```
int singleNumber(int* nums, int numsSize) {
// idea : iterate over all elements in array for every bit, get sum of all 1s,
// bits with sum not multiple of 3, are 1's bit in the single occurance element
// e.g. , 5,5,5,2 - 101 101 101 010 - 313 - answer is 3%3,1%3,3%3 - 010
int x, res = 0;
for(int i = 0; i < sizeof(int)*8; i++){
x = 1 << i;
int sum = 0;
for(int j = 0; j < numsSize; j++){
if(x & nums[j]) sum++;
}
if(sum % 3) res = res | x;
}
return res;
}
```