Find the element that appears once where every other element appears twice | GeeksforGeeks

This video is contributed by Shubham Kumar

Please Like, Comment and Share the Video among your friends.

Also, Subscribe if you haven't already! :)

[Jechjames]

Could someone actually explain how this works. For example I have an array [1, 2, 2, 1, 3] We are setting the res value to ar[0]. So when the loop starts it compare res = ar[0] to ar[i]. So for example it would compare 1 and 2. Then it will go to then it will compare 1 and 2. Then it will compare 1 and 2. Them will compare 1 and 1 which will return 0? So for the above example how would it isolate 3??

[Tensor]

What about this
sum(set(arr))*2 – sum(arr)

[wecan]

int findSingle(int ar[], int ar_size)
{
// Do XOR of all elements and return
int res = ar[0];
for (int i = 1; i < ar_size; i++)
res = res ^ ar[i];

return res;
}

[LinhHoang-zimt]

Damn! This seemingly simple question is actually not simple. I wouldn't see the XOR in a million year. And yes I was a digital hardware design before. I'd used XOR for simple checksum and stuff.

[BPJennieYeager]

May I ask why it requires O(1) extra space? How to calculate it?

[II_xD_II]

can we sort the array and then see if the adjacent element is == to the selected element

[SumitKumar-fngj]

But what about if there is more than 1 number which appear exactly once.

[pranavkurkure]

What if the non repetiting element is not at the first position in array?

[HansNiemann]

Having looked at the code what would be XOR of a number with a different number

[anujjhamb]

It fails when either there are multiple numbers with single occurrence or when the number occurs with odd frequency.

[shivamagarwal]

its wrong res should not be equal to arr[0]