diff --git a/3sum/reach0908.js b/3sum/reach0908.js
new file mode 100644
index 000000000..18895e583
--- /dev/null
+++ b/3sum/reach0908.js
@@ -0,0 +1,46 @@
+/**
+ * 시간복잡도: O(n²)
+ * 공간복잡도: O(1) (결과 배열 제외)
+ * 풀이 방법: 정렬 후 투 포인터 방식
+ * @param {number[]} nums
+ * @return {number[][]}
+ */
+const threeSum = function (nums) {
+  const sortedNums = nums.sort((a, b) => a - b);
+  const result = [];
+
+  for (let i = 0; i < sortedNums.length; i += 1) {
+    // 첫 번째 요소의 중복 제거
+    if (i > 0 && sortedNums[i] === sortedNums[i - 1]) {
+      continue;
+    }
+
+    let left = i + 1;
+    let right = sortedNums.length - 1;
+
+    while (left < right) {
+      const threeSum = sortedNums[i] + sortedNums[left] + sortedNums[right];
+
+      if (threeSum > 0) {
+        right -= 1;
+      } else if (threeSum < 0) {
+        left += 1;
+      } else {
+        result.push([sortedNums[i], sortedNums[left], sortedNums[right]]);
+
+        // 중복 제거
+        while (left < right && sortedNums[left] === sortedNums[left + 1]) {
+          left += 1;
+        }
+        while (left < right && sortedNums[right] === sortedNums[right - 1]) {
+          right -= 1;
+        }
+
+        left += 1;
+        right -= 1;
+      }
+    }
+  }
+
+  return result;
+};
diff --git a/climbing-stairs/reach0908.js b/climbing-stairs/reach0908.js
new file mode 100644
index 000000000..aa49b68c3
--- /dev/null
+++ b/climbing-stairs/reach0908.js
@@ -0,0 +1,18 @@
+/**
+ * @description
+ * time complexity: O(n)
+ * space complexity: O(n)
+ * runtime: 0ms
+ * 풀이 방법: 기본적인 DP 풀이 방법
+ * @param {number} n
+ * @return {number}
+ */
+const climbStairs = function (n) {
+  const dp = [1, 2];
+
+  for (let i = 2; i < n; i += 1) {
+    dp[i] = dp[i - 1] + dp[i - 2];
+  }
+
+  return dp[n - 1];
+};
diff --git a/product-of-array-except-self/reach0908.js b/product-of-array-except-self/reach0908.js
new file mode 100644
index 000000000..498cf2e57
--- /dev/null
+++ b/product-of-array-except-self/reach0908.js
@@ -0,0 +1,28 @@
+/**
+ * @description
+ * time complexity: O(n)
+ * space complexity: O(n)
+ * runtime: 5ms
+ * 풀이 방법 : 40분고민하다가 다른사람 풀이보고 누적합을 이용하면 된다해서 구현함
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var productExceptSelf = function (nums) {
+  const result = new Array(nums.length).fill(1);
+
+  let prefix = 1;
+
+  for (let i = 0; i < nums.length; i += 1) {
+    result[i] = prefix;
+    prefix *= nums[i];
+  }
+
+  let postfix = 1;
+
+  for (let i = nums.length - 1; i >= 0; i -= 1) {
+    result[i] *= postfix;
+    postfix *= nums[i];
+  }
+
+  return result;
+};
diff --git a/valid-anagram/reach0908.js b/valid-anagram/reach0908.js
new file mode 100644
index 000000000..7d325ffc7
--- /dev/null
+++ b/valid-anagram/reach0908.js
@@ -0,0 +1,40 @@
+/**
+ * @description
+ * time complexity: O(nlogn) split 시 새로운 배열을 생성하고 sort 시 nlogn 시간 소요
+ * space complexity: O(n) split 시 새로운 배열을 생성함
+ * runtime: 32ms
+ * 풀이 방법: 두 문자열을 정렬하여 비교하는 방법
+ * @param {string} s
+ * @param {string} t
+ * @return {boolean}
+ */
+const isAnagram = function (s, t) {
+  return s.split("").sort().join("") === t.split("").sort().join("");
+};
+
+/**
+ * @description
+ * time complexity: O(n)
+ * space complexity: O(n)
+ * runtime: 15ms
+ * 풀이 방법: 해쉬맵을 통해 카운트를 추가하거나 제거하는 방식, 유니코드도 대응가능
+ * @param {string} s
+ * @param {string} t
+ * @return {boolean}
+ */
+const isAnagramSolution2 = function (s, t) {
+  if (s.length !== t.length) return false;
+
+  const map = new Map();
+
+  for (let i = 0; i < s.length; i += 1) {
+    map.set(s[i], (map.get(s[i]) || 0) + 1);
+    map.set(t[i], (map.get(t[i]) || 0) - 1);
+  }
+
+  for (const value of map.values()) {
+    if (value !== 0) return false;
+  }
+
+  return true;
+};
diff --git a/validate-binary-search-tree/reach0908.js b/validate-binary-search-tree/reach0908.js
new file mode 100644
index 000000000..811abc058
--- /dev/null
+++ b/validate-binary-search-tree/reach0908.js
@@ -0,0 +1,26 @@
+/**
+ * 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 {boolean}
+ */
+var isValidBST = function (root, low = -Infinity, high = Infinity) {
+  if (!root) {
+    return true;
+  }
+
+  if (root.val <= low || root.val >= high) {
+    return false;
+  }
+
+  return (
+    isValidBST(root.left, low, root.val) &&
+    isValidBST(root.right, root.val, high)
+  );
+};