diff --git a/course-schedule/wogha95.js b/course-schedule/wogha95.js
new file mode 100644
index 000000000..746bc08b5
--- /dev/null
+++ b/course-schedule/wogha95.js
@@ -0,0 +1,49 @@
+/**
+ * TC: O(V + E)
+ * SC: O(V + E)
+ * N: numCourses(all of vertex), P: prerequisites(all of edge)
+ */
+
+/**
+ * @param {number} numCourses
+ * @param {number[][]} prerequisites
+ * @return {boolean}
+ */
+var canFinish = function (numCourses, prerequisites) {
+  const STEP = {
+    before: 0,
+    ing: 1,
+    after: 2,
+  };
+  const stepBoard = Array.from({ length: numCourses }, () => STEP.before);
+  const board = Array.from({ length: numCourses }, () => []);
+
+  for (const [a, b] of prerequisites) {
+    board[a].push(b);
+  }
+
+  for (let index = 0; index < numCourses; index++) {
+    if (isCycle(index)) {
+      return false;
+    }
+  }
+  return true;
+
+  function isCycle(current) {
+    if (stepBoard[current] === STEP.end) {
+      return false;
+    }
+    if (stepBoard[current] === STEP.ing) {
+      return true;
+    }
+
+    stepBoard[current] = STEP.ing;
+    for (const next of board[current]) {
+      if (isCycle(next)) {
+        return true;
+      }
+    }
+    stepBoard[current] = STEP.end;
+    return false;
+  }
+};
diff --git a/invert-binary-tree/wogha95.js b/invert-binary-tree/wogha95.js
new file mode 100644
index 000000000..066c45ab5
--- /dev/null
+++ b/invert-binary-tree/wogha95.js
@@ -0,0 +1,33 @@
+/**
+ * 양쪽 자식 노드 주소를 교환하고 dfs로 순회합니다.
+ *
+ * TC: O(N)
+ * 모든 트리를 순회합니다.
+ *
+ * SC: O(N)
+ * 최악의 경우 (한쪽으로 치우친 트리) N만큼 CallStack이 생깁니다.
+ *
+ * N: tree의 모든 node 수
+ */
+
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @return {TreeNode}
+ */
+var invertTree = function (root) {
+  if (!root) {
+    return root;
+  }
+  [root.left, root.right] = [root.right, root.left];
+  invertTree(root.left);
+  invertTree(root.right);
+  return root;
+};
diff --git a/jump-game/wogha95.js b/jump-game/wogha95.js
new file mode 100644
index 000000000..1d102141b
--- /dev/null
+++ b/jump-game/wogha95.js
@@ -0,0 +1,37 @@
+/**
+ * TC: O(N)
+ * SC: O(1)
+ * N: nums.length
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {boolean}
+ */
+var canJump = function (nums) {
+  if (nums.length === 1) {
+    return true;
+  }
+
+  let maximumIndex = 0;
+
+  for (let index = 0; index < nums.length; index++) {
+    const jumpLength = nums[index];
+
+    if (jumpLength === 0) {
+      continue;
+    }
+
+    if (maximumIndex < index) {
+      return false;
+    }
+
+    maximumIndex = Math.max(maximumIndex, index + nums[index]);
+
+    if (maximumIndex >= nums.length - 1) {
+      return true;
+    }
+  }
+
+  return false;
+};
diff --git a/search-in-rotated-sorted-array/wogha95.js b/search-in-rotated-sorted-array/wogha95.js
new file mode 100644
index 000000000..4d8362737
--- /dev/null
+++ b/search-in-rotated-sorted-array/wogha95.js
@@ -0,0 +1,65 @@
+/**
+ * 모든 케이스가 많지 않다고 생각되어 분기처리하였습니다..
+ * 더 간결한 풀이법을 고려해보는 중..
+ *
+ * TC: O(log N)
+ * 이진탐색을 이용하여 순회합니다.
+ *
+ * SC: O(1)
+ * 이진탐색에 이용되는 투 포인터의 공간복잡도를 갖습니다.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} target
+ * @return {number}
+ */
+var search = function (nums, target) {
+  if (nums.length === 1) {
+    return target === nums[0] ? 0 : -1;
+  }
+
+  let left = 0;
+  let right = nums.length - 1;
+
+  while (left < right) {
+    const center = Math.floor((left + right) / 2);
+    if (target === nums[left]) {
+      return left;
+    }
+    if (target === nums[center]) {
+      return center;
+    }
+    if (target === nums[right]) {
+      return right;
+    }
+
+    if (nums[left] <= nums[center] && nums[center] < nums[right]) {
+      if (target < nums[left] || nums[right] < target) {
+        return -1;
+      } else if (nums[left] < target && target < nums[center]) {
+        right = center;
+      } else if (nums[center] < target && target < nums[right]) {
+        left = center + 1;
+      }
+    } else if (nums[right] < nums[left] && nums[left] <= nums[center]) {
+      if (nums[right] < target && target < nums[left]) {
+        return -1;
+      } else if (nums[left] < target && target < nums[center]) {
+        right = center;
+      } else {
+        left = center + 1;
+      }
+    } else if (nums[center] < nums[right] && nums[right] < nums[left]) {
+      if (nums[center] < target && target < nums[right]) {
+        left = center + 1;
+      } else if (nums[right] < target && target < nums[left]) {
+        return -1;
+      } else {
+        right = center;
+      }
+    }
+  }
+
+  return -1;
+};