Single Number II

#bit counting at every position and consider those bits whose count%3==1 because if the number occured once then either a bit at any position can be 1 or 4(if all elements have also that bit)
#T.C=32*n n=total number of elements for each element.
class Solution:
    def singleNumber(self, nums: List[int]) -> int:
        result = 0

        for bit in range(32):
            count = 0
            for num in nums:
                if (num >> bit) & 1:
                    count += 1

            if count % 3 != 0: #Right shift 'num' by 'bit' positions will make that bit at 1st place from right Then AND with 1 to check if that bit is set (1) or not (0)
                result |= (1 << bit) # if set then do or in the result at that position.

        # handle negative number (32-bit signed)
        if result >= 2**31:
            result -= 2**32

        return result



#Time complexity O(n), no space used
class Solution:
    def singleNumber(self, nums: List[int]) -> int:
        ones=0 #it will store only those elements which occurs once but not third time element
        twos=0 #it will store only those elements which occurs twice
        for i in nums:
            ones= (ones^i) & ~twos
            twos=(twos^i) & ~ones
        return ones #finally ones will have a element which will have the frequency of 1