diff --git a/contains-duplicate/dusunax.py b/contains-duplicate/dusunax.py
new file mode 100644
index 000000000..fb9255c70
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+++ b/contains-duplicate/dusunax.py
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+'''
+# Leetcode 217. Contains Duplicate
+
+use set to store distinct elements 🗂️
+
+## Time and Space Complexity
+
+```
+TC: O(n)
+SC: O(n)
+```
+
+### TC is O(n):
+- iterating through the list just once to convert it to a set.
+
+### SC is O(n):
+- creating a set to store the distinct elements of the list.
+'''
+
+class Solution:
+    def containsDuplicate(self, nums: List[int]) -> bool:
+        return len(nums) != len(set(nums))
+
diff --git a/house-robber/dusunax.py b/house-robber/dusunax.py
new file mode 100644
index 000000000..38c238942
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+++ b/house-robber/dusunax.py
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+'''
+# Leetcode 198. House Robber
+
+use **dynamic programming** to solve this problem. (bottom-up approach) 🧩
+
+choose bottom-up approach for less space complexity.
+
+## DP relation
+
+```
+dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
+```
+
+- **dp[i - 1]:** skip and take the value from the previous house
+- **dp[i - 2]:** rob the current house, add its value to the maximum money from two houses before
+
+## Time and Space Complexity
+
+```
+TC: O(n)
+SC: O(n)
+```
+
+### TC is O(n):
+- iterating through the list just once to calculate the maximum money. = O(n)
+
+### SC is O(n):
+- using a list to store the maximum money at each house. = O(n)
+
+'''
+
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        if len(nums) == 1:
+            return nums[0]
+
+        dp = [0] * len(nums)
+        dp[0] = nums[0]
+        dp[1] = max(nums[0], nums[1])
+
+        for i in range(2, len(nums)):
+            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
+            
+        return dp[-1]
diff --git a/longest-consecutive-sequence/dusunax.py b/longest-consecutive-sequence/dusunax.py
new file mode 100644
index 000000000..b912c4807
--- /dev/null
+++ b/longest-consecutive-sequence/dusunax.py
@@ -0,0 +1,38 @@
+'''
+# Leetcode 128. Longest Consecutive Sequence
+
+keep time complexity O(n) by iterating through the set and accessing elements in O(1) time. ⚖️
+
+## Time and Space Complexity
+```
+TC: O(n)
+SC: O(n)
+```
+
+### TC is O(n):
+- iterating through the set. O(n)
+- accessing elements in the set. O(1)
+- while loop incrementing `current_num` while `current_num + 1 in nums_set`. O(1)
+
+### SC is O(n):
+- creating a set from the list of numbers. O(n)
+'''
+
+class Solution:
+    def longestConsecutive(self, nums: List[int]) -> int:
+        nums_set = set(nums) # O(n)
+        longest_sequence = 0
+
+        for num in nums_set: # O(n)
+            if (num - 1) not in nums_set:
+                current_num = num
+                current_sequence = 1
+            
+                while current_num + 1 in nums_set: # O(1) 
+                    current_num += 1
+                    current_sequence += 1
+
+                longest_sequence = max(current_sequence, longest_sequence)
+
+        return longest_sequence
+
diff --git a/top-k-frequent-elements/dusunax.py b/top-k-frequent-elements/dusunax.py
new file mode 100644
index 000000000..1e8ee7a4c
--- /dev/null
+++ b/top-k-frequent-elements/dusunax.py
@@ -0,0 +1,69 @@
+'''
+# Leetcode 347. Top K Frequent Elements
+
+use **Counter** to count the frequency of each element in the list 🚀
+
+- **solution 1**: use **sorted()** to sort the elements by their frequency in descending order.
+- **solution 2**: use **bucket sort** to sort the elements by their frequency in descending order. (efficient!)
+
+## Time and Space Complexity
+
+### solution 1: topKFrequent()
+
+```
+TC: O(n log n)
+SC: O(n)
+```
+
+#### TC is O(n log n):
+- iterating through the list just once to count the frequency of each element. = O(n)
+- sorting the elements by their frequency in descending order. = O(n log n)
+
+#### SC is O(n):
+- using a Counter to store the frequency of each element. = O(n)
+- sorted() creates a new list that holds the elements of frequency_map. = O(n)
+- result list that holds the top k frequent elements. = O(k)
+
+### solution 2: topKFrequentBucketSort()
+
+```
+TC: O(n)
+SC: O(n)
+```
+
+#### TC is O(n):
+- iterating through the list just once to count the frequency of each element. = O(n)
+- creating **buckets** to store the elements by their frequency. = O(n)
+- iterating through the buckets in reverse order to get only the top k frequent elements. = O(n)
+
+#### SC is O(n):
+- using a Counter to store the frequency of each element. = O(n)
+- using buckets to store the elements by their frequency. = O(n)
+- result list that holds only the top k frequent elements. = O(k)
+'''
+
+class Solution:
+    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
+        frequency_map = Counter(nums) # TC: O(n), SC: O(n)
+        sorted_frequencies = sorted(frequency_map.items(), key=lambda x: x[1], reverse=True) # TC: O(n log n), SC: O(n)
+        
+        result = [] # SC: O(k)
+        for e in sorted_frequencies: # TC: O(k)
+            result.append(e[0])
+
+        return result[0:k]
+    
+    def topKFrequentBucketSort(self, nums: List[int], k: int) -> List[int]:
+        frequency_map = Counter(nums)
+        n = len(nums)
+        buckets = [[] for _ in range(n + 1)]
+
+        for num, freq in frequency_map.items():
+            buckets[freq].append(num)
+
+        result = []
+        for i in range(len(buckets) - 1, 0, -1):
+            for num in buckets[i]:
+                result.append(num)
+                if len(result) == k:
+                    return result
diff --git a/valid-palindrome/dusunax.py b/valid-palindrome/dusunax.py
new file mode 100644
index 000000000..52ee0fd1a
--- /dev/null
+++ b/valid-palindrome/dusunax.py
@@ -0,0 +1,31 @@
+'''
+# Leetcode 125. Valid Palindrome
+
+use `isalnum()` to filter out non-alphanumeric characters 🔍
+
+## Time and Space Complexity
+
+```
+TC: O(n)
+SC: O(n)
+```
+
+### TC is O(n):
+- iterating through the string just once to filter out non-alphanumeric characters. O(n)
+
+### SC is O(n):
+- `s.lower()` creates a new string. O(n)
+- creating a new string `converted_s` to store the filtered characters. O(n)
+- `converted_s[::-1]` creates a new reversed string. O(n)
+'''
+
+class Solution:
+    def isPalindrome(self, s: str) -> bool:
+        if s == " ":
+            return True
+
+        s = s.lower()
+        converted_s = ''.join(c for c in s if c.isalnum())
+
+        return converted_s == converted_s[::-1]
+