268 Missing Number

Missing Number

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.

 

Example 1:

**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.

Example 2:

**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.

Example 3:

**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.

 

Constraints:

n == nums.length
1 <= n <= 104
0 <= nums[i] <= n
All the numbers of nums are **unique**.

 

Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?

int missingNumber(int* nums, int numsSize) {
    int all_xor = 0 ;
    printf("%d", 1 ^ 1);
    for (int i = 0 ; i < numsSize; i ++) {
        all_xor = all_xor ^ *(nums + i) ^ i;
    }   

    // 3^0  ^ 0^1 ^ 1^2 ^ 3
    return all_xor ^ numsSize;
}