diff --git a/best-time-to-buy-and-sell-stock/JoyJaewon.py b/best-time-to-buy-and-sell-stock/JoyJaewon.py
new file mode 100644
index 000000000..33f5258de
--- /dev/null
+++ b/best-time-to-buy-and-sell-stock/JoyJaewon.py
@@ -0,0 +1,34 @@
+'''
+Brainstorm
+- Brute force: double for-loop O(N^2)
+- Efficient: single pass O(N)
+
+Plan
+1. Initialize the variables
+2. Iterate through each price
+    2-1. Update the min price
+    2-2. Calculate and update max profit
+3. Return the max profit
+'''
+
+
+def maxProfit(prices):
+    min_price = float('inf')
+    max_profit = 0
+    for price in prices:
+        if price < min_price:
+            min_price = price
+        elif price - min_price > max_profit:
+            max_profit = price - min_price
+    return max_profit
+
+#TC: O(N), SC: O(1)
+
+# Normal Case
+print(maxProfit([7, 1, 5, 3, 6, 4]) == 5)  
+print(maxProfit([7, 6, 4, 3, 1]) == 0)  
+
+# Edge Case
+print(maxProfit([]) == 0)  
+print(maxProfit([1]) == 0)  
+
diff --git a/contains-duplicate/JoyJaewon.py b/contains-duplicate/JoyJaewon.py
new file mode 100644
index 000000000..cae5bc2af
--- /dev/null
+++ b/contains-duplicate/JoyJaewon.py
@@ -0,0 +1,31 @@
+'''
+Brainstrom
+- brute force: double for-loop (O^2)
+- efficient approach: use set O(N)
+
+Plan
+1. Initialize the set
+2. Iterate over the array
+    2-1. For each element, check if it's in the set. If it is, return True
+    2-2. If not, add the num to the set
+3. Return False since no duplicates are found
+'''
+
+def containsDuplicate(nums):
+    num_set=set()
+    for num in nums:
+        if num in num_set:
+            return True
+        num_set.add(num)
+    return False
+
+#TC: O(N), SC: O(N)
+
+#Normal case
+print(containsDuplicate([1, 2, 3, 1]) == True)
+print(containsDuplicate([1, 2, 3, 4]) == False)
+
+#Edge case
+print(containsDuplicate([1]) == False)
+print(containsDuplicate([]) == False) 
+
diff --git a/two-sum/JoyJaewon.py b/two-sum/JoyJaewon.py
new file mode 100644
index 000000000..cc5966f0d
--- /dev/null
+++ b/two-sum/JoyJaewon.py
@@ -0,0 +1,33 @@
+'''
+Brainstorm
+- Brute force: double for-loop / recursion - O(N^2)
+- Two Pointer: O(nlogn) - sorting
+- Efficient approach: use hashmap to go through the array once O(N)
+
+Plan
+1. Initialize the dictionary
+2. Iterate over the array
+    2-1. For each number, calculate the complement
+    2-2. check if the complement exists in the dictionary
+        2-3. Return the indices if it exists
+    2-4. If not, add the current number and its index to the dictionary
+'''
+
+def twoSum(nums, target):
+    memo = {}  
+    for i in range(len(nums)):
+        diff = target - nums[i]
+        if diff in memo:           
+            return [memo[diff], i]
+        memo[nums[i]] = i 
+
+#TC:O(N), SC:O(N)
+
+#Normal case
+print(twoSum([2, 7, 11, 15], 9) == [0, 1])
+print(twoSum([3, 2, 4], 6) == [1, 2])
+
+#Edge case
+print(twoSum([1, 2], 3) == [0, 1])
+print(twoSum([1, 5, 10], 20) == None) 
+
diff --git a/valid-anagram/JoyJaewon.py b/valid-anagram/JoyJaewon.py
new file mode 100644
index 000000000..e0dacab6a
--- /dev/null
+++ b/valid-anagram/JoyJaewon.py
@@ -0,0 +1,39 @@
+'''
+Brainstorm
+- brute force: double for-loop O(N^2)
+- efficient: use hash map O(N)
+
+plan
+1. check the string lengths. Return false if different
+2. Initialize the dictionary
+3. Check the dictionary
+    3-1. If all counts are zero, strings are anagrams. Return True
+    3-2. If not, return False
+'''
+def isAnagram(s, t):
+    if len(s) != len(t):
+        return False
+    
+    count = {}
+    for char in s:
+        count[char] = count.get(char, 0) + 1  
+    
+    for char in t:
+        if char in count:
+            count[char] -= 1
+            if count[char] < 0:
+                return False
+        else:
+            return False
+    
+    return all(x == 0 for x in count.values())
+
+#TC: O(N), SC: O(N) 
+
+#Normal case
+print(isAnagram("anagram", "nagaram") == True) 
+print(isAnagram("rat", "car") == False) 
+
+#Edge case
+print(isAnagram("", "") == True) 
+
diff --git a/valid-palindrome/JoyJaewon.py b/valid-palindrome/JoyJaewon.py
new file mode 100644
index 000000000..bf7f7cfa3
--- /dev/null
+++ b/valid-palindrome/JoyJaewon.py
@@ -0,0 +1,39 @@
+'''
+Brainstorm
+- Brute force: double for-loop O(N^2)
+- Efficient: two pointer O(N)
+
+Plan
+1. Preprocess the input data (lowercase, remove non-alphanumeric char)
+2. Use two pointers to check for palindrome
+'''
+#Version 1
+def isPalindrome(s):
+    filtered_chars = [c.lower() for c in s if c.isalnum()]
+    left, right = 0, len(filtered_chars) - 1
+
+    while left < right:
+        if filtered_chars[left] != filtered_chars[right]:
+            return False
+        left, right = left + 1, right - 1
+    
+    return True
+
+
+#Version 2
+def isPalindrome2(s):
+    filtered_chars = [c.lower() for c in s if c.isalnum()]
+    return filtered_chars == filtered_chars[::-1]
+
+
+#TC: O(N), SC: O(N)
+
+#Normal case
+print(isPalindrome("A man, a plan, a canal: Panama") == True) 
+print(isPalindrome("race a car") == False) 
+
+#Edge case
+print(isPalindrome("") == True) 
+print(isPalindrome(" ") == True) 
+print(isPalindrome("a.") == True) 
+