diff --git a/counting-bits/hyejjun.js b/counting-bits/hyejjun.js
new file mode 100644
index 000000000..48f97d9c4
--- /dev/null
+++ b/counting-bits/hyejjun.js
@@ -0,0 +1,31 @@
+/**
+ * @param {number} n
+ * @return {number[]}
+ */
+var countBits = function (n) {
+
+  let result = [];
+
+  for (let i = 0; i <= n; i++) {
+
+    let binary = i.toString(2);
+    let sum = 0;
+
+    for (let char of binary) {
+      sum += Number(char);
+    }
+    result.push(sum);
+  }
+
+  return result;
+
+};
+
+
+console.log(countBits(2));
+console.log(countBits(5));
+
+/*
+시간 복잡도: O(n * log n)
+공간 복잡도: O(n)
+*/
diff --git a/decode-ways/hyejjun.js b/decode-ways/hyejjun.js
new file mode 100644
index 000000000..83b4d5f0d
--- /dev/null
+++ b/decode-ways/hyejjun.js
@@ -0,0 +1,42 @@
+/**
+ * @param {string} s
+ * @return {number}
+ */
+
+var numDecodings = function (s) {
+    if (s == null || s.length === 0) {
+        return 0;
+    }
+
+    const n = s.length;
+    const dp = new Array(n + 1).fill(0);
+    dp[0] = 1;
+    dp[1] = s[0] === '0' ? 0 : 1;
+
+    for (let i = 2; i <= n; i++) {
+        const oneDigit = parseInt(s.substring(i - 1, i));
+        const twoDigits = parseInt(s.substring(i - 2, i));
+
+        // Check if single digit decoding is possible
+        if (oneDigit >= 1 && oneDigit <= 9) {
+            dp[i] += dp[i - 1];
+        }
+
+        // Check if two digit decoding is possible
+        if (twoDigits >= 10 && twoDigits <= 26) {
+            dp[i] += dp[i - 2];
+        }
+    }
+
+    return dp[n];
+};
+
+
+console.log(numDecodings("12"));
+console.log(numDecodings("226"));
+console.log(numDecodings("06"));
+
+/*
+시간 복잡도: O(n)
+공간 복잡도: O(n)
+*/
diff --git a/valid-anagram/hyejjun.js b/valid-anagram/hyejjun.js
new file mode 100644
index 000000000..1888ce81e
--- /dev/null
+++ b/valid-anagram/hyejjun.js
@@ -0,0 +1,33 @@
+/**
+ * @param {string} s
+ * @param {string} t
+ * @return {boolean}
+ */
+var isAnagram = function (s, t) {
+  if (s.length !== t.length) return false;
+
+  let countS = {};
+  let countT = {};
+
+  for (let i = 0; i < s.length; i++) {
+    countS[s[i]] = (countS[s[i]] || 0) + 1;
+    countT[t[i]] = (countT[t[i]] || 0) + 1;
+  }
+
+  for (let key in countS) {
+    if (countS[key] !== countT[key]) {
+      return false;
+    }
+  }
+
+  return true;
+
+};
+
+console.log(isAnagram("anagram", "nagaram"));
+console.log(isAnagram("rat", "car"));
+
+/*
+시간 복잡도: O(n)
+공간 복잡도: O(n)
+*/