arcbjorn

# thoughtbook

Sunday, March 5, 2023

# Single number III

Leetcode 260 problem.

Given an integer array nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once. You can return the answer in any order.

## Go

• Time complexity: $O(n)$ - n is a length of a number array
• Auxiliary space: $O(1)$ - constant amount of space
func singleNumber(nums []int) []int {
xor := 0

for _, value := range nums {
xor = xor ^ value
}

lastBit := xor & (-xor)
uniqueNumbers := make([]int, 2)

for _, value := range nums {

if value & lastBit == 0 {
uniqueNumbers[0] = uniqueNumbers[0] ^ value
} else {
uniqueNumbers[1] = uniqueNumbers[1] ^ value
}
}
return uniqueNumbers
}

## C++

• Time complexity: $O(n)$ - n is a length of a number array
• Auxiliary space: $O(1)$ - constant amount of space
class Solution {
public:
vector<int> singleNumber(vector<int>& nums) {
unsigned int all_xor = 0;
int length = nums.size();

// XOR all nums
for(int i = 0; i < length; i++){
all_xor = all_xor ^ nums[i];
}

// right set bit of XOR
all_xor = all_xor & (-all_xor);

// group with XOR bit position and group with unset bit
vector<int> uniqueNumbers (2, 0);

for(int i = 0; i < length; i++){
if(all_xor & nums[i]){
uniqueNumbers[0] = uniqueNumbers[0] ^ nums[i];
} else {
uniqueNumbers[1] = uniqueNumbers[1] ^ nums[i];
}
}

return uniqueNumbers;
}
};