feat(work/Easy/Array): add solutions for counting elements and missing number problems

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

@@ -0,0 +1,90 @@
+/*
+ * 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;
+};