Rearrange Array so that Arr[i] becomes Arr[Arr[i]] with O(1) extra space

Rearrange a given array so that Arr[i] becomes Arr[Arr[i]] with O(1) extra space.
Example:
Input : [1, 0]
Return : [0, 1]
Let's say N = size of the array. Then, the following holds true :
All elements in the array are in the range [0, N-1]
N * N does not overflow for a signed integer

[@neerajgarg9096]

Hey Alisha Great Explanation but have one doubt that its like we have to mug up this approach so if i try to solve this after 1 week i definitely forget it so how to remember these type of questions?

[@NeelakshiSachdeva]

Yes same

[@AvinashKumar-vv4vd]

Really good explanation 👍

[@reshmah1497]

Useful video...explanation is understandable

[@dheerajmct]

You r simply the best

[@rahultomar8349]

thanks and keep making the such awesome videos.

[@asad_mirza]

Muskuraiye ...you have a new subscriber 😁

[@ceb-rj4qw]

SUper

[@amberraghuwanshi859]

WOW

[@AmanRaj-ux8cw]

I'm searching a good video for array explanation and I really find good content. Thank you so much for sharing this. 😀

[@billgatesbisht6665]

new approach .... SC O(1)... TC O(n) ....
void arrange(long long arr[], int n) {
for(int i=0;i

[@piyushsrivastava5883]

for the people why she is taking n as parameter of dividing or taking modulo because as the question suggested the element is less than n so the number should be in between .might clear

[@SachinNagareiGTB]

can you give me explanation on this= sum(arr,n-1)+arr[n-1],what exactly going on here

[@siddharthb2226]

Awesome explanation! I have started solving InterviewBit problems, hopefully u have ciovered them all

[@HarendraKumar-yr4gt]

Best explanation ever, thanks, please keep making such awesome videos! learning a lot

[@avishjain5799]

What is the intuition behind this approach.

[@rohankumarshah5679]

i thought why there's need to put modulo n in arr[arr[i]]%n *n since all the numbers in arr are less than n
but
now i realized as we multiply no's are changing at arr[i] so we have to first modulo n then just do multiply with n

[@suyashpurwar631]

Should have focused on the mathematical intuition

[@learncreative4732]

How to get logic these type of question 😢

[@rashedin6356]

approach should be cyclic sort

[@RajeshwariGorla-jk1pr]

how to convert into long??

[@deepakgaroda975]

Your way of explaining is awesome...

[@harshvardhan7161]

constant space**

[@muskancreativity7]

Best explanation 💜

[@janaSdj]

wowwww....

[@dspeaks9756]

Thanks Alisha....what an easy explanation.....this gave me learning without wasting much time

[@anshikaaa3716]

Great explanation! Immediately subscribed :)

[@AmarjeetKumar-to9ub]

🔥++

[@venkatanagaprathapyelugula7738]

Thanks for explanation...Thanks my dear sister.

[@niranjanaware5930]

plz continue creating such great sols..thanks

[@satyareddy8671]

I love the way you explained mam.

[@harshitsharma1880]

clean and clear explanation thanks for good content..

[@milindbadekar3503]

Thank u.Keep working .Dont stop.

[@jashwanthpeddisetty3264]

awesome explanation

[@thegreekgoat98]

Nice explanation

[@ShivamGupta-po7rg]

thank you ; )

[@raj_kundalia]

thank you

[@rajaspaunikar9695]

ALISHA MAHARANI KI JAI!!!

[@SANCHITJAIN_]

great explanation...loved it !!

[@SatishKanaujiyaeb]

This question is like hit or miss. If you hit, you will solve this otherwise (:-

[@066lakshmipriya3]

Why should we take %n before multiplying with n?

[@rohankumarshah5679]

initially all the numbers in array are 0 to n-1 ie less than n so no need to modulo, right ??
just do arr[i] = arr[arr[i]] * n;
but,
when we have done one operation what if we are accessing the number at arr[i] that was earlier multiplied with n by us,
therefore we need to initially modulo n then multiply with n.