Problem was with ( ) == a[i]
I also made the same mistake
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its the brackets that’s giving you ac. (a[i] & a[j]) == a[i]
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== has higher preference than & operator
That’s why you got WA in first code.
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got it. operator precedence, my bad sorry ! 
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So what’s your question exactly ?
If you are asking that why second one got AC then its because (i<j) was given in question and you need to put the operands of bitwise &, in round brackets.
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This is the reason you should turn on compiler warnings. I wrote the first code, and got a warning.
My Geany Compiler Flags are:
Compile (F8):
g++ -std=c++17 -Wshadow -Wall -o "%e" "%f" -O2 -Wno-unused-result
Build (F9):
g++ -std=c++17 -Wshadow -Wall -o "%e" "%f" -g -fsanitize=address -fsanitize=undefined -D_GLIBCXX_DEBUG
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‘==’ is done first and then ‘&’. The problem is just with the brackets.
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Yes but it worked for the samples. That is why I submitted the code in the first place.
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In sample all the number were same in the array
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Then can you explain why it worked for the sample test case?
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Just generate some test cases and check which ones have different output on the codes.
So it was intentionally done 
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in the first sample all the numbers were the same and in the second one, you don’t get to the point of comparison
I guess so 
I think first loop should be till n-1 and the brackets .
Do the dry run you will understand .
Ps: we are checking for pairs hence why should we check for last element ?
No, j has to be less than n
I said first loop brother
I know that. If i is n-1, j will be n but it should be less than n it won’t work.
