diff --git a/contains-duplicate/f-exuan21.java b/contains-duplicate/f-exuan21.java
new file mode 100644
index 000000000..acb71d219
--- /dev/null
+++ b/contains-duplicate/f-exuan21.java
@@ -0,0 +1,16 @@
+// time : O(n)
+// space : O(n)
+
+class Solution {
+    public boolean containsDuplicate(int[] nums) {
+        Set<Integer> set = new HashSet<>();
+        for(int i : nums) {
+            if(set.contains(i)) {
+                return true;
+            }
+            set.add(i);
+        }
+        return false;
+    }
+}
+
diff --git a/kth-smallest-element-in-a-bst/f-exuan21.java b/kth-smallest-element-in-a-bst/f-exuan21.java
new file mode 100644
index 000000000..95433a739
--- /dev/null
+++ b/kth-smallest-element-in-a-bst/f-exuan21.java
@@ -0,0 +1,50 @@
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode() {}
+ *     TreeNode(int val) { this.val = val; }
+ *     TreeNode(int val, TreeNode left, TreeNode right) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+class Solution {
+
+    private int count = 0;
+    private TreeNode resultNode = null;
+
+    public int kthSmallest(TreeNode root, int k) {
+        searchNode(root, k);
+        return resultNode.val;
+    }
+
+    public void searchNode(TreeNode node, int k) {
+
+        if(resultNode != null) return;
+
+        if(node.left != null) {
+            searchNode(node.left, k); 
+        }
+
+        count++;
+
+        if(count == k) {
+            resultNode = node;
+            return;
+        }
+
+        if(node.right != null) {
+            searchNode(node.right, k);
+        }
+    }
+}
+
+// n : 노드 개수
+// time : O(n) 최악의 경우 모든 노드를 탐색해야함
+// space : O(n) 최악의 경우 한 쪽으로 노드가 치우쳐져 있음 
+// -> 재귀 호출이 이루어지므로 스택에 쌓임 -> 한 쪽으로 쏠려 있으면 트리의 높이가 n이 됨 (트리의 최대 높이가 스택의 최대 깊이)
diff --git a/number-of-1-bits/f-exuan21.java b/number-of-1-bits/f-exuan21.java
new file mode 100644
index 000000000..9c06c4329
--- /dev/null
+++ b/number-of-1-bits/f-exuan21.java
@@ -0,0 +1,16 @@
+// time : O(1) 
+// space : O(1)
+
+class Solution {
+    public int hammingWeight(int n) {
+        int count = 0;
+
+        while(n != 0) {
+            if((n&1) == 1) count++;
+            n = n >> 1;
+        }
+
+        return count;
+    }
+}
+
diff --git a/top-k-frequent-elements/f-exuan21.java b/top-k-frequent-elements/f-exuan21.java
new file mode 100644
index 000000000..412b6fe93
--- /dev/null
+++ b/top-k-frequent-elements/f-exuan21.java
@@ -0,0 +1,39 @@
+
+class Solution {
+    public int[] topKFrequent(int[] nums, int k) {
+        Map<Integer, Integer> map = new HashMap<>();
+        for(int num : nums) {
+            map.put(num, map.getOrDefault(num, 0) + 1);
+        }
+        
+        PriorityQueue<Map.Entry<Integer, Integer>> queue = new PriorityQueue<>(
+            (a, b) -> Integer.compare(b.getValue(), a.getValue())
+        );
+
+        for(Map.Entry<Integer, Integer> entry : map.entrySet()) {
+            queue.offer(entry);
+        }
+
+        int[] res = new int[k];
+
+        for(int i = 0; i < k; i++) {
+            res[i] = queue.poll().getKey();
+        }
+
+        return res;
+    }
+}
+
+// n : nums의 길이
+// m : nums에서 서로 다른 숫자의 개수
+
+// time : O(n) + O(m*logm) + O(k*logm) = O(n + m*logm + k*logm)
+// 최악의 경우, nums 가 다 unique 하기 때문에 n == m == k 가 됨
+// 따라서, O(n*logn)
+
+// space : O(m) + O(m) + O(k) = O(m + k)
+// 최악의 경우 n == m == k 가 됨
+// 따라서, O(n)
+
+
+