Can i get any efficient way to solve this

#1

can i get any efficient way to solve this

#2

Hi @navin

You can use bitwise xor of all the elements of the array. As we know that a xor a = 0. And all elements which are present twice will give final xor as 0 and when that single number is xored with 0 we get final answer of xor of all numbers as that number which is present only 1 time.

Hope it Helps.

#3

sorry my doubt was in Playing with bits challenge