thank you for your reply. okk i will mention my issue on twitter
yes agreed. I was just being safe.Hardly matters.
The maximum s(x) of will be 81 (when all the 9 digits are 9).
So, we iterate s(x) from 0 to 81 and for each s(x), we find s(s(x)) (which is nothing but summing the digits of s(x) numbers together).
x + s(x) + s(s(x)) = n. Rewriting this,
x = n - s(x) - s(s(x)).
Now, we know all the terms in the right and hence we find x. For the x, we check if the sum of digits of the x is s(x), (which we were already using in the eqn.).
If it was equal, x is one such number that satisfies the equation.
The intuition is realizing that for every s(x), there can be only one x that can satisfy the given equation.
Max s(x) will be 81 not 89, but thatās exactly what I would do
Yep. That was a typo !
i have not got any mail regarding waiting list
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Cool, BTW i hope you know that the edits history are visible publicly
@tarunpatel1860 If you only consider the row operations, then each row will have the same value in each of its columns and this value will depend upon how many times the rth row was included in the query i.e frequency[row]. Now each column operation would also increase the value of each row by 1. So basically the questions boils down to finding the highest repeating row value and column value in the queries and the answer would be their addition.
max(freq[row_in_query]) + max(freq[col_in_query]).
Time complexity - O(m+n)
Space complexity - O(m+n)
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Yeah there is a waitlist too. I had solved 4/5 and received a mail 2 days back saying that I was put into waitlist and today I received a mail saying that I have cleared the entrance test and I was given the link to schedule an interview.
@sanchit2812 Considering the constraints, you could just simulate the whole process. I used circular doubly linked list to do alternate eliminations until I was left with only 2 nodes. The people corresponding to these nodes would be our answer.
same with be buddy. Solved 4 on first day still didnāt received an email.I called IB Academy they told us this time around 10000 students appeared for the online test ,Now me-āaapna chutiya kat gaya ab lag raha haiā . Worst case : āAap academy mein nhi daaloge toh hum khud hi referrals ka intezaam kar lengešā
10000!!!
Itne log kaha se aa gaye bhai, kuchh jada hi marketing nahi kar di ye log ne is bar.
aur nhi toh kya bhaiā¦ inko pata tha popularity badh gayi hai (through number of registrations for online entrance examā¦must have reached 10k+ )ā¦Isliye Ek interview bhi ghused diya inneā¦Ek baat aur jisko DP aati haiā¦ 4/5 questions kar pa raha haiā¦ Toh unko academy ki zarurat kya haiā¦ Sale ye dalal(middle man) ka kaam shuru kar diya inneā¦ Ek toh best out of best lengeā¦ O bache toh acha results denge hi naā¦ Jo banda 4 star coder hai Codechef meinā¦ Usko zarurat hi kya hai IB kiā¦ Off-campus apply karna chahiye unko.
@abhishek97 If its easy then everyone taking referral and applied through off campus .Do you really think that if someone is applied for IB doesnt applied off campus.Basically everyone wanted a good job but there is a gap between them which IB academy fills out .Do you know I applied around many companies still dont get any result.And its the case for everyone .Its easy to say that 4* or 5* should need to apply off campus rather join IB but its hard its really hard .If someone helps us to cross that bridge than why not high rates coders try out .If you are capable enough then you are selected .
When are they declaring result?
what is the last date for you to schedule interview ?
One thing I want to add , my response was influenced by our Institute surrounding (by god level coders), because those people already have multiple offers all above 8L>=inhand_salary . IB academy is truly helping those who come from tier-2,3 institutes , in many cases these people are rejected on the basis of Resume, even if they clear the Coding/Technical Round(s).
If ( 3>= star && ā Tier-2,3 Institutes )
Go ahead JOIN! For you people amazon will be cakewalk 
after joining IB.