diff --git a/work/3/Easy/Array/1365.how-many-numbers-are-smaller-than-the-current-number.py b/work/3/Easy/Array/1365.how-many-numbers-are-smaller-than-the-current-number.py
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+"""
+1365. How Many Numbers Are Smaller Than the Current Number
+Difficulty: Easy
+https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/
+
+──────────────────────────────────────────────────
+
+Given the array nums, for each nums[i] find out how many numbers in
+the array are smaller than it. That is, for each nums[i] you have to
+count the number of valid j's such that j != i and nums[j] < nums[i].
+
+Return the answer in an array.
+
+
+
+Example 1:
+
+Input: nums = [8,1,2,2,3]
+Output: [4,0,1,1,3]
+Explanation:
+For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and
+3).
+For nums[1]=1 does not exist any smaller number than it.
+For nums[2]=2 there exist one smaller number than it (1).
+For nums[3]=2 there exist one smaller number than it (1).
+For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).
+
+Example 2:
+
+Input: nums = [6,5,4,8]
+Output: [2,1,0,3]
+
+Example 3:
+
+Input: nums = [7,7,7,7]
+Output: [0,0,0,0]
+
+
+
+Constraints:
+
+        • 2 <= nums.length <= 500
+
+        • 0 <= nums[i] <= 100
+"""
+
+from collections import Counter
+
+
+class Solution:
+    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
+        freq = Counter(nums)
+        sorted_nums = sorted(freq.keys())
+        smaller = {}
+        count = 0
+        for num in sorted_nums:
+            smaller[num] = count
+            count += freq[num]
+
+        return [smaller[num] for num in nums]
diff --git a/work/3/Easy/Array/268.missing-number.py b/work/3/Easy/Array/268.missing-number.py
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+"""
+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?
+"""
+
+
+class Solution:
+    def missingNumber(self, nums: List[int]) -> int:
+        n = len(nums)
+        expected_sum = (n * (n + 1)) // 2
+        actual_sum = sum(nums)
+        return expected_sum - actual_sum