refactor(work): restructure solution directories with bucket level and add .gitkeep placeholders

work/1/Easy/Array/1.two-sum.py

@@ -0,0 +1,60 @@
+"""
+1. Two Sum
+Difficulty: Easy
+https://leetcode.com/problems/two-sum/
+
+──────────────────────────────────────────────────
+
+You are given an array of integers nums and an integer target, return
+indices of the two numbers such that they add up to target.
+
+You may assume that each input would have exactly one solution, and
+you may not use the same element twice.
+
+You can return the answer in any order.
+
+ 
+
+Example 1:
+
+Input: nums = [2,7,11,15], target = 9
+Output: [0,1]
+Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].
+
+Example 2:
+
+Input: nums = [3,2,4], target = 6
+Output: [1,2]
+
+Example 3:
+
+Input: nums = [3,3], target = 6
+Output: [0,1]
+
+ 
+
+Constraints:
+
+	• 2 <= nums.length <= 10^4
+
+	• -10^9 <= nums[i] <= 10^9
+
+	• -10^9 <= target <= 10^9
+
+	• Only one valid answer exists.
+
+ 
+
+Follow-up: Can you come up with an algorithm that is less than O(n^2)
+time complexity?
+"""
+
+class Solution:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        seen = {}
+        for i, num in enumerate(nums):
+            complement = target - num
+            if complement in seen:
+                return [seen[complement], i]
+            seen[num] = i
+        return []