diff --git a/contains-duplicate/sun912.py b/contains-duplicate/sun912.py
new file mode 100644
index 000000000..c86fb22de
--- /dev/null
+++ b/contains-duplicate/sun912.py
@@ -0,0 +1,8 @@
+"""
+    Time complexity: O(n)
+"""
+
+class Solution:
+    def containsDuplicate(self, nums: List[int]) -> bool:
+        nums_set = set(nums)
+        return len(nums_set) != len(nums)
diff --git a/kth-smallest-element-in-a-bst/sun912.py b/kth-smallest-element-in-a-bst/sun912.py
new file mode 100644
index 000000000..bbfceff85
--- /dev/null
+++ b/kth-smallest-element-in-a-bst/sun912.py
@@ -0,0 +1,27 @@
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+
+"""
+    TC: O(n) in worst case
+    SC: O(n) in worst case
+"""
+class Solution:
+    def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
+        n = 0
+        stack = []
+        current_node = root
+
+        while current_node or stack:
+            while current_node:
+                stack.append(current_node)
+                current_node = current_node.left
+
+            current_node = stack.pop()
+            n += 1
+            if n == k:
+                return current_node.val
+            current_node = current_node.right
diff --git a/number-of-1-bits/sun912.py b/number-of-1-bits/sun912.py
new file mode 100644
index 000000000..0df9a5d29
--- /dev/null
+++ b/number-of-1-bits/sun912.py
@@ -0,0 +1,13 @@
+"""
+ TC: O(log n)
+ SC: O(1)
+"""
+
+class Solution:
+    def hammingWeight(self, n: int) -> int:
+        result = 0
+        while n > 0:
+            result += n % 2
+            n = n//2
+
+        return result
\ No newline at end of file
diff --git a/palindromic-substrings/sun912.py b/palindromic-substrings/sun912.py
new file mode 100644
index 000000000..325f61e65
--- /dev/null
+++ b/palindromic-substrings/sun912.py
@@ -0,0 +1,26 @@
+"""
+    TC: O(n^2)
+
+    TC in first try was O(n^3).
+    It is the improved codes from neetcodes
+    key point is that each character in loop is considered to middle one.
+    Ant then, check left end and right end.
+
+"""
+class Solution:
+    def countSubstrings(self, s: str) -> int:
+        result = 0
+        for i in range(len(s)):
+            result += self.countPalindrom(s, i, i)
+            result += self.countPalindrom(s, i, i+1)
+        return result
+
+
+    def countPalindrom(self, s, left_end, right_end):
+        result = 0
+        while left_end >= 0 and right_end < len(s) and s[left_end] == s[right_end]:
+            result += 1
+            left_end -= 1
+            right_end += 1
+
+        return result
diff --git a/top-k-frequent-elements/sun912.py b/top-k-frequent-elements/sun912.py
new file mode 100644
index 000000000..5fe78ea9c
--- /dev/null
+++ b/top-k-frequent-elements/sun912.py
@@ -0,0 +1,24 @@
+"""
+    TC : O(n)
+    SC: O(n)
+    used bucket sort
+    index -> count number
+    list value -> frequent nums
+"""
+
+class Solution:
+    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
+        count = {}
+        freq = [[] for i in range(len(nums)+1)]
+
+        for num in nums:
+            count[num] = 1 + count.get(num, 0)
+        for key, val in count.items():
+            freq[val].append(key)
+
+        result = []
+        for i in range(len(freq) - 1, 0, -1):
+            for n in freq[i]:
+                result.append(n)
+                if len(result) == k:
+                    return result