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# Missing Number LeetCode Solution

## Problem – 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?

### Missing Number LeetCode Solution in Java

``````public int missingNumber(int[] nums) {

int xor = 0, i = 0;
for (i = 0; i < nums.length; i++) {
xor = xor ^ i ^ nums[i];
}

return xor ^ i;
}``````

### Missing Number LeetCode Solution in C++

``````class Solution {
public:
int missingNumber(vector<int>& nums) {
int result = nums.size();
int i=0;

for(int num:nums){
result ^= num;
result ^= i;
i++;
}

return result;
}
};``````

### Missing Number LeetCode Solution in Python

``````class Solution(object):
def missingNumber(self, nums):
return sum(range(len(nums)+1)) - sum(nums)``````
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