# Contest - Day 1 - Single Number

Leetcode is running a 30 day “Covid 19” style contest for everyone on self quarantine and social distancing. Let’s jump in!

## Single Number

### Problem

Given a non-empty array of integers, every element appears twice except for one. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Examples:

``````Input: [2,2,1]
Output: 1

Input: [4,1,2,1,2]
Output: 4``````

### Thinking

This is very easy using the `collections.Counter` in python stlib. That might be cheating but I’m going to allow it. Stand on the shoulders of well tested (code) giants, I say.

However, the question posits there is a way to do it without using extra memory. So, storing counters is likely not the optimal solution.

This led me to thinking about truth tables. What boolean operation will indicate once but not twice? XOR. For example:

``````0 ^ 1 => 1
0 ^ 1 ^ 1 => 0.``````

If the result of our XOR logic is the input value, there is only one occurrence of it. If it’s zero, then there was either none, or more than 1 of them.

### Corner Cases

I don’t think I missed any this time.

### Improvements

I don’t think I missed any this time.

### Solution

Counting

``````import collections

class Solution:
def singleNumber(self, nums: List[int]) -> int:
counts = collections.Counter(nums)
return counts.most_common()[-1][0]``````

XOR

``````class Solution:
def singleNumber(self, nums: List[int]) -> int:
r = 0
for n in nums:
r ^= n
return r``````

### Score

Counting

``````Runtime: 84 ms, faster than 85.38% of Python3 online submissions for Single Number.
Memory Usage: 16.3 MB, less than 6.56% of Python3 online submissions for Single Number.``````

XOR

``````Runtime: 88 ms, faster than 65.17% of Python3 online submissions for Single Number.
Memory Usage: 15.6 MB, less than 6.56% of Python3 online submissions for Single Number.``````

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