diff --git a/product-of-array-except-self/higeuni.js b/product-of-array-except-self/higeuni.js
new file mode 100644
index 000000000..3c8a8417b
--- /dev/null
+++ b/product-of-array-except-self/higeuni.js
@@ -0,0 +1,24 @@
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ * 
+ * Time Complexity: O(n)
+ * Space Complexity: O(n)
+ */
+var productExceptSelf = function(nums) {
+  const answer = []
+  const zeros = nums.filter(n => n === 0).length;
+  if (zeros > 1) return new Array(nums.length).fill(0);
+  
+  const productOfNums = nums.reduce((acc, cur) => cur === 0 ? acc : acc * cur, 1);
+  
+  nums.forEach(num => {
+    if (num === 0) {
+      answer.push(productOfNums);
+    } else {
+      answer.push(zeros ? 0 : productOfNums / num);
+    }
+  });
+  return answer
+};
+
diff --git a/reverse-bits/higeuni.js b/reverse-bits/higeuni.js
new file mode 100644
index 000000000..51cb54c04
--- /dev/null
+++ b/reverse-bits/higeuni.js
@@ -0,0 +1,36 @@
+/**
+ * @param {number} n - a positive integer
+ * @return {number} - a positive integer
+ * 
+ * Time Complexity: O(n)? n이 32이기 때문에 O(1)? 
+ * Space Complexity: O(1)
+ */
+
+// 일반적인 풀이
+var reverseBits = function(n) {
+  const binary = n.toString(2);
+  const reversedBinary = binary.split('').reverse().join('').padEnd(32, '0');
+  const answer = parseInt(reversedBinary, 2);
+
+  return answer;
+  // return parseInt(n.toString(2).split('').reverse().join('').padEnd(32, '0'), 2);
+};
+
+/**
+ * 
+ * Time Complexity: O(1)
+ * Space Complexity: O(1)
+*/
+
+// 비트 연산을 이용한 풀이
+var reverseBits2 = function(n) {
+  let result = 0;
+  for(let i = 0; i < 32; i++) {
+    // result를 왼쪽으로 시프트하고 n의 마지막 비트를 더함
+    result = (result << 1) | (n & 1);
+    // n을 오른쪽으로 시프트
+    n = n >> 1;
+  }
+  return result >>> 0;
+}
+
diff --git a/two-sum/higeuni.js b/two-sum/higeuni.js
new file mode 100644
index 000000000..466f99ddd
--- /dev/null
+++ b/two-sum/higeuni.js
@@ -0,0 +1,22 @@
+/**
+ * @param {number[]} nums
+ * @param {number} target
+ * @return {number[]}
+ * 
+ * Time Complexity: O(n)
+ * Space Complexity: O(n)
+ */
+var twoSum = function(nums, target) {
+  const map = new Map();
+    
+  for (let i = 0; i < nums.length; ++i) {
+    const diff = target - nums[i];
+
+    if (map.has(diff)) {
+      return [map.get(diff), i];
+    }
+
+    map.set(nums[i], i);
+  }
+};
+