diff --git a/combination-sum/byol-han.js b/combination-sum/byol-han.js
new file mode 100644
index 000000000..bf80724b1
--- /dev/null
+++ b/combination-sum/byol-han.js
@@ -0,0 +1,25 @@
+/**
+ * @param {number[]} candidates
+ * @param {number} target
+ * @return {number[][]}
+ */
+var combinationSum = function (candidates, target) {
+  const result = [];
+
+  function backtrack(remaining, combination, start) {
+    if (remaining === 0) {
+      result.push([...combination]);
+      return;
+    }
+    if (remaining < 0) return;
+
+    for (let i = start; i < candidates.length; i++) {
+      combination.push(candidates[i]);
+      backtrack(remaining - candidates[i], combination, i); // 같은 숫자 다시 사용 가능
+      combination.pop(); // backtrack
+    }
+  }
+
+  backtrack(target, [], 0);
+  return result;
+};
diff --git a/decode-ways/byol-han.js b/decode-ways/byol-han.js
new file mode 100644
index 000000000..f80602bf4
--- /dev/null
+++ b/decode-ways/byol-han.js
@@ -0,0 +1,27 @@
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var numDecodings = function (s) {
+  if (!s || s[0] === "0") return 0;
+
+  const n = s.length;
+  const dp = Array(n + 1).fill(0);
+
+  dp[0] = 1; // 빈 문자열은 1가지 방법
+  dp[1] = 1; // 첫 글자가 0이 아니면 1가지 방법
+
+  for (let i = 2; i <= n; i++) {
+    const oneDigit = parseInt(s.slice(i - 1, i));
+    const twoDigits = parseInt(s.slice(i - 2, i));
+
+    if (oneDigit >= 1 && oneDigit <= 9) {
+      dp[i] += dp[i - 1];
+    }
+    if (twoDigits >= 10 && twoDigits <= 26) {
+      dp[i] += dp[i - 2];
+    }
+  }
+
+  return dp[n];
+};
diff --git a/maximum-subarray/byol-han.js b/maximum-subarray/byol-han.js
new file mode 100644
index 000000000..cb5a5ec33
--- /dev/null
+++ b/maximum-subarray/byol-han.js
@@ -0,0 +1,19 @@
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var maxSubArray = function (nums) {
+  // 초기값 설정: 현재까지의 최대합과 전체 최대합을 배열의 첫 번째 값으로 초기화
+  let currentSum = nums[0];
+  let maxSum = nums[0];
+
+  // 두 번째 원소부터 순회
+  for (let i = 1; i < nums.length; i++) {
+    // 이전까지의 합에 현재 원소를 더할지, 아니면 현재 원소부터 새로 시작할지 결정
+    currentSum = Math.max(nums[i], currentSum + nums[i]);
+    // 전체 최대값 갱신
+    maxSum = Math.max(maxSum, currentSum);
+  }
+
+  return maxSum;
+};
diff --git a/number-of-1-bits/byol-han.js b/number-of-1-bits/byol-han.js
new file mode 100644
index 000000000..c221e7cd5
--- /dev/null
+++ b/number-of-1-bits/byol-han.js
@@ -0,0 +1,30 @@
+/**
+ * @param {number} n
+ * @return {number}
+ */
+//1. divide the number by 2 and count the remainder
+var hammingWeight = function (n) {
+  let count = 0;
+  while (n > 0) {
+    if (n % 2 === 1) {
+      count++;
+    }
+    n = Math.floor(n / 2);
+  }
+  return count;
+};
+
+//2. Count the number of set bits (1s) in the binary representation of n
+var hammingWeight = function (n) {
+  return n.toString(2).split("1").length - 1;
+};
+
+//3. bit manipulation
+var hammingWeight = function (n) {
+  let count = 0;
+  while (n > 0) {
+    count += n & 1; // 마지막 비트가 1이면 count++
+    n = n >>> 1; // 오른쪽으로 한 비트 이동 (2로 나눔)
+  }
+  return count;
+};
diff --git a/valid-palindrome/byol-han.js b/valid-palindrome/byol-han.js
new file mode 100644
index 000000000..7e26307c1
--- /dev/null
+++ b/valid-palindrome/byol-han.js
@@ -0,0 +1,32 @@
+/**
+ * @param {string} s
+ * @return {boolean}
+ */
+
+// 1. Two Pointers
+// time complexity: O(n)
+// space complexity: O(1)
+var isPalindrome = function (s) {
+  let sRefine = s.toLowerCase().replace(/[^a-z0-9]/g, "");
+  let left = 0;
+  let right = sRefine.length - 1;
+
+  while (left < right) {
+    if (sRefine[left] !== sRefine[right]) {
+      return false;
+    }
+    left++;
+    right--;
+  }
+
+  return true;
+};
+
+// 2. String Manipulation
+// time complexity: O(n)
+// space complexity: O(n)
+var isPalindrome = function (s) {
+  let refined = s.toLowerCase().replace(/[^a-z0-9]/g, "");
+  let reversed = refined.split("").reverse().join("");
+  return refined === reversed;
+};