Rearrange Array Elements by Sign | 2 Varieties of same Problem

[@takeUforward]

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[@UECHarshaVardhanRavipalli]

Thanks for the consistency bhai!!!😇

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@take U forward bhai aise hi aapne sikha diya gara toh baaki ka pata nahi but main pakka google mein aaunga ,Hope You complete this course as soon posssible. Your video lectures are best!

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this is undoubtfully an unmatchable dsa course

[@BootCampResources]

Can you please explain in place approach using two pointer?

[@AshishPrajapati-j7r]

I believe this can be one solution ``` const putInAlternativesOrder = (list) => { let pointer = 0; let positive_flag = true; for (let index = 0; index 0) { pointer++; positive_flag = false; } else if (list[index] > 0) { [list[index], list[pointer]] = [list[pointer], list[index]]; pointer++; positive_flag = false; } } else { if (list[pointer] < 0) { pointer++; positive_flag = true; } else if (list[index] < 0) { [list[index], list[pointer]] = [list[pointer], list[index]]; pointer++; positive_flag = true; } } } return list; }; ```

[@manthenanagendra1077]

Thank you striver because of you i was able to solve all 4 questions in todays leetcode contest..,you will definitely reach great heights for all the work you are doing

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Trust me I have watched all most all your videos but right now I feel the quality content, sound, depth of explanation, everything has reached an ultimate level. Ty for such an awesome content. Hope one day I will also have knowledge and confidence in DSA same as you.😀

[@takeUforward]

Yes, with time I have understood, what exactly people need.

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@@takeUforward NICE BRO

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Review about striver....tree,graph and dp playlist??.....please !!

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@@captain-ne8qy wherever you go, the topics mentioned are must, bare minimum pick any sheet and solve 80-85% of the questions. My personal view on this, if you do TakeuForward you would grasp the concept well, it would definitely take some time, but you will be crystal clear and sound in the concept as well as solving questions. I think the sheet is designed in such a way and every question seems linked and connected. PS. This is totally based on my own understanding. Otherwise you can use KZbin and watch other channels too or search for individual questions and then do the topics .

[@Maneeshce2007]

I find comments are the beginning of every video so true ... This is is unbelievable and the explanations are so crisp and clear. I am enjoying DS for the first time.

[@sayantanpoddar5428]

UNDERSTOOD!! superb explaination ....PLEASE UPLOAD STRING VIDEO SERIES

[@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....

[@ParasSharma-mf8or]

0:00 Introduction of course 0:42: Rearrange array elements by sign(variety 1) 2:33 Brute force 4:30 Pseudo code (Brute force) 5:43 Optimal approach 8:36 Code-compiler (Optimal) 9:45 Variety 2 13:38 Pseudo code (Variety 2) 15:25 Code-compiler (Variety 2) 21:01 Outro

[@stith_pragya]

Understood.....Thank You So Much for this wonderful video....🙏🏻🙏🏻🙏🏻🙏🏻

[@vishalsharma9660]

Is bande ke course ke har video par like banta hain...so thoroughly explained.

[@346_PrachiRaut]

this man is literally an angel for devs

[@chillin_08]

How can the lectures be this amazing ✨.. You are a saviour to us for suree!!!

[@manasranjanmahapatra3729]

UnderStood. Started Preparing Again. Just wanted to tell you I was a in Support Shifted to Development last year. Your content helped me a lot ❤💪.

[@Mythri333]

The best channel among all channels in world ❤..Thank you so much 😊

[@Mb-nr5nz]

Thanks for giving such clear explanation, i understood each step you said from brute to optimal , question done 👍

[@manvendrasinghrathore8100]

One word for you : King of explanation/knowledge

[@sarangkumarsingh7901]

Understood Sir...............Awesome Lecture...............

[@cinime]

Understood! Super awesome explanation as always, thank you so much for your effort!!

[@saswatrath4646]

Bhaiya simplified appproach of your variety two Just the thing changed in the question is we have to insert negative at even indexes and positive at odd indexes a problem from code studio. #include void rearrange(vector &arr) { // Write your code here. int n = arr.size(); vector pos, neg; for (int i = 0; i < n; i++) { if (arr[i] < 0) { neg.push_back(arr[i]); } else { pos.push_back(arr[i]); } } int index = 0, i = 0, j = 0; while (i < pos.size() && j < neg.size()) { arr[index++] = neg[i++]; arr[index++] = pos[j++]; } while (i < neg.size()) { arr[index++] = neg[i++]; } while(j < pos.size()){ arr[index++] = pos[j++]; } // return; }

[@haryanvibhajan001]

You don't need motivation , you are MOTIVATION!!!🏴‍☠🏴‍☠

[@kingbadshah452]

thanks striver understood everything

[@brajeshmohanty2558]

Itna consistent ke se ho bhai thanks per mai morning Mai dekhuna ga🙏 love you

[@arijit9862]

Hey Striver! Here, "vector ans(n , 0)" this will also take a O(N) time, right? so technically it is O(2*N), isn't It ??

[@takeUforward]

technically #include also takes a lot of time, usually, these things are not considered as we speak during interviews.

[@md.ualiurrahmanrahat2400]

The time complexity explanation for variety 2 problem was lit!!!

[@prathameshjadhav2942]

Great explain

[@aakashpatel1250]

Best explanation ever 🤩🙌

[@samuelfrank1369]

Understood. Thanks a lot. Please upload more videos Bhaiyaaa

[@aryansinha1818]

What an awesome mic, really worth it!!

[@sukhpreetsingh5200]

Amazing explanation

[@ankush8243]

Thanks Bro💙 Keep teaching like this!! Appreciate your efforts!!

[@Raj10185]

My solution for the second variety :- #include using namespace std; int main() { vector arr = {-1,2,3,4,-3,1}; vector pos ; vector neg; for(auto it : arr) { if(it>= 0) pos.push_back(it); else neg.push_back(it); } if(pos.size()==neg.size()) { int k =0; int j =0; for(int i =0;i

[@anuplohar23]

Hats off to your efforts Sir ❤❤

[@Josuke217]

Found the optimal solution on my own :) Great series Striver!

[@shauryajain5358]

Understood and thanks for another great lecture! Just a small confusion , I submitted the optimal solution for variety 1 on Leetcode but it took more time than the brute force solution. Why is that?

[@divyanshujaiswal4682]

for variety 2 i use 2 stack to store +ve and _ve elements and then pop one by one untill both the stacks are empty and store into an ans array.

[@rohankadam7376]

Super easy explanation, Thank you !!

[@aniketgarg9575]

Such a wonderful explanation thank you sir 🙏

[@infernogamer52]

Understood bhaiya!

[@lifehustlers164]

Completed 9/28!!!

[@srilathareddy9450]

Excellent explanation, understood

[@hareshnayak7302]

Understood,Thanks srtiver for this amazing video.

[@SahilKumar-vd7vj]

I think in this video you tell us the problem of alternative numbers - [-1,2,3,4,-3,1] the answer of this - [3,-1,4,-3,4,1] your answer is the very complex that you explain in this video, and when i ask to the chat gpt the answer will be very easy and understandable also, i am not comparing with AI but please do answer as much as that we remember long , and in your answer it is very difficult to remember the approach.

[@NazeerBashaShaik]

Great explanation, thank you.

[@dEviL_bIsWaJiT]

Variety-2 - Alternate solution(C++) vectorpos; vectorneg; for(int i=0;i=0){ pos.push_back(arr[i]); } else{ neg.push_back(arr[i]); } } int i=0; int j=0; arr.clear(); while(i

[@jineshnadar6409]

Variety 2 without using the extra spaces for pos and neg. class Solution: def rearrangeArray(self, nums): ls = [ 0 ] * len( nums ) pos = 0 neg = 1 minCount = 0 for i in range( len( nums )): if nums[ i ]> 0: minCount +=1 negNumsAreMin = False if minCount > len( nums ) - minCount: negNumsAreMin = True minCount = len( nums ) - minCount count = 0 for i in range( len( nums )): if nums[ i ] > 0: ls[ pos ] = nums[ i ] if negNumsAreMin: if ( count < minCount ): pos +=2 else: pos += 1 count +=1 else: pos += 2 else: ls[ neg ] = nums[ i ] if negNumsAreMin == False: count +=1 if ( count < minCount ): neg +=2 else: neg += 1 else: neg += 2 return ls s = Solution() print(s.rearrangeArray([-1,-2,-3,-4,1,2,3,4,5,6,7]))

[@aliakbaransaria3-925]

Very good explanation keep it up

[@RealRaghu09]

We can also do swaping with odd index element with even index element if element are positive or negative Time Complexity is O(N)

[@VENUGOPALRAO-w6h]

thanks you for your explanation I understood it but sir can this be solved using pointers

[@mehaksharma7959]

i have a question do I need to first try to solve the problem by myself then watch video or should i watch videa then solve the problem myself?

[@ravansingh6325]

I think in second variety 2 pointer solution will be the optimal one.

[@MohitKumar-o3l1u]

Understood.

[@ijatin1213]

Variety 2: Done using 3 For Loops vector alternateNumbers(vector&a) { vector pos ; vector neg; for(int i = 0; i< a.size(); i++){ if(a[i] >= 0) { pos.push_back(a[i]); } else{ neg.push_back(a[i]); } } for(int i = 0 ; i < pos.size() ; i++){ a[2*i] = pos[i]; } for (int i = 0; i < neg.size(); i++) { a[2 * i + 1] = neg[i]; } return a; };

[@The_yellow_flash-07]

it was brilliant.

[@DevashishJose]

Understood, Thank you so much

[@jatinkaushik2607]

Instead of collecting the nums in two array, can we do like catch the count of pos and negative numbers and use loop as per greater count conditions? this one is also optimised instead of using space of two array we are counting on reference variables.

[@takeUforward]

We can, it is pretty much the same if you consider tc and sc

[@Aditi_krishna]

is it possible to optimise space complexity using swaping????????????

[@rwordspecialist6734]

Using pointers??

[@rachitsingh2013]

Do you, at any point in the near future, think of creating a backend development course?

[@xoxoxo7313]

Great explanation Striver !!!

[@gourabbhattacharjee2128]

Any Java people, here's a short code using functional programming.. public int[] rearrangeAltPosNeg(int[] arr) { var res = new int[arr.length]; var pos = Arrays.stream(arr).filter(num -> num >= 0).toArray(); var neg = Arrays.stream(arr).filter(num -> num < 0).toArray(); var cmnSize = Math.min(pos.length, neg.length); for (var i = 0; i < cmnSize; i++) { res[2 * i] = pos[i]; res[2 * i + 1] = neg[i]; } var remArr = pos.length > neg.length ? pos : neg; System.arraycopy(remArr, cmnSize, res, cmnSize * 2 , arr.length - cmnSize * 2); return res; }

[@priyanshughosh6143]

both the problem links are pointing to same question. plz provide us with the correct links

[@konankikeerthi]

Understood bro. Thank you

[@GhostVaibhav]

Understood🔥

[@priyankagarg3891]

Can it be solved in O(1) space complexity?

[@gossgowtham]

Understood both!

[@pushkardeore4324]

in the first variety, what if we first put all the positive numbers before the negatives in O(N) complexity like we sorted ones and zeros, and then again rearrange them in the alternate manner with another O(N/2) complexity? so that tc=O(2N) and sc=O(1)

[@charan123rams3]

Thank you so much sir 🥰

[@isheep9025]

self note: the following does the same in 0(1) extra space but does not preserve the original order if positive and negative elements class Solution { public: vector rearrangeArray(vector& nums) { int temp=-1; for(int i=0;i

[@saikumargatla4706]

Hey striver what if zero is present in array will it be considered pos or neg .or else it is stated that array will have no zeros

[@smritiii7697]

0 is also +ve ig

[@abhijeetmishra3804]

understood. but bhaiya how to remember this many algos like array has then linked list stack queue tree graph and all this how will we remember so many codes and algorithm during interview?

[@Se7ven_Plays]

Love you bro I'll watch it later

[@Curiosity_hub2404]

can we really rearrage it i th same array...is there any way to do that?

[@AbhishekPandey-dj2eo]

Keep the good work going on

[@mr_noob6361]

Hii sir .....i am not like a wizard coder as like you ..but i made 2nd variety in o(n) pass here is my code.... arr1=[3,1,-2,-5,2,-4] #not using any extra space # for i in range(len(arr1)): # if(i%2==0):#even index # if(arr1[i]>0): # continue # else: # #kya pata woh negative hoo # for j in range(i+1,len(arr1)): # if arr1[j]>0: # arr1[j],arr1[i]=arr1[i],arr1[j] # break # else: # if(arr1[i]

[@ronitakhariya4094]

Hello Great job in writing these codes. i saw both of them and noticed that yes the first code is not taking any extra space so sc is O(1) but the time complexity for worst case will be o(n^2) if all your values are negative or all are positive since its not gonna break from the nested for loop and around O(n^2) if all elements are negative except the last one and so on. In your second solution you changed time complexity from O(2N) to O(N) however when calculating time complexity for larger inputs we tend to ignore the constant part in big O notation for analysing scalability and efficiency. so it does't exactly optimise the solution. I wrote this comment to just clear doubts in new students and not to criticise your solution.

[@mr_noob6361]

@@ronitakhariya4094 😊😊 thanks

[@smartcreation2671]

is Big O(2n) == Big O(n)

[@gagan_.hegde14]

no

[@Mrindia-k8n]

00:06 Rearrange array Elements by sign 02:18 Rearrange positive and negative numbers maintaining their order 04:14 The solution involves rearranging elements in an array by separating positive and negative elements. 06:34 Implementing the optimal approach for array reversal 08:53 If positive is not equal to negative, there can be two cases: either the number of positives are greater than negatives or the number of negatives are greater than positives. 11:01 There can be two cases: either the number of positives is greater than negatives, or the number of negatives is greater than positives. 13:18 Algorithm for arranging positive and negative numbers in alternate order 15:31 Separate positive and negative elements in an array 17:30 Extract positive and negative dots from given data. 19:30 The worst complexity of the problem is O(n).

[@momentcoder]

Understood ✅

[@swastiktiwari7066]

First variety doing in place without creating answer vector: class Solution: def rearrangeArray(self, nums: List[int]) -> List[int]: n = len(nums) w = 20 # num bits for number, w+1 (0-idx) is sign bit inneg = 1

[@Editzx.10]

Understood Champ!

[@samridhiarora4014]

00:06 Rearrange array Elements by sign 02:18 Rearrange positive and negative numbers maintaining their order 04:14 The solution involves rearranging elements in an array by separating positive and negative elements. 06:34 Implementing the optimal approach for array reversal 08:53 If positive is not equal to negative, there can be two cases: either the number of positives are greater than negatives or the number of negatives are greater than positives. 11:01 There can be two cases: either the number of positives is greater than negatives, or the number of negatives is greater than positives. 13:18 Algorithm for arranging positive and negative numbers in alternate order 15:31 Separate positive and negative elements in an array 17:30 Extract positive and negative dots from given data. 19:30 The worst complexity of the problem is O(n). 21:19 Follow me on all social media handles

[@placement123]

can it be done without extraaa space??

[@amithkotian7556]

Good one.. understood

[@bharathi66]

Understood Bhaiya

[@rumiNITPatna]

amazing video!

[@_hulk748]

Understood sir❤🙏🙇‍♂

[@tamilmukbang3789]

understood thank you bro

[@heyOrca2711]

Understood Sir!

[@AniketKumar-hf2bo]

UNDERSTOOD ❤❤❤❤👌👌

[@ksankethkumar7223]

Now the problem is changed same as variety 1 in CodeStudio too!!

[@takeUforward]

OOps :|

[@gautamsaxena4647]

understood bhaiya

[@mehaksharma7959]

and one more thing I try to solve the problem after a few days the one thing that i remember is brute force , i sometimes forget the optimal solution like for algos like kadanes

[@kk420100]

i always use this approach..it's a generic approach - Step1. Solve a problem using B Force Step2. For the Better Solution try using 1 for loop Step3. Better solution 2 - Try using hashing Step4. Optimal Solution if any for this question, your default approach should be converting 2 or more for loop to 1 for loop. Here that will be also the optimal solution. In this case, it's just a re-write of BF

[@per.seus._]

under❤stood

[@oyeesharme]

thanks bhaiya

[@shaiksoofi3741]

Thank you

[@prabhagaikwad4849]

Understood as always 😍

[@Ayushsharma-qc7qx]

Understand

[@dummyaccount5960]

This is the question of easy type. Donno why it is in the category of Medium in A2Z DSA Sheet...

[@vineetmittal3837]

For first variety of problem my solution would be : int j = 0; for (int i=0; i < n; i++){ while (j = 0) || (j%2 != 0 && arr[j] < 0)){ j++; } else { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; break; } } } This is not using any extra space and time complexity is O(n). Please correct me if there is any mistake.

[@tejas7379]

I think time complexity is O(n^2)

[@tanishkumar6682]

understood

[@Adityakumar-dl4vc]

Thank

[@AbhishekKumar-cv1dh]

Understood !! 🔥🔥

[@AbhishekBhattacharjee-j2m]

UNDERSTOOD