The correct output for the above tc is -
1
1 2
1 2 1
1 2 1 3
1 2 1 3 1
1 2 1 3 1 2
1 2 1 3 1 2 1
1 2 1 3 1 2 1 4
1 2 1 3 1 2 1 4 1
1 2 1 3 1 2 1 4 1 2
Your output for N =12 is
1 2 3 4 3 2 1 2 3 4 3 2
where everything appears twice in the array (1 and 4 appear twice each; 2 and 3 appear 4 times each)
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Thank you!!
@knight_rider45 i think it is because of the constraint n<=1000 and the pattern:
If you see the pattern carefully then it goes like:
Position at which 1st occurrence of following numbers spotted:
1: 1
2: 2
3: 4
4: 8
5: 16
6: 32
and so on it follows powers of two.
So, 11 will occur at 1024 position
I guess that’s why there is no use to go beyond .
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I took the output for vector size for 10,11 and 12. You can also calculate using a general formula
No. of terms in nth term = 2*(no of terms in n-1th term) + 1
Brother plz help me
1 2 3 2 1 2 3
If I take 1 2 3 2 as subarray then the frequency of 2 will be 2 but it is correct. I can not understand how it is correct brother . can you please explain this matters??
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frequency of any number is odd for a given subarray then it is correct and in the subarray
1 2 3 2, 1 and 3 have odd occurrences. you just need to avoid the subarray in which all the numbers have even frequencies
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We are not supposed to have a subarray where all the elements has even frequencies…
So in 1 2 3 2… 2 have even occurence, but 3 and 1 have odd number of occurence
you can also see in subarray 1 2 3 2 1 ,
only 3 has odd number of occurences… it is still valid because all elements are not supposed to have even frequencies at the same time
Good observation,thanks.
please tell me why answer is the first bit one of gray code on Tester’s Solution ?
True , Even the editorial sucks 
If you have any constructive criticism as to how the editorial can be improved, please let me know 
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