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137. Single Number II
Description
Given an integer array nums
where every element appears three times except for one, which appears exactly once. Find the single element and return it.
You must implement a solution with a linear runtime complexity and use only constant extra space.
Example 1:
Input: nums = [2,2,3,2] Output: 3
Example 2:
Input: nums = [0,1,0,1,0,1,99] Output: 99
Constraints:
1 <= nums.length <= 3 * 10^{4}
2^{31} <= nums[i] <= 2^{31}  1
 Each element in
nums
appears exactly three times except for one element which appears once.
Solutions
Solution 1: Bitwise Operation
We can enumerate each binary bit $i$, and for each binary bit, we calculate the sum of all numbers on that bit. If the sum of the numbers on that bit can be divided by 3, then the number that only appears once on that bit is 0, otherwise it is 1.
The time complexity is $O(n \times \log M)$, where $n$ and $M$ are the length of the array and the range of elements in the array, respectively. The space complexity is $O(1)$.
Solution 2: Digital Circuit
We can use a more efficient method that uses digital circuits to simulate the above bitwise operation.
Each binary bit of an integer can only represent 2 states, 0 or 1. However, we need to represent the sum of the $i$th bit of all integers traversed so far modulo 3. Therefore, we can use two integers $a$ and $b$ to represent it. There are three possible cases:
 The $i$th bit of integer $a$ is 0 and the $i$th bit of integer $b$ is 0, which means the modulo 3 result is 0;
 The $i$th bit of integer $a$ is 0 and the $i$th bit of integer $b$ is 1, which means the modulo 3 result is 1;
 The $i$th bit of integer $a$ is 1 and the $i$th bit of integer $b$ is 0, which means the modulo 3 result is 2.
We use integer $c$ to represent the number to be read in, and the truth table is as follows:
$a_i$  $b_i$  $c_i$  New $a_i$  New $b_i$ 

0  0  0  0  0 
0  0  1  0  1 
0  1  0  0  1 
0  1  1  1  0 
1  0  0  1  0 
1  0  1  0  0 
Based on the truth table, we can write the logical expression:
\[a_i = a_i' b_i c_i + a_i b_i' c_i'\]and:
\[b_i = a_i' b_i' c_i + a_i' b_i c_i' = a_i' (b_i \oplus c_i)\]The final result is $b$, because when the binary bit of $b$ is 1, it means that the number appears only once.
The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

class Solution { public int singleNumber(int[] nums) { int a = 0, b = 0; for (int c : nums) { int aa = (~a & b & c)  (a & ~b & ~c); int bb = ~a & (b ^ c); a = aa; b = bb; } return b; } }

class Solution { public: int singleNumber(vector<int>& nums) { int a = 0, b = 0; for (int c : nums) { int aa = (~a & b & c)  (a & ~b & ~c); int bb = ~a & (b ^ c); a = aa; b = bb; } return b; } };

''' A 32bit number can be created to count the number of occurrences of 1 in each digit. If a certain digit is 1, then if the integer appears three times, the remainder of 3 is 0, so that the numbers of each digit position are added up then take the remainder of 3, and the final number remaining is a single number. ''' class Solution: def singleNumber(self, nums: List[int]) > int: ans = 0 for i in range(32): cnt = sum(num >> i & 1 for num in nums) if cnt % 3: if i == 31: # int overflow， or just throw exception ans = 1 << i else: ans = 1 << i return ans ############ class Solution: def singleNumber(self, nums: List[int]) > int: a = b = 0 for c in nums: aa = (~a & b & c)  (a & ~b & ~c) bb = ~a & (b ^ c) a, b = aa, bb return b

func singleNumber(nums []int) int { a, b := 0, 0 for _, c := range nums { aa := (^a & b & c)  (a & ^b & ^c) bb := ^a & (b ^ c) a, b = aa, bb } return b }

function singleNumber(nums: number[]): number { let a = 0; let b = 0; for (const c of nums) { const aa = (~a & b & c)  (a & ~b & ~c); const bb = ~a & (b ^ c); a = aa; b = bb; } return b; }

impl Solution { pub fn single_number(nums: Vec<i32>) > i32 { let mut ans = 0; for i in 0..32 { let count = nums .iter() .map(v (v >> i) & 1) .sum::<i32>(); ans = count % 3 << i; } ans } }

class Solution { func singleNumber(_ nums: [Int]) > Int { var a = nums.sorted() var n = a.count for i in stride(from: 0, through: n  2, by: 3) { if a[i] != a[i + 1] { return a[i] } } return a[n  1] } }

function singleNumber(nums) { let ans = 0; for (let i = 0; i < 32; i++) { const count = nums.reduce((r, v) => r + ((v >> i) & 1), 0); ans = count % 3 << i; } return ans; }