leetcode/work/Easy/Array/268.missing-number.js

/*
 * 268. Missing Number
 * Difficulty: Easy
 * https://leetcode.com/problems/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 <= 10^4
 * 
 * 	• 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) {
  const numSet = new Set(nums);
  const expectedCount = nums.length + 1;

  for (let i = 0; i < expectedCount; i++){
    if(!numSet.has(i)){
      return i;
    }
  }
  return -1;
};