Missing_Number.js

// 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?

/**
 * @param {number[]} nums
 * @return {number}
 */
var missingNumber = function (nums) {
  nums.sort((a, b) => a - b);
  let mis = 0;
  for (let key of nums) if (mis === key) mis = key + 1;
  return mis;
};