Solution: Missing Number

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Leetcode Problem #268 (Easy): Missing Number

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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:

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:

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

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Idea:

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

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

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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:

(Jump to: Problem Description || Solution Idea)

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

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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();
    }
}

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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);
    }
};

Comments

[Nithish Kumar]

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 :)

[seanpgallivan]

The code is very readable, which is good, but indexOf() is an O(n) time function by itself, so the overall time complexity becomes O(n^2), and the space complexity becomes O(n) since you're creating a new array (original).

[Nithish Kumar]

Thank you for explaining... Even though I'm not good at Big O notation, I can understand ur explanation.