diff --git a/contains-duplicate/GangBean.java b/contains-duplicate/GangBean.java
new file mode 100644
index 000000000..a66343f82
--- /dev/null
+++ b/contains-duplicate/GangBean.java
@@ -0,0 +1,12 @@
+import java.util.Arrays;
+import java.util.stream.Collectors;
+
+class Solution {
+    public boolean containsDuplicate(int[] nums) {
+        /***
+            compare length of array and length of set.
+            O(n) given that n is length of array nums
+        */
+        return nums.length > Arrays.stream(nums).boxed().collect(Collectors.toSet()).size();
+    }
+}
diff --git a/house-robber/GangBean.java b/house-robber/GangBean.java
new file mode 100644
index 000000000..c5138dd02
--- /dev/null
+++ b/house-robber/GangBean.java
@@ -0,0 +1,28 @@
+class Solution {
+    public int rob(int[] nums) {
+        /**
+            r[0] = a[0]
+            r[1] = max(a[1], r[0])
+            r[2] = max(r[1], a[2] + r[0])
+            r[3] = max(r[2], a[3] + r[1])
+            ...
+            r[k] = max(r[k-1], a[k] + r[k-2]) O(1)
+        */
+        int[] r = new int[nums.length];
+
+        for (int i = 0; i < nums.length; i++) { // O(N)
+            if (i == 0) {
+                r[i] = nums[i];
+                continue;
+            }
+            if (i == 1) {
+                r[i] = Math.max(nums[i], r[i-1]);
+                continue;
+            }
+            r[i] = Math.max(r[i-1], nums[i] + r[i-2]);
+        }
+
+        return r[nums.length - 1];
+    }
+}
+
diff --git a/longest-consecutive-sequence/GangBean.java b/longest-consecutive-sequence/GangBean.java
new file mode 100644
index 000000000..c0b7beb86
--- /dev/null
+++ b/longest-consecutive-sequence/GangBean.java
@@ -0,0 +1,23 @@
+import java.util.*;
+import java.util.stream.Collectors;
+
+class Solution {
+    public int longestConsecutive(int[] nums) {
+        Set<Integer> set = Arrays.stream(nums).boxed().collect(Collectors.toSet());
+
+        int maxLength = 0;
+        for (int num: nums) {
+            // 각 숫자에 대해 최초 값이 가능하면, 즉 num-1이 존재하지 않으면 최대 length 구하기
+            if (set.contains(num - 1)) continue;
+            int length = 1;
+            int start = num;
+            while (set.contains(start + 1)) {
+                length++;
+                start++;
+            }
+            maxLength = Math.max(length, maxLength);
+        }
+
+        return maxLength;
+    }
+}
diff --git a/palindromic-substrings/GangBean.java b/palindromic-substrings/GangBean.java
new file mode 100644
index 000000000..ebb22ac35
--- /dev/null
+++ b/palindromic-substrings/GangBean.java
@@ -0,0 +1,29 @@
+class Solution {
+    public int countSubstrings(String s) {
+        /** 
+            각 문자를 중간으로 갖는 palindrome 여부 체크
+            + 두개의 문자를 중간으로 갖는 palindrome 여부 체크
+        */
+        int count = 0;
+        int length = s.length();
+        for (int i = 0; i < length; i++) { // O(N)
+            int start = i;
+            int end = i;
+            while (0 <= start && end < length && start <= end && s.charAt(start) == s.charAt(end)) { // O(N)
+                count++;
+                start--;
+                end++;
+            }
+
+            start = i;
+            end = i + 1;
+            while (0 <= start && end < length && start <= end && s.charAt(start) == s.charAt(end)) { // O(N)
+                count++;
+                start--;
+                end++;
+            }
+        }
+
+        return count; // O(N^2)
+    }
+}
diff --git a/top-k-frequent-elements/GangBean.java b/top-k-frequent-elements/GangBean.java
new file mode 100644
index 000000000..17a8cecac
--- /dev/null
+++ b/top-k-frequent-elements/GangBean.java
@@ -0,0 +1,47 @@
+import java.util.*;
+import java.util.stream.Collectors;
+
+class Solution {
+    public int[] topKFrequent(int[] nums, int k) {
+        /**
+            k번째로 많이 등장한 숫자를 구하는 문제.
+            구해야 하는 것.
+            1. 등장횟수의 순위
+            2. k번째 순위에 해당하는 수의 값
+
+            1. 등장 횟수를 구하는 방법: 전체를 순회해 숫자별 등장 횟수를 저장 -> O(N)
+            2. 등장 횟수의 순위를 구하는 방법: 등장 횟수에 대해 sort 
+                -> O(MlogM) where N = M(M+1) / 2
+                -> logN = logM + log(M+1) + C
+                -> M = N^(1/2)
+                -> O(MlogM) = O(N^1/2logN) (??? 여기가 맞는지 모르겠네요..)
+            3. 역 인덱스를 구해 매핑되는 수를 구함 -> O(N)
+
+            전체: O(N)
+         */
+        Map<Integer, Integer> numToCount = new HashMap<>();
+        for (int num : nums) { // O(N)
+            numToCount.put(num, numToCount.getOrDefault(num, 0) + 1);
+        }
+
+        Map<Integer, List<Integer>> countToNum = new HashMap<>();
+        for (Map.Entry<Integer, Integer> entry: numToCount.entrySet()) {
+            List<Integer> numList = countToNum.getOrDefault(entry.getValue(), new ArrayList<>());
+            numList.add(entry.getKey());
+            countToNum.put(entry.getValue(), numList);
+        } // O(logN): because sum of all counts is equal to N
+
+        List<Integer> countRank = countToNum.keySet().stream().sorted(Collections.reverseOrder()).collect(Collectors.toList()); 
+        // O(MlogM) where N = M(M+1) / 2
+
+        int[] ret = new int[k];
+        int idx = 0;
+        int rank = 0;
+        while (idx < k) { // O(k) <= O(N)
+            for (int num : countToNum.get(countRank.get(rank++))) {
+                ret[idx++] = num;
+            }
+        }
+        return ret;
+    }
+}
diff --git a/valid-palindrome/GangBean.java b/valid-palindrome/GangBean.java
new file mode 100644
index 000000000..121372857
--- /dev/null
+++ b/valid-palindrome/GangBean.java
@@ -0,0 +1,20 @@
+class Solution {
+    public boolean isPalindrome(String s) {
+        /**
+         * Step 1: Convert the input string to lowercase. -> O(n)
+         * Step 2: Remove all non-alphanumeric characters using regex. -> O(n)
+         * Step 3: Use two pointers to compare characters from both ends of the string. -> O(n)
+         * Total time complexity: O(n), where n is the length of the string.
+         */
+        s = s.toLowerCase();
+        s = s.replaceAll("[^a-z0-9]", "");
+        int left = 0;
+        int right = s.length() - 1;
+        if (right <= 0) return true;
+        while (left < right) {
+            if (s.charAt(left++) != s.charAt(right--)) return false;
+        }
+        return true;
+    }
+}
+