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# Single Number II LeetCode Solution

## Problem – Single Number II LeetCode Solution

Given an integer array `nums` where every element appears three times except for one, which appears exactly onceFind 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 * 104`
• `-231 <= nums[i] <= 231 - 1`
• Each element in `nums` appears exactly three times except for one element which appears once.

## Single Number II LeetCode Solution in Python

``````class Solution:
def singleNumber(self, nums):
single = 0
for i in range(32):
count = 0
for num in nums:
if num & (1<<i) == (1<<i): count += 1
single |= (count%3) << i

return single if single < (1<<31) else single - (1<<32)
``````

## Single Number II LeetCode Solution in C++

``````   int singleNumber(vector<int>& s)
{
vector<int> t(32);////Made a array contain 32 elements.
int sz = s.size();
int i, j, n;
for (i = 0; i < sz; ++i)
{
n = s[i];
for (j = 31; j >= 0; --j)
{
t[j] += n & 1;//Find the last digit.
n >>= 1;
if (!n)
break;
}
}
int res = 0;
for (j = 31; j >= 0; --j)
{
n = t[j] % 3;//"3" represents k times.
if (n)
res += 1 << (31 - j);
}
return res;
}``````

## Single Number II LeetCode Solution in Java

``````public int singleNumber(int[] nums) {
int ones = 0, twos = 0, threes = 0;

for (int i = 0; i < nums.length; i++) {
// twos holds the num that appears twice
twos |= ones & nums[i];

// ones holds the num that appears once
ones ^= nums[i];

// threes holds the num that appears three times
threes = ones & twos;

// if num[i] appears three times
// doing this will clear ones and twos
ones &= ~threes;
twos &= ~threes;
}

return ones;
}
``````
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