diff --git a/find-minimum-in-rotated-sorted-array/YuuuuuuYu.java b/find-minimum-in-rotated-sorted-array/YuuuuuuYu.java
new file mode 100644
index 0000000000..535f45c286
--- /dev/null
+++ b/find-minimum-in-rotated-sorted-array/YuuuuuuYu.java
@@ -0,0 +1,26 @@
+/**
+ * Runtime: 0ms
+ * Time Complexity: O(log n)
+ *
+ * Memory: 43.81MB
+ * Space Complexity: O(1)
+ *
+ * Approach: 이진 탐색
+ */
+class Solution {
+    public int findMin(int[] nums) {
+        int left = 0;
+        int right = nums.length-1;
+
+        while (left < right) {
+            int mid = left + (right-left)/2;
+            if (nums[mid] <= nums[right]) {
+                right = mid;
+            } else {
+                left = mid + 1;
+            }
+        }
+
+        return nums[left];
+    }
+}
diff --git a/maximum-depth-of-binary-tree/YuuuuuuYu.java b/maximum-depth-of-binary-tree/YuuuuuuYu.java
new file mode 100644
index 0000000000..46eb3ff124
--- /dev/null
+++ b/maximum-depth-of-binary-tree/YuuuuuuYu.java
@@ -0,0 +1,20 @@
+/**
+ * Runtime: 0ms
+ * Time Complexity: O(n)
+ *
+ * Memory: 44.28MB
+ * Space Complexity: O(h)
+ * - h: 트리의 높이 or 최대 깊이
+ *
+ * Approach: 재귀를 이용한 DFS 접근법
+ * - 루트부터 시작하여 재귀가 시작할 때마다 1로 계산
+ * - 트리의 끝에 도달하면 0을 리턴하여 차례대로 남은 깊이를 계산
+ */
+class Solution {
+    public int maxDepth(TreeNode root) {
+        if (root == null)
+            return 0;
+
+        return 1 + Math.max(maxDepth(root.left), maxDepth(root.right));
+    }
+}
diff --git a/merge-two-sorted-lists/YuuuuuuYu.java b/merge-two-sorted-lists/YuuuuuuYu.java
new file mode 100644
index 0000000000..e0181563c6
--- /dev/null
+++ b/merge-two-sorted-lists/YuuuuuuYu.java
@@ -0,0 +1,34 @@
+/**
+ * Runtime: 0ms
+ * Time Complexity: O(n)
+ * - list1.length + list2.length
+ *
+ * Memory: 44.28MB
+ * Space Complexity: O(1)
+ *
+ * Approach: 빈 node를 만들고 list1, list2의 현재 값을 비교하여 더 작은 값을 추가
+ *
+ */
+class Solution {
+    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
+        if (list1 == null) return list2;
+        else if (list2 == null) return list1;
+
+        ListNode result = new ListNode();
+        ListNode currentNode = result;
+        while (list1 != null && list2 != null) {
+            if (list1.val > list2.val) {
+                currentNode.next = list2;
+                list2 = list2.next;
+            } else {
+                currentNode.next = list1;
+                list1 = list1.next;
+            }
+
+            currentNode = currentNode.next;
+        }
+
+        currentNode.next = list1 != null ? list1 : list2;
+        return result.next;
+    }
+}
diff --git a/word-search/YuuuuuuYu.java b/word-search/YuuuuuuYu.java
new file mode 100644
index 0000000000..bcc64a9b17
--- /dev/null
+++ b/word-search/YuuuuuuYu.java
@@ -0,0 +1,67 @@
+/**
+ * Runtime: 128ms
+ * Time Complexity: O(m x n x 4^l)
+ * - m: board의 행 길이
+ * - n: board의 열 길이
+ * - l: word의 길이
+ *
+ * Memory: 42.69MB
+ * Space Complexity: O(l)
+ *
+ * Approach: DFS + 백트래킹
+ * 1) board를 순회하며 word의 첫 글자와 일치하는 글자를 찾음
+ * 2) 일치하는 글자를 찾으면 DFS를 통해 상하좌우로 다음 글자를 찾음
+ * 3) 성능 최적화
+ * - 백트래킹을 위해 방문한 글자는 임시로 다른 문자('#')로 변경 (비트마스크)
+ * - charAt 캐싱
+ */
+class Solution {
+    int[] dx = {-1, 1, 0, 0};
+    int[] dy = {0, 0, -1, 1};
+
+    public boolean exist(char[][] board, String word) {
+        int m = board.length;
+        int n = board[0].length;
+        char startChar = word.charAt(0);
+
+        for (int i=0; i<m; i++) {
+            for (int j=0; j<n; j++) {
+                if (board[i][j] == startChar) {
+                    if (dfs(board, i, j, 0, word)) {
+                        return true;
+                    }
+                }
+            }
+        }
+
+        return false;
+    }
+
+    boolean dfs(char[][] board, int x, int y, int index, String word) {
+        if (index == word.length()-1) {
+            return true;
+        }
+
+        char temp = board[x][y];
+        char nextChar = word.charAt(index+1);
+        board[x][y] = ' ';
+
+        for (int i=0; i<4; i++) {
+            int nx = x+dx[i];
+            int ny = y+dy[i];
+
+            if (nx >= 0 && nx < board.length &&
+                    ny >= 0 && ny < board[0].length &&
+                    board[nx][ny] == nextChar) {
+
+                if (dfs(board, nx, ny, index+1, word)) {
+                    board[x][y] = temp;
+                    return true;
+                }
+            }
+        }
+
+        board[x][y] = temp;
+        return false;
+    }
+}