diff --git a/combination-sum/juhui-jeong.ts b/combination-sum/juhui-jeong.ts
new file mode 100644
index 0000000000..57a5e7ae46
--- /dev/null
+++ b/combination-sum/juhui-jeong.ts
@@ -0,0 +1,28 @@
+// 시간 복잡도: O(2^(target / minCandidate))
+// 공간 복잡도: O(target / minCandidate)
+function combinationSum(candidates: number[], target: number): number[][] {
+  const result: number[][] = [];
+
+  const dfs = (index: number, remain: number, path: number[]) => {
+    if (remain === 0) {
+      result.push([...path]);
+      return;
+    }
+
+    if (remain < 0 || index === candidates.length) {
+      return;
+    }
+
+    const num = candidates[index];
+
+    path.push(num);
+    dfs(index, remain - num, path);
+    path.pop();
+
+    dfs(index + 1, remain, path);
+  };
+
+  dfs(0, target, []);
+
+  return result;
+}
diff --git a/number-of-1-bits/juhui-jeong.ts b/number-of-1-bits/juhui-jeong.ts
new file mode 100644
index 0000000000..4a6744aaee
--- /dev/null
+++ b/number-of-1-bits/juhui-jeong.ts
@@ -0,0 +1,27 @@
+//풀이 1
+// 시간 복잡도: O(k)
+// 공간 복잡도: O(1)
+// 속도: 2ms
+function hammingWeight(n: number): number {
+  let count = 0;
+  while (n) {
+    n = n & (n - 1);
+    count++;
+  }
+  return count;
+}
+/*
+속도는 빠르지만 복잡도가 높기 때문에 적합하지 않음.
+Brian Kernighan 알고리즘을 활용한 비트 계산으로 재풀이
+
+// 시간 복잡도: O(log n)
+// 공간 복잡도: O(log n)
+// 속도: 0ms
+function hammingWeight(n: number): number {
+  const bitNumber = n.toString(2);
+  const bitString = String(bitNumber);
+
+  const bitArray = bitString.split('').map((s) => Number(s));
+  return bitArray.reduce((a, b) => a + b);
+}
+*/
diff --git a/valid-palindrome/juhui-jeong.ts b/valid-palindrome/juhui-jeong.ts
new file mode 100644
index 0000000000..a59e912e40
--- /dev/null
+++ b/valid-palindrome/juhui-jeong.ts
@@ -0,0 +1,24 @@
+// 시간 복잡도: O(n)
+// 공간 복잡도: O(n)
+function isPalindrome(s: string): boolean {
+  //1. s에서 문자와 숫자가 아닌 다른 것들은 제거
+  //2. 대문자 -> 소문자, 띄어쓰기, 공백 제거
+  let originalString = s;
+  let alphabeticString = originalString.replace(/[^a-zA-Z0-9]/g, '');
+  const filteredString = alphabeticString.toLowerCase().trim();
+
+  //3. 제거 후 length === 0 일 경우 true로 결과 값 반환.
+  if (filteredString.length === 0) {
+    return true;
+  }
+
+  let count = 0;
+  let roundedStringLength = Math.round(filteredString.length / 2);
+  for (let i = 0; i < roundedStringLength; i++) {
+    // 양 끝의 글자가 동일한지 체크하고
+    if (filteredString[i] === filteredString[filteredString.length - i - 1]) {
+      count += 1;
+    }
+  }
+  return count === roundedStringLength ? true : false;
+}