BS-8. Single Element in Sorted Array

[@takeUforward]

The brute force can be better by just doing a XOR, but the reason we did that brute was to understand the binary search approach!

[@sagarpatle461]

Outstanding 😊

[@sudhanshushekhar4222]

Understood

[@DeepakKumar-uq9js]

Thanks bhai Ji😃

[@samuelfrank1369]

Will not the time Complexity differ? Brute force by xor will be O(n) where as this will be O(logn). Or are you with respect to the conditions that are being checked for logn times will be equivalent to O(n) time Complexity?

[@gauravpandey-fx9hk]

Brute force Simplified: int n = nums.length; if(n == 1) return nums[0]; // O(n/2) for(int i = 1;i

[@ratnadeepsaha7675]

8 video at once. U r a legend for the community. Salute.

[@yashmundada2483]

In short, If I'm on (even, odd), the element occurs after me, so eliminate everything before me (the left half) If I'm on (odd, even), the element occurred before me, so eliminate everything after me (the right half) Great video as always!

[@thefourhourtalk]

thanks buddy

[@nehalchaudhary3319]

These made this ques just a piece of cake, thanks 👍🏻

[@shambhaviaggarwal9977]

Thanks

[@PrakharSrivastava-k2d]

Another way to implement this without reducing the search space would be to use the condition (r>l+1), this way you are always ensuring that the search space is atleast of size 3. So now in the end, your anwer would be either arr[l] or arr[r] and you can check for both of them. Also you'll have to place a check for when the array size is 1. By using this technique, you won't have to write code for edge case and also you don't have to think about reducing search space. Although striver's approach of reducing search space was also amazing.

[@lakshyagupta9435]

The way you explained the approach is just awesome.

[@PrioritizingPeace]

weekend 27 July 2024 - Streak-1 I previously studied the binary search concept. I've now started Striver's playlist and just completed the first eight videos. Let's keep going!

[@shivajirao999]

wrote the code in one go, without any error, all the test cases passed, the satisfaction level is insane

[@sayanpradhan1366]

apne is question ka logic khud banaya tha ya Striver bhaiyya ka video dekhne ke bad kiya tha ?

[@raghavmanish24]

it's obvious sir , how one can not understand this simplest explanation.......thankuu so much sir

[@adityanigam8373]

Striver you are the best, you clear even the smallest doubt, I always had a doubt to whether to take low< high or low

[@AdityaKumar-be7hx]

Finished all 8 videos Striver :) When can we get rest of the videos? Thanks for putting in so much of effort to make these awesome DSA playlists available for free. All these (graph, DP, trie, tree, recursion, etc) are truly the best.

[@YogeshKumar-px5bd]

The solution is great. More focussed on writing clean and readable code but not so much intuitive at first.

[@harshsolanki1058]

Two pointer method also gets the code done in O(log - base 2 - n). Keeping pointers low=0 and high=n-1 and doing simultaneous search and increasing or decreasing pointers by 2 @takeUforward

[@VasanthChoudary-uc5cz]

would'nt it take o(n/2)??

[@shibainu7500]

Two pointer is O(N) because you are traversing each element atleast once even though the number of iterations are n/2 In binary search, we completely reject half of the search space and that's why it is O(logN)

[@arijitroy8652]

We can write the same for loop loop ie. for(i=1; i

[@ujjawal_]

00:00 Problem Explanation 02:42 Bruteforce (Approach 1) 04:52 Edge Cases 05:45 Binary Search (Approach 2) 20:27 Code

[@ratnadeepsaha7675]

Good work 👍

[@senseiAree]

I'm sorry for not being able to continue for some days. I had additional workload at my office which halted my learning curve but I made sure my daily streak is maintained in Leetcode and Coding Ninjas.

[@toxaim9936]

i did it by length till mid approach(different way to look at same thing): 1) if length from low to mid is odd then storing only pairs will overflow the last pair such that num[m]==num[m+1] , so move right 2) if len from low to mid is even then storing only pairs will exhaust the length such that num[m-1]==num[m], so we move right

[@chiragsharma8905]

00:06 Find the single element in a sorted array. 02:25 Identifying the single element in a sorted array using Brute Force 04:53 Apply binary search to optimize the code 07:16 Identifying the half and location of the single element. 09:30 Write a lot of edge cases and eliminate conditional statements 11:45 Performing binary search to find the single element in a sorted array 14:04 Identify if you are on the left half or the right half and eliminate accordingly. 16:19 Binary Search to find the single element in a sorted array. 18:42 In binary search, we eliminate the left or right half of the search space based on whether we are standing at an odd or even index. 20:42 The main focus of this video is on code readability, consistent use of variables, and understanding the concept of elimination in binary search. Crafted by Merlin AI.

[@baderajesh23]

Eliminating left or right part based on even,odd logic is awesome :)

[@Satya-g5t]

I really appreciate your enthusiasm and solutions.

[@neerajkumar-ts6om]

Ok so now onwards I will try to solve the edge cases before in the problem, this will allow me to focus on the main logical part and reduce stress.

[@bapanapallihemanth705]

You are the king 👑 of coding striver

[@AbhishekSingh-cq2jx]

understood better than Scaler paid cource really Thank You

[@abhirajrohilla]

class Solution { public: int singleNonDuplicate(vector& nums) { int ans=0; for(int i=0;i

[@aayushgakhar3525]

it has complexity of n

[@vijaynag7723]

its really great series ,Thanks Striver. Aap nhi hote to humara kya hota !!!!!!!!!!!!!!!!

[@Manasidas99]

Great video help me a lot I can't explain how much help it Thank you, sir.

[@maaz.i7]

Can be done by XOR Just keep doing XOR from start to end of array, the element which appears only once will be the final result of XOR, rest will become 0 because A XOR A = 0 and 0 XOR A = A

[@Shunya_Advait]

Understood Sir, thanks a lot for this amazing video.

[@Vamshi9876]

Thanks. Great way of explaining complex questions.

[@MJBZG]

another solution is take xor of all the elements, TC ---> O(N), SC = O(1)

[@PankajSingh-kz7or]

we can also elimate a particular half on the basis of size of the array because single element will always be present in odd size array

[@charan123rams3]

Sir your really great and inspiring us to learn more about the coding,thank you so much 😢

[@bhupendramaurya6587]

bhaiya, you are explaining the concept too good, Thank you so much.

[@abhishekkumar-fe8lw]

We can also trim search further by putting left=2 , and right =n-3.

[@sayakghosh5104]

Mind blowing explanation.

[@chethanprabhu4475]

I think we can consider low = 0 and high = arr.length - 1. Always there will be mininum 3 elements in array and hence mid can never be equal to low or high.

[@sushantguria8820]

Understood , amazing explanation as always.

[@cinime]

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

[@djsanket25]

sir please create a course on English speaking , i loved your way of speaking . you course will be brilliant one. it will be very helpfull❤❤❤❤

[@iamprateek3220]

Another solution: int singleNonDuplicate(vector& arr) { int n= arr.size(); int l=0, r=n-1; if(n==1){ return arr[0]; } while(l

[@gauravmasand]

Hey Striver this is one better code like i use mod operator to change the value of mid and i find ur approch is bit of lengthy and this has also O(log n) time and O(1) space complexity class Solution: def singleNonDuplicate(self, nums: List[int]) -> int: if (len(nums)==1): return nums[0] left, right = 0, len(nums) - 1 while left < right: mid = (left + right) // 2 # Ensure mid is even so that it pairs with mid + 1 if mid % 2 == 1: mid -= 1 # If the pair is correct, the single element is to the right if nums[mid] == nums[mid + 1]: left = mid + 2 else: right = mid # left will be pointing to the single element return nums[left]

[@udatyadeb101]

understood, for java people who are getting stuck in 25th test case, instead of using == operator, use .equals and it will pass.

[@tech1hutz]

Really Great explanation bhaiya ❤ pls can you also make a series for greedy algo questions and its approach obviously after completing this ongoing binary search series.....Its much needed coz your way of explaining approaching a problem really helps in building concepts as well as clear understanding of any problem.Thank you so much for all the series you made ...

[@neerajgupta9151]

yes bhaiya, plz make one on greedy

[@TrendyGamer007]

@takeUforward @striver please upload the remaining binary search videos as most of us have already finished watching all 8 videos … and the content was superb 👍🏻👍🏻👍🏻.

[@sufyainposharkar2080]

Here's the code with same logic but simpler, class Solution { public: int singleNonDuplicate(vector& nums) { int low = 0; int high = nums.size() - 1; while (low < high) { int mid = low + (high - low) / 2; // Ensure mid is even for comparison with mid + 1 if (mid % 2 == 1) mid--; // If pair is found, the unique element is in the right half if (nums[mid] == nums[mid + 1]) { low = mid + 2; } else { // Otherwise, it's in the left half high = mid; } } return nums[low]; } };

[@yashraj5898]

understood ! great work buddy !

[@ashishranjan8350]

Hey striver please upload rest of the videos in this series.

[@mohdbilal2]

simple brute force is take xor of all element with xorr=0; ultimately you will get single element

[@takeUforward]

Yes but the reason of saying this brute force was to explain the thought process of binary search

[@mohdbilal2]

@@takeUforward Okay, Thanks brother 🥰 You are doing a great work.

[@hiteshpanchal5772]

Striver bhai maza hi aa gaya ...............one request to you is plzz bhai upload video alternate day😍

[@jeehub041]

Sir series me maja aa raha .... Ab aage ka videos bhi upload kar do please🙏🙏🙏

[@manavsingh5919]

Thank you Striver...Understood everything

[@prathmeshparab7046]

Amazing explanation ❣❣

[@58harshverma57]

Quite interesting question!

[@AYJ959]

Simple O(n) in java return arr.stream().reduce(0, (a,b) -> (a^b));

[@rushidesai2836]

Weirdly good question.

[@VasanthChoudary-uc5cz]

you can also optimise the brute force by using two pointer technique.

[@traymk]

Kya mast thumbnail hai.. 🔥

[@dayashankarlakhotia4943]

Good explanation with dry run understood

[@abid3801]

Striver one dumb question. When you started practicing DSA .Did you able to come up these tricks on your own at least for some?

[@Thorfinnn1]

Never bro it take several times to build a avg logic

[@suyashsonar305]

This was a good question to understand BS But if we have time constraints we should go for the liner XOR operation of each element in the array and store the xor. XOR will ultimately eliminate the 2 times appearing numbers and we will end up getting the single occurring number in the array. Great solution by Striver as alway ❤❤

[@montypannu1975]

Yes but it will have T.C of O(N) and with binary search we get O(logN)

[@pokigamerop]

My O(1) solution: That Single element is me :)

[@sandeep4915]

that trimming❤❤

[@ayushgaurabh8604]

superb explanation

[@rkgk5445]

The way you say Single 😀, I don't think it's your problem Sir... right now Although, kudos to your way of explanation...👏

[@kingbadshah452]

thanks striver understood everything

[@inderjeet09]

Please bro make a playlist on bit manipulation also thats a very difficult topic for us . Only you can make that easy.

[@chphanisimha7802]

Such an incredible work!!

[@prayasgautam9595]

To further optimise this BS approach, we can add a condition of (!(mid & 1) && nums[mid] != nums[mid - 1] && nums[mid] != nums[mid + 1]) as the single element will always be at an even index.

[@adarshkumarrao3478]

UNDERSTOOD

[@bhavyasrikanchi5753]

excellenttttt explaination..!!!!! Understood

[@be-sreenu2202]

it might be possible using XOR but TC-O(N)

[@sagarshah5341]

MIND BLOWN AT @7:14 DAyummmmmmmm!!!!!!!

[@atharva_g_vlogs]

UNDERSTOOD SIR

[@rahulkumarbarik7584]

That single element is me 🥲

[@utsavseth6573]

Shandaar.

[@arpitamanderna9111]

Hey striver !! please upload rest of the videos too!! you are the best!!

[@mayankshakya9200]

When to trim the right half and left half is not clear properly plz emphasise on that part only plzzzz and rest is just amazing love form our side

[@takeUforward]

Here it was easy, so did not focus much, if you take a pen and paper, it is an easy problem! With problems we have, stay rest assured.

[@himaniupadhyay8201]

Thanks Lord Striver

[@ashwath1914]

If we start with low = 0, high = n-1, the edge cases should be written inside and the code will look like this: int singleNonDuplicate(vector& arr) { int n = arr.size(); int l = 0, h = n-1, ans = -1; if(n == 1) return arr[0]; while(l

[@aakashsharma780]

Understood @striver ❤🙌🥳

[@samuelfrank1369]

Understood. Thanks a lot.

[@jayantgahlot2653]

Better Code: while(lo

[@VenusaiYalamanchili]

How am I gonna get that observation of even odd, or odd ,even is it easy or am I the only one who felt amazed when Striver said that?

[@abhijitkumarsinha]

count me also

[@hidden_star14]

You're not alone. I feel so dumb, I can't tell you! 😭

[@ShortsChachu]

​@@hidden_star14Same bro 😢

[@NazeerBashaShaik]

Understood, thank you.

[@sunnypunia6485]

bhaiya please complete all lectures of all questions in a to z dsa as soon as possible it will be very helpful

[@anweshandhara2059]

XORRing entire collection

[@Bigg_boss_trolls]

just awesome.

[@abhinavkumar1976]

Bhai ne 1 din me RONIN bna diya code studios par . Send more videos !

[@hemantjagtap7340]

Striver, when will you upload the remaining vedios of Binary Search playlist ?

[@divyadwivedi1527]

Plz upload rest of the video as soon as possible

[@oyeesharme]

understood bhaiya

[@mohitsingh13]

Understood ❤

[@Learnprogramming-q7f]

Thank you Bhaiya

[@shahbazaalam7559]

What if a no repeated odd no of times so In this case this technique of dealing with indexes will me valid or not? For example {1,1,1,2,3,3} Here find the single element

[@DeepakPatel-d5v]

Thanks a lot Bhaiya

[@AshmitIITH]

This problem can also be done using a hashmap or dictionary but with O(n) tc. Still it can be better approach that brute force approach and we can tell this method in the interview ig

[@adbhutakalpniya]

bhaiya bawal ho aap

[@YourCodeVerse]

Understood✅🔥🔥

[@bishalsarkar9205]

Understood.

[@sanbidrc]

Simpler solution: public int singleNonDuplicate(int[] nums) { int l = 0; int h = nums.length-1; while(l < h){ int m = l + (h-l)/2; if(m%2 == 0){ if(nums[m] == nums[m+1]){ l = m + 2; } else{ h = m; } } else{ if(nums[m] == nums[m-1]){ l = m + 1; } else{ h = m; } } } return nums[l]; }

[@harshilpatel3205]

Understood sir 😇😊