refactor: improve comments for clarity and consistency in twoSum implementations

Author: hongseoupyun

two-sum/hongseoupyun.py

@@ -3,7 +3,8 @@
 # For each numbers in given array, calculate complement (target - num)
 # If complement is in hash table(seen), return array of index of the currnet number and index of complement in hash table(seen)
 # If complement is not in has table(seen), store the current number in hash table(seen) and continue checking the numbers
-#
+# The algorithm uses a single loop that goes through each element in nums exactly one time.
+# If the array contains n elements, the loop runs n times → O(n).
 class Solution:
     def twoSum(self, nums: List[int], target: int) -> List[int]:
         seen = {}
@@ -13,4 +14,3 @@ def twoSum(self, nums: List[int], target: int) -> List[int]:
                 return [seen[complement], i]
             seen[num] = i               
         return []
-      

two-sum/hongseoupyun.ts

@@ -1,9 +1,10 @@
-// # Create a empty hash table(seen) and store numbers as a key and their indices as a value
-// # Iterate through the given array(nums) 
-// # For each numbers in given array, calculate complement (target - num)
-// # If complement is in hash table(seen), return array of index of the currnet number and index of complement in hash table(seen)
-// # If complement is not in has table(seen), store the current number in hash table(seen) and continue checking the numbers
-//
+// Create a empty hash table(seen) and store numbers as a key and their indices as a value
+// Iterate through the given array(nums) 
+// For each numbers in given array, calculate complement (target - num)
+// If complement is in hash table(seen), return array of index of the currnet number and index of complement in hash table(seen)
+// If complement is not in has table(seen), store the current number in hash table(seen) and continue checking the numbers
+// The algorithm uses a single loop that goes through each element in nums exactly one time.
+// If the array contains n elements, the loop runs n times → O(n).
 
 function twoSum2(nums: number[], target: number): number[] {
     const seen:{[key:number] : number} = {};