diff --git a/src/my_project/interviews/amazon_high_frequency_23/round_2/k_free_subsets.py b/src/my_project/interviews/amazon_high_frequency_23/round_2/k_free_subsets.py
index 6934705b..6188ef27 100644
--- a/src/my_project/interviews/amazon_high_frequency_23/round_2/k_free_subsets.py
+++ b/src/my_project/interviews/amazon_high_frequency_23/round_2/k_free_subsets.py
@@ -54,4 +54,55 @@ def countTheNumOfKFreeSubsets(self, nums: List[int], k: int) -> int:
                 res *= chain_res
                 i = j
         
-        return res
\ No newline at end of file
+        return res
+    
+
+'''
+Detailed Algorithm Explanation
+Part 1: Why Group by num % k?
+Two numbers can have a difference of exactly k only if they have the same remainder when divided by k.
+
+Mathematical proof:
+
+If a - b = k, then a = b + k
+Therefore: a % k = (b + k) % k = b % k
+Example: nums = [2, 3, 5, 8], k = 5
+
+num | num % 5 | group
+----|---------|-------
+2   |    2    | Group A
+3   |    3    | Group B  
+5   |    0    | Group C
+8   |    3    | Group B
+
+
+Why this matters: Elements from different groups can never differ by k, so they're independent. We can combine any subset from Group A with any subset from Group B.
+
+Part 2: Building Chains
+Within each group, we sort and find chains where consecutive elements differ by exactly k.
+
+Example with Group B: [3, 8]
+
+Sorted: [3, 8]
+Check: 8 - 3 = 5 ✓
+Chain: 3 → 8
+
+Another example: nums = [1, 6, 11, 21], k = 5 (all have remainder 1)
+
+Sorted: [1, 6, 11, 21]
+Check: 6-1=5 ✓, 11-6=5 ✓, 21-11=10 ✗
+Chains: [1 → 6 → 11], [21]
+
+
+Part 3: House Robber DP - The Core Logic
+For a chain like [3 → 8], we can't pick both 3 and 8 (they differ by k). This is the House Robber problem: count all subsets where we don't pick adjacent elements.
+
+DP State Variables
+take = number of valid subsets that INCLUDE the current element
+skip = number of valid subsets that EXCLUDE the current element
+
+DP Transitions
+new_take = skip      # To take current, we MUST have skipped previous
+new_skip = take + skip  # To skip current, we can take or skip previous
+
+'''    
\ No newline at end of file
diff --git a/src/my_project/interviews/amazon_high_frequency_23/round_2/two_sum_round_5.py b/src/my_project/interviews/amazon_high_frequency_23/round_2/two_sum_round_5.py
new file mode 100644
index 00000000..e840ea34
--- /dev/null
+++ b/src/my_project/interviews/amazon_high_frequency_23/round_2/two_sum_round_5.py
@@ -0,0 +1,16 @@
+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 []
diff --git a/src/my_project/interviews/amazon_high_frequency_23/round_2/valid_palindrome_round_7.py b/src/my_project/interviews/amazon_high_frequency_23/round_2/valid_palindrome_round_7.py
new file mode 100644
index 00000000..43525f5d
--- /dev/null
+++ b/src/my_project/interviews/amazon_high_frequency_23/round_2/valid_palindrome_round_7.py
@@ -0,0 +1,23 @@
+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='[^a-zA-Z0-9]', repl='', string=s)
+
+        # Determine if it 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 
diff --git a/src/my_project/interviews/top_150_questions_round_21/word_pattern.py b/src/my_project/interviews/top_150_questions_round_21/word_pattern.py
new file mode 100644
index 00000000..4a509a9b
--- /dev/null
+++ b/src/my_project/interviews/top_150_questions_round_21/word_pattern.py
@@ -0,0 +1,21 @@
+from typing import List, Union, Collection, Mapping, Optional
+from abc import ABC, abstractmethod
+
+class Solution:
+    def wordPattern(self, pattern: str, s: str) -> bool:
+
+        dic_answer = dict()
+
+        words = s.split(' ')
+
+        if len(words) != len(pattern) or len(set(words)) != len(set(pattern)):
+            return False 
+        else:         
+            for k, v in enumerate(words):
+                if v in dic_answer:
+                    if dic_answer[v] != pattern[k]:
+                        return False 
+                else:
+                    dic_answer[v] = pattern[k]
+
+            return True 
\ No newline at end of file
diff --git a/tests/test_150_questions_round_21/test_word_pattern_round_21.py b/tests/test_150_questions_round_21/test_word_pattern_round_21.py
new file mode 100644
index 00000000..9521aa7b
--- /dev/null
+++ b/tests/test_150_questions_round_21/test_word_pattern_round_21.py
@@ -0,0 +1,20 @@
+import unittest
+from src.my_project.interviews.top_150_questions_round_21\
+.word_pattern import Solution
+
+class WordPatternTestCase(unittest.TestCase):
+
+    def test_is_word_pattern(self):
+        solution = Solution()
+        output = solution.wordPattern(pattern="abba", s="dog cat cat dog")
+        self.assertTrue(output)
+
+    def test_is_no_word_pattern_one(self):
+        solution = Solution()
+        output = solution.wordPattern(pattern="abc", s="dog cat cat fish")
+        self.assertFalse(output)
+
+    def test_is_no_word_pattern_two(self):
+        solution = Solution()
+        output = solution.wordPattern(pattern="aa", s="dog cat cat fish")
+        self.assertFalse(output)
\ No newline at end of file