if we already have a mask as res( first xor calculation )why we calculate that again. It is same as mask
Calculating mask
Hello @amanpunetha,
When you perform the first xor i.e. you are left with a^b where a and b are the required unique numbers.
The idea used here is to divide the entire sequence into two groups, one containing a and other containing b.
To separate them into two groups,
we are finding the position of first set bit from right in the xor i.e. a^b.
Then, mask is created by right shifting the 1 by ‘pos’ positions.
Answer to your question:
No, the mask and xor will not necessarily be the same always.
Example:
xor= 100100
mask=000100
Suggestion:
Re-watch the video now.
Hope, this would help.
Give a like if you are satisfied.
but if i iterate again to loop again perform
if((result_xor | sum[i]) ==sum[i]){
result1=result1^ sum[i];
}
else{
result2=result2 ^ result[2];
}
simply i can find my ans .
not to calculate mask again
Hello @amanpunetha,
Consition: result_xor | sum[i]) ==sum[i] will never satisfy.
Please, give an example in support of this.
array : 5 3 1 2 5 1 7 2
result_xor = 4
result1=0,result2=0
if(result_xor | sum[i]) ==sum[i]):
so: 4 | 5 ==5 it will xor with result 1 same for 5 and 7 further
and 4 | 1 !=1 it will xor with result 2 same for 1 2 3 further
result1 will give 7
result2 will give 3
Hello @amanpunetha,
Your approach will fail for test cases like:
For the sequence:
1 2 2 3 3 4
Unique numbers:
1 = 0 0 1
4 = 1 0 0
xor = 5 (1 0 1)
5 | 4= 1 0 1 (this is not equals to 4 i.e. sum[i])
5 | 1= 1 0 1 (even, this is not equal to 1 i.e. sum[i])
How would you separate them now?
thank you 
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