feat(work): add solutions for LeetCode 1365 and 268 problems

work/3/Easy/Array/1365.how-many-numbers-are-smaller-than-the-current-number.py

@@ -0,0 +1,60 @@
+"""
+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]