diff --git a/121 Leetcode/121 leetcode.py b/121 Leetcode/121 leetcode.py
index ea6114e9..cccd16b8 100644
--- a/121 Leetcode/121 leetcode.py	
+++ b/121 Leetcode/121 leetcode.py	
@@ -1,5 +1,5 @@
 class Solution:
-    def maxProfit(self, prices: List[int]) -> int:
+    def maxProfit(self, prices: list[int]) -> int:
         maxProfit = 0
         currentMax = 0
 
diff --git a/Python/Dynamic_Programming/CountMinSteps_Iterative.py b/Python/Dynamic_Programming/CountMinSteps_Iterative.py
index 30b35469..ed540552 100644
--- a/Python/Dynamic_Programming/CountMinSteps_Iterative.py
+++ b/Python/Dynamic_Programming/CountMinSteps_Iterative.py
@@ -5,7 +5,7 @@
 # 2.) If n is divisible by 2, divide by 2.( if n % 2 == 0, then n = n / 2 ) ,
 # 3.) If n is divisible by 3, divide by 3. (if n % 3 == 0, then n = n / 3 ). 
 
-# Now solve this iteratively :
+# Now solve this iteratively 
 
 from sys import stdin
 from sys import maxsize as MAX_VALUE
diff --git a/Python/Leetcode_with_Python/Leetcode_#1_Two_sum.py b/Python/Leetcode_with_Python/Leetcode_#1_Two_sum.py
new file mode 100644
index 00000000..5fbda3bd
--- /dev/null
+++ b/Python/Leetcode_with_Python/Leetcode_#1_Two_sum.py
@@ -0,0 +1,79 @@
+# 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].
+
+
+class Solution:
+    def twoSum(self, num:list[int], target: int)-> list[int]:
+        for i in range(len(num)):
+            for j in range(i+1, len(num)):
+                if num[i] + num[j] == target:
+                    return [i, j]
+                
+# general_solution = Solution()
+# print(general_solution.twoSum([2,7,1,11], 9))
+
+# This code taking way too much runtime because of O(n^2)
+                
+# We will try another approch which will be Hashmap
+                
+
+class Solution2:
+    def Two_Sum(self, num: list[int], target:int)->list[int]:
+        saw = {}
+
+        for i, num in enumerate(num):
+            if target - num in saw:
+                return [saw[target-num], i]
+            elif num not in saw:
+                saw[num] = i
+            
+
+# hashmap = Solution2()
+# print(hashmap.Two_Sum([2,1,11,23,7], 9))
+                
+
+# Used binary search Algorithm 
+                
+
+
+
+class Solution3:
+    def Twosum(self, num: list[int], target: int) -> list[int]:
+        # Create a list of tuples (original number, index)
+        num_with_indices = [(num[i], i) for i in range(len(num))]
+        print(num_with_indices)
+        # Sort the list of tuples based on the numbers
+        num_with_indices.sort()
+        # num_with_indices.sort()
+        
+        # print(num_with_indices)
+        low = 0
+        high = len(num) - 1
+
+        while low <= high:
+            current_sum = num_with_indices[low][0] + num_with_indices[high][0]
+
+            if current_sum < target:
+                low += 1
+            elif current_sum > target:
+                high -= 1
+            else:
+                return [num_with_indices[low][1], num_with_indices[high][1]]
+
+        return [-1, -1]  # Return [-1, -1] if no solution is found
+
+
+
+
+
+hashmap = Solution3()
+print(hashmap.Twosum([2, 3, 23, 7], 9))
\ No newline at end of file
diff --git a/Python/Searching-Algorithms/binary_search.py b/Python/Searching-Algorithms/binary_search.py
index ea000683..b52f4de2 100644
--- a/Python/Searching-Algorithms/binary_search.py
+++ b/Python/Searching-Algorithms/binary_search.py
@@ -1,5 +1,5 @@
 class Solution:
-    def search(self, nums: List[int], target: int) -> int:
+    def search(self, nums: list[int], target: int) -> int:
         low = 0
         high = len(nums) - 1
         mid = 0