diff --git a/contains-duplicate/ppxyn1.py b/contains-duplicate/ppxyn1.py
index 7c128588e3..3af34fcf90 100644
--- a/contains-duplicate/ppxyn1.py
+++ b/contains-duplicate/ppxyn1.py
@@ -1,5 +1,6 @@
 # idea: Hash
 
+# Ans 1
 class Solution:
     def containsDuplicate(self, nums: List[int]) -> bool:
         count_dict={}
@@ -12,8 +13,10 @@ def containsDuplicate(self, nums: List[int]) -> bool:
         return False
     
 
-'''
-Trial and error
-Printing I/O inside the loop may cause Output Limit Exceeded
-'''
+# Ans 2
+# Time Complexity: O(n) (Set)
+class Solution:
+    def containsDuplicate(self, nums: List[int]) -> bool:
+        return len(nums) != len(set(nums))
+
 
diff --git a/longest-consecutive-sequence/ppxyn1.py b/longest-consecutive-sequence/ppxyn1.py
index 6358515f3a..1b233c5dba 100644
--- a/longest-consecutive-sequence/ppxyn1.py
+++ b/longest-consecutive-sequence/ppxyn1.py
@@ -1,4 +1,6 @@
 # idea: Hash
+# start after checking left neightbour 
+# Time complexity = O(n)
 class Solution:
     def longestConsecutive(self, nums: List[int]) -> int:
         if not nums:
@@ -8,7 +10,7 @@ def longestConsecutive(self, nums: List[int]) -> int:
         max_len = 1  
 
         for num in num_set:
-            if num - 1 not in num_set:
+            if num - 1 not in num_set: 
                 current = num
                 tmp = 1
                 while current + 1 in num_set:
diff --git a/top-k-frequent-elements/ppxyn1.py b/top-k-frequent-elements/ppxyn1.py
index e6402c0bdb..cda401b8ea 100644
--- a/top-k-frequent-elements/ppxyn1.py
+++ b/top-k-frequent-elements/ppxyn1.py
@@ -1,5 +1,6 @@
 # idea: dictonary
 
+# Ans 1
 class Solution:
     def topKFrequent(self, nums: List[int], k: int) -> List[int]:
         count_dict = {}
@@ -15,7 +16,19 @@ def topKFrequent(self, nums: List[int], k: int) -> List[int]:
             ans.append(sorted_items[i][0])
         return ans
     
-'''
-Similar way : Using Counter() function
-'''
+
+
+# With Couter 
+# Time Complexity: O(nlog(n))
+
+from collections import Counter
+class Solution:
+    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
+        count_dict = Counter(nums)
+        count_dict = sorted(count_dict.items(), key=lambda x: x[1], reverse=True)
+        ans = []
+        for i in range(k):
+            ans.append(count_dict[i][0])
+        return ans
+    
 
diff --git a/two-sum/ppxyn1.py b/two-sum/ppxyn1.py
index 8ba1cf394d..f0deb83201 100644
--- a/two-sum/ppxyn1.py
+++ b/two-sum/ppxyn1.py
@@ -1,5 +1,6 @@
 # idea: For each number n in nums, check if (target - n) exists in the remaining elements.
 
+# Ans 1
 class Solution:
     def twoSum(self, nums: List[int], target: int) -> List[int]:
         for idx, num in enumerate(nums):
@@ -8,27 +9,28 @@ def twoSum(self, nums: List[int], target: int) -> List[int]:
                 return [idx, nums.index(required_num, idx+1)]
 
 
-'''
-Trial and error
-idea : two pointer
-I struggled to handle the indices of the original array after sorting it.
-The code below fails when negative numbers are involved.
-I realized it would be tricky to solve this problem with the two-pointer.
-'''
 
-# class Solution:
-#     def twoSum(self, nums: List[int], target: int) -> List[int]:
-#         sorted_nums = sorted(nums)
-#         left, right = 0, len(nums) - 1
+# Ans 2
+# idea : Two-pointer
+# Time Complexity : O(n log n)
+
+class Solution:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        # Keep original indices !
+        new_nums = [(idx, num) for idx, num in enumerate(nums)]
+        new_nums.sort(key=lambda x: x[1])
+
+        left, right = 0, len(new_nums)-1
         
-#         while left < right:
-#             s = sorted_nums[left] + sorted_nums[right]
-#             if s == target:
-#                 left_idx = nums.index(sorted_nums[left])
-#                 right_idx = nums.index(sorted_nums[right], left_idx + 1)
-#                 return [left_idx, right_idx]
-#             elif s < target:
-#                 left += 1
-#             else:
-#                 right -= 1
+        while left < right:
+            idx_left, idx_right = new_nums[left][0], new_nums[right][0]
+            val = new_nums[left][1] + new_nums[right][1]
+            if val == target:
+                return [idx_left, idx_right]
+            elif val < target:
+                left +=1 
+            else:
+                right -=1
+        return []
+