is not a leader In this context, the leaders (130 and 125 minutes) are the runners who are faster than everyone who finishes after them. This mirrors the concept of "leaders" in an array, where each element must be greater than or equal to all the elements that come after it.

[@Mythri333]

Thank you so much ❤❤❤❤❤

[@ArvindSingh-wj7vy]

understood boss 😎

[@Gaurav_Tripathi_]

Understood Bhaiya.!!

[@faizanahmed9304]

Understood bhaiya

[@HARSHA_27]

Understood!!🙇‍♂

[@sumitworld918]

Understood 💯❤

[@samuelfrank1369]

Understood. Thanks a lot

[@curs3m4rk]

This was very easy

[@tahaansari5621]

Woke up for 2nd Ramzan Sehri and saw ur video and clicked. The question was relatively easy so I understood in the first go.

[@md_seraj786_]

Understood : 💜

[@AbhinavPundhir]

class Solution{ //Function to find the leaders in the array. public: vector leaders(int a[], int n){ vector vec; vec.push_back(a[n-1]); for(int i=n-2; i>=0; i--){ if(a[i]>=vec.back()){ vec.push_back(a[i]); } } reverse(vec.begin(), vec.end()); return vec; } };

[@rumiNITPatna]

thank u striver!

[@lifehustlers164]

Completed 12/28!!!

[@banothutharun2743]

great sir

[@BG-lj6fw]

thanks a ton. understood!!

[@swapnilhajare5557]

Understood 👍🏻

[@srushtibhagchandani7255]

Understoood!!

[@harshalparanjiya5850]

let's go --> 1 january 🔥🔥

[@nikhilgautam007]

Thanks buddy 🙏

[@RaunitJaiswal-s9v]

int max_from_right = arr[n - 1]; leaders.push_back(max_from_right); // The rightmost element is always a leader // Traverse the array from right to left for (int i = n - 2; i >= 0; i--) { if (arr[i] >= max_from_right) { max_from_right = arr[i]; leaders.push_back(max_from_right); } } This approach has a time complexity of O(n) since we traverse the array only once.

[@tamilmukbang3789]

understood thank you bro

[@utsavseth6573]

Thanks Raj.

[@RajGupta-i3w]

Just a small thing that I want to share what I observed in this question is that, even if question is asking to return the answer array in sorted way, we dont need to apply sorting because since we are taking elements from right to left and according to question a leader must be greater than all elements occuring at right, so the answer array will automatically will be in a sorted order. so Overall Time complexity will only be just O(N). I hope I am able to explain my observation.

[@AbjSir]

u can do easily in O(n) if you use list and then store elements as list.push_front()

[@atozhungama2321]

understood👍

[@ankushladani496]

Thank you Bhaiya

[@ali_dev7415]

GFG solution in C++:- class Solution{ //Function to find the leaders in the array. public: vector leaders(int a[], int n){ // Code here vector ans; int maxx=a[n-1]; ans.push_back(a[n-1]); for(int i=n-2;i>=0;i--){ if(a[i]>=maxx){ ans.push_back(a[i]); maxx=a[i]; } } reverse(ans.begin(), ans.end()); return ans; } };

[@per.seus._]

understood❤

[@gautamsaxena4647]

understood bhaiya

[@stranger1967]

understood sir ❤❤

[@asheshkaran5843]

Understiood!!

[@viralmoments571]

if not work write if statement as follow if(arr[i]>=maxi)

[@rajatpatra2593]

Understood ❤️

[@ultimategamer5136]

UNDERSTOOD

[@sayakghosh5104]

Understood!

[@rahil1575]

in the if loop we can directly assign maxi=a[i]

[@shaheenparween4326]

understood✔

[@SiddheshKukade]

Thank you so much

[@oyeshxrme]

thanks bhaiya

[@amitpatil7659]

Understood :)

[@anurag_vish]

Understood....

[@samarthpai5359]

understood!

[@ayushgupta0]

Understood.

[@mrsttechno5974]

Understand sir

[@rohanraj2604]

Leetcode Question No. 1299 is just a variation of this question try it out for those who do not find the leetcode question link in the A2Z sheet.

[@naveensaicremsiyadlapalli3769]

for better approach can we use next greater element concept using stack ???

[@takeUforward]

Not needed unnecessary space