Leetcode --- 268. Missing Number
===

## Description
Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.

### Examples

* Ex1:
Input: nums = [3,0,1]
Output: 2
Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.

* Ex2:
Input: nums = [0,1]
Output: 2
Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.

* Ex3:
Input: nums = [9,6,4,2,3,5,7,0,1]
Output: 8
Explanation: n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.

* Ex4:
Input: nums = [0]
Output: 1
Explanation: n = 1 since there is 1 number, so all numbers are in the range [0,1]. 1 is the missing number in the range since it does not appear in nums.

### Constraints
* n == nums.length
* 1 <= n <= 10^4^
* 0 <= nums[i] <= n
* All the numbers of nums are unique.

## Solutions

### Approach 1: hashmap

class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        hashmap = {}
        for i in range(len(nums)):
            hashmap[nums[i]] =1
        
        for i in range(len(nums)+1):
            if i not in hashmap:
                return i

#### Complxity Analysis
* Time complexity:
O(n)+ O(n+1) = > ==O(n)==
* Space complexity:
dict : ==O(n)==

#### submissions
Runtime: 201ms
Memory: 15.7MB

### Approach 2:bit manipulation

class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        l = len(nums)
        tmp = 0
        i = 0
        while i < l:
            tmp = tmp ^ i ^ nums[i]
            i += 1
        return tmp ^ i

Key concept:
兩個相同的值被XOR會抵消掉，最後剩下唯一missing value為答案
*Note*
* 0 ^ any = any
* 1 ^ any = 反向any
* any ^ any = 0

#### Complexity analysis
* Time complexity
O(n)
* Space complexity
O(1)

#### submissions
Runtime: 181ms
Memory: 15.3MB

### Approach 3 : sum

class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        l = len(nums)
        return int((1+l)*l/2 - sum(nums))

#### Complexity analysis
* Time Complexity
sum API :O(n) => O(n)
* Space complexity
O(1)

#### Submissions
Runtime: 128ms
Memory: 15.5MB

###### tags: `Leetcode`