*This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful,* *please like**this post and/or* *upvote**my solution post on Leetcode's forums.*

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Leetcode Problem #268 (*Easy*): Missing Number

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*Description:*

*Description:*

(*Jump to*: *Solution Idea* || *Code*: *JavaScript* | *Python* | *Java* | *C++*)

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.

Follow up: Could you implement a solution using only`O(1)`

extra space complexity and`O(n)`

runtime complexity?

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*Examples:*

*Examples:*

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.

Example 4: 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.

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*Constraints:*

*Constraints:*

`n == nums.length`

`1 <= n <= 10^4`

`0 <= nums[i] <= n`

- All the numbers of
`nums`

are unique.

####
*Idea:*

*Idea:*

(*Jump to*: *Problem Description* || *Code*: *JavaScript* | *Python* | *Java* | *C++*)

The sum of the numbers from **1** to **N** is the **N**th **triangular number**, defined as **N * (N + 1) / 2**. It stands to reason, then, that we can simply find the difference between the **N**th triangular number and the sum of **nums**, which should be our missing number.

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*Javascript Code:*

*Javascript Code:*

(*Jump to*: *Problem Description* || *Solution Idea*)

```
const missingNumber = nums =>
nums.length * (nums.length + 1) / 2 - nums.reduce((a,c) => a + c)
```

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*Python Code:*

*Python Code:*

(*Jump to*: *Problem Description* || *Solution Idea*)

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

####
*Java Code:*

*Java Code:*

(*Jump to*: *Problem Description* || *Solution Idea*)

```
class Solution {
public int missingNumber(int[] nums) {
return nums.length * (nums.length + 1) / 2 - Arrays.stream(nums).sum();
}
}
```

####
*C++ Code:*

*C++ Code:*

(*Jump to*: *Problem Description* || *Solution Idea*)

```
class Solution {
public:
int missingNumber(vector<int>& nums) {
return nums.size() * (nums.size() + 1) / 2 - accumulate(nums.begin(), nums.end(), 0);
}
};
```

## Top comments (3)

Hey sean , Thank you for sharing this problem. I really like it. After seeing the problem des. , i came up with this solution. Is this a better approach?

const findMissing = arr => {

const orginal = Array.from({length:arr.length+1} , (_,i) => i);

return orginal.filter(v => arr.indexOf(v) === -1)[0]

}

Thanks in advance :)

The code is very readable, which is good, but

indexOf()is anO(n) timefunction by itself, so the overalltime complexitybecomesO(n^2), and thespace complexitybecomesO(n)since you're creating a new array (original).Thank you for explaining... Even though I'm not good at Big O notation, I can understand ur explanation.