The int that appears an odd number of times.

[MeeranRizvi]

Hello Guys,

Here i am having a array

array=[1,2,3,2,4,2,4,3,5,5,1]

I just need to find out which number that appears an odd no of times

For this array the output should be:

Output:
2

How should do that?

[Mekire]

Please provide us with your attempt.

[MeeranRizvi]

i did with dictionary

a = [1,2,3,2,4,2,4,3,5,5,1]
result = dict((i, a.count(i)) for i in a)
print result

I just got the counts but how to print only the int that appears an odd no of times

Output:
{1: 2, 2: 3, 3: 2, 4: 2, 5: 2}

[Larz60+]

hint

if number % 2:

[Skaperen]

1 appears 1 time and 1 is odd

---

what if ...

a = [1,2,3,2,4,2,3,5,3]

lots of oddities!

[MeeranRizvi]

Yeah You are right skaperen!
Now i changed my input to a new one,i written code also it works as expect

a = [1,2,3,2,4,2,4,3,5,5,1]
result = dict((i, a.count(i))for i in a)
for i in result:
if a.count(i)%2!=0:
 print "The int that occur odd no of times:",i

Output:
The int that occur odd no of times: 2

[wavic]

I think '&' operator will be faster

In [1]: %time 5 % 2 == 1
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 4.05 µs
Out[1]: True

In [2]: %time 5 & 1 == 1
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 3.1 µs
Out[2]: True