adding updates

src/my_project/interviews/amazon_high_frequency_23/common_algos/two_sum_round_7.py

@@ -0,0 +1,17 @@
+from typing import List, Union, Collection, Mapping, Optional
+from abc import ABC, abstractmethod
+
+class Solution:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+
+        answer = dict()
+
+        for k, v in enumerate(nums):
+
+            if v in answer:
+                return [answer[v], k]
+            else: 
+                answer[target - v] = k 
+
+        return []
+

src/my_project/interviews/amazon_high_frequency_23/common_algos/valid_palindrome_round_7.py

@@ -0,0 +1,22 @@
+from typing import List, Union, Collection, Mapping, Optional
+from abc import ABC, abstractmethod
+import re
+
+class Solution:
+    def isPalindrome(self, s: str) -> bool:
+
+        # To lowercase
+        s = s.lower()
+
+        # Remove non-alphanumeric characters
+        s = re.sub(pattern=r'[^a-zA-Z0-9]', repl='', string=s)
+
+        # Determine if s is palindrome or not
+        len_s = len(s)
+
+        for i in range(len_s//2):
+
+            if s[i] != s[len_s - 1- i]:
+                return False 
+
+        return True 

src/my_project/interviews/amazon_high_frequency_23/round_2/number_of_ways_to_select_building.py

@@ -10,33 +10,23 @@ def numberOfWays(self, s: str) -> int:
         then extend them with the appropriate third character.
 
         Time: O(n), Space: O(1)
-        """
-        count_0 = 0  # Count of '0' seen so far
-        count_1 = 0  # Count of '1' seen so far
-        count_01 = 0  # Count of "01" patterns (0 followed by 1)
-        count_10 = 0  # Count of "10" patterns (1 followed by 0)
-
-        result = 0
-
-        for char in s:
-            if char == '0':
-                # Current '0' can complete "010": we need "01" before this '0'
-                result += count_01
-
-                # Current '0' can start new "01" patterns: pair with each previous '0'
-                # Wait no, "01" means 0 then 1, so we can't create "01" with another 0
-
-                # Current '0' extends all previous '1' to form "10" pattern
-                count_10 += count_1
-
-                count_0 += 1
-            else:  # char == '1'
-                # Current '1' can complete "101": we need "10" before this '1'
-                result += count_10
-
-                # Current '1' extends all previous '0' to form "01" pattern
-                count_01 += count_0
-
-                count_1 += 1
-
-        return result
+        """        
+        total_zeros = s.count('0')
+        total_ones = len(s) - total_zeros
+
+        left_zeros = left_ones = 0
+        ways = 0
+
+        for ch in s:
+            if ch == '1':
+                # middle is '1' -> need '0' on both sides -> "010"
+                right_zeros = total_zeros - left_zeros
+                ways += left_zeros * right_zeros
+                left_ones += 1
+            else:  # ch == '0'
+                # middle is '0' -> need '1' on both sides -> "101"
+                right_ones = total_ones - left_ones
+                ways += left_ones * right_ones
+                left_zeros += 1
+
+        return ways

src/my_project/interviews/amazon_high_frequency_23/round_6/number_of_ways_to_select_building.py

@@ -0,0 +1,32 @@
+from typing import List, Union, Collection, Mapping, Optional
+from abc import ABC, abstractmethod
+
+class Solution:
+    def numberOfWays(self, s: str) -> int:
+        """
+        Count valid 3-building selections forming "010" or "101" patterns.
+
+        Key insight: Track how many 2-character patterns we can form,
+        then extend them with the appropriate third character.
+
+        Time: O(n), Space: O(1)
+        """        
+        total_zeros = s.count('0')
+        total_ones = len(s) - total_zeros
+
+        left_zeros = left_ones = 0
+        ways = 0
+
+        for ch in s:
+            if ch == '1':
+                # middle is '1' -> need '0' on both sides -> "010"
+                right_zeros = total_zeros - left_zeros
+                ways += left_zeros * right_zeros
+                left_ones += 1
+            else:  # ch == '0'
+                # middle is '0' -> need '1' on both sides -> "101"
+                right_ones = total_ones - left_ones
+                ways += left_ones * right_ones
+                left_zeros += 1
+
+        return ways