Single Number II
Implement singleNumber
This time the pattern is triples, not pairs — every value in the array repeats exactly three times except for one lone holdout that shows up just once. Identify it.
Plain XOR can't help here, since XOR-ing a value with itself three times just leaves the value unchanged (it doesn't cancel like it does for pairs). Instead, look at each of the 32 bit positions on its own: a value repeated three times always contributes a multiple of 3 to that position's total, so any position whose total isn't a multiple of 3 must be getting its extra contribution from the one unpaired value.
Example 1:
Input: nums = [2,2,3,2]
Output: 3
Example 2:
Input: nums = [0,1,0,1,0,1,99]
Output: 99
Example 3:
Input: nums = [5,5,5,7]
Output: 7
+ 10 hidden test cases run on Submit.
Constraints:
- ●
1 ≤ nums.length ≤ 3 × 10⁴ - ●
-3 × 10⁴ ≤ nums[i] ≤ 3 × 10⁴ - ●
Every value in nums appears exactly three times, except for one value which appears exactly once
nums =
[2, 2, 3, 2]