diff --git a/combination-sum/renovizee.java b/combination-sum/renovizee.java
new file mode 100644
index 000000000..bd61a5e24
--- /dev/null
+++ b/combination-sum/renovizee.java
@@ -0,0 +1,18 @@
+
+
+// tag renovizee 3week
+// https://github.com/DaleStudy/leetcode-study/issues/254
+// https://leetcode.com/problems/valid-palindrome/ #39 #Medium
+class Solution {
+    // Solv1 :
+    // 시간복잡도 : O(n)
+    // 공간복잡도 : O(n)
+    public List<List<Integer>> combinationSum(int[] candidates, int target) {
+
+    }
+}
+
+//-------------------------------------------------------------------------------------------------------------
+// Java 문법 피드백
+//
+//-------------------------------------------------------------------------------------------------------------
diff --git a/decode-ways/renovizee.java b/decode-ways/renovizee.java
new file mode 100644
index 000000000..abe02623e
--- /dev/null
+++ b/decode-ways/renovizee.java
@@ -0,0 +1,18 @@
+
+
+// tag renovizee 3week
+// https://github.com/DaleStudy/leetcode-study/issues/268
+// https://leetcode.com/problems/valid-palindrome/ #91 #Medium
+class Solution {
+    // Solv1 :
+    // 시간복잡도 : O(n)
+    // 공간복잡도 : O(n)
+    public int numDecodings(String s) {
+
+    }
+}
+
+//-------------------------------------------------------------------------------------------------------------
+// Java 문법 피드백
+//
+//-------------------------------------------------------------------------------------------------------------
diff --git a/maximum-subarray/renovizee.java b/maximum-subarray/renovizee.java
new file mode 100644
index 000000000..ab522e3f2
--- /dev/null
+++ b/maximum-subarray/renovizee.java
@@ -0,0 +1,18 @@
+
+
+// tag renovizee 3week
+// https://github.com/DaleStudy/leetcode-study/issues/275
+// https://leetcode.com/problems/valid-palindrome/ #53 #Medium
+class Solution {
+    // Solv1 :
+    // 시간복잡도 : O(n)
+    // 공간복잡도 : O(n)
+    public int maxSubArray(int[] nums) {
+
+    }
+}
+
+//-------------------------------------------------------------------------------------------------------------
+// Java 문법 피드백
+//
+//-------------------------------------------------------------------------------------------------------------
diff --git a/number-of-1-bits/renovizee.java b/number-of-1-bits/renovizee.java
new file mode 100644
index 000000000..22de6f39c
--- /dev/null
+++ b/number-of-1-bits/renovizee.java
@@ -0,0 +1,45 @@
+
+
+// tag renovizee 3week
+// https://github.com/DaleStudy/leetcode-study/issues/232
+// https://leetcode.com/problems/number-of-1-bits #191 #Easy
+class Solution {
+    // Solv2 :
+    // 시간복잡도 : O(1)
+    // 공간복잡도 : O(1)
+    public int hammingWeight(int n) {
+        int result = 0;
+        for (int i = 0; i < 32; i++) {
+            if (((n >> i) & 1) == 1) {
+                result++;
+            }
+        }
+        return result;
+
+    }
+//    // Solv1 :
+//    // 시간복잡도 : O(log n)
+//    // 공간복잡도 : O(1)
+//    public int hammingWeight(int n) {
+//        int result = 0;
+//        int current = n;
+//        while (current >= 2) {
+//            if ((current % 2) == 1) {
+//                result++;
+//            }
+//            current = current / 2;
+//        }
+//        if (current == 1) {
+//            result++;
+//        }
+//        return result;
+//
+//    }
+}
+
+//-------------------------------------------------------------------------------------------------------------
+// Java 문법 피드백
+// 1) String s=Integer.toBinaryString(n); 숫자를 이진수 string으로 변환하는 방법
+// 2) 숫자를 비트 연산하는 방법 n >> i 는 정수 n을 i 만큼 오른쪽으로 shift 함, ex) 1011 -> 0101
+// 3) & 은 비트에서 and 연산이고 & 1은 마지막 비트 검사로 특수하게 사용됨, 둘다 1인 경우만 1
+//-------------------------------------------------------------------------------------------------------------
diff --git a/valid-palindrome/renovizee.java b/valid-palindrome/renovizee.java
new file mode 100644
index 000000000..d0ffc402b
--- /dev/null
+++ b/valid-palindrome/renovizee.java
@@ -0,0 +1,36 @@
+
+
+// tag renovizee 3week
+// https://github.com/DaleStudy/leetcode-study/issues/220
+// https://leetcode.com/problems/valid-palindrome/ #125 #Easy
+class Solution {
+    // Solv1 :
+    // 시간복잡도 : O(n)
+    // 공간복잡도 : O(n)
+    public boolean isPalindrome(String s) {
+//        replaceAll(...): 문자열 전체를 한 번 순회 → O(n)
+//        trim(): 공백을 양쪽 끝에서만 탐색 → O(n) 이라고 보지만 보통 무시 가능한 수준
+//        toLowerCase(): 모든 문자를 소문자로 바꿈 → O(n)
+        String cleanString = s.replaceAll("[^a-zA-Z0-9]", "").trim().toLowerCase();
+
+        int left = 0;
+        int right = cleanString.length() - 1;
+
+        //O(n)
+        while (left < right) {
+            if (cleanString.charAt(left) != cleanString.charAt(right)) {
+                return false;
+            }
+            left++;
+            right--;
+        }
+        return true;
+
+    }
+}
+
+//-------------------------------------------------------------------------------------------------------------
+// Java 문법 피드백
+// 1) char[] 대문자 Char 가 아니고 소줌ㄴ자
+// 2) ~.equals는 char에서 제공되지 않음
+//-------------------------------------------------------------------------------------------------------------