diff --git a/contains-duplicate/ekgns33.java b/contains-duplicate/ekgns33.java
new file mode 100644
index 000000000..2c74dbea7
--- /dev/null
+++ b/contains-duplicate/ekgns33.java
@@ -0,0 +1,49 @@
+import java.util.HashSet;
+import java.util.Set;
+
+/*
+start 7:06~7:14 PASS
+input : integer array
+output : return if input contains duplicate
+constraint : 
+1) empty array?
+    nope. at least one 
+2) size?
+    [1, 10^5]
+3) sorted?
+    nope.
+4) range of elements in the array?
+    [-10^9, 10^9] >> max 2 * 10*9 +1
+
+brute force:
+ds : array. algo : just nested for-loop
+iterate through the array, for index i 0 to n. n indicates the size of input
+    nested loop fo r index j from i+1 to n
+        if nums[j] == nums[i] return true;
+
+return false;
+
+time : O(n^2), space : O(1)
+
+better:
+ds : hashset. algo : one for-loop
+iterate through the array:
+    if set contains current value:
+        return true;
+    else
+        add current value to set
+
+return false;
+time : O(n) space :O(n) 
+
+ */
+class Solution {
+  public boolean containsDuplicate(int[] nums) {
+    Set<Integer> set = new HashSet<>();
+    for(int num : nums) {
+      if(set.contains(num)) return true;
+      set.add(num);
+    }
+    return false;
+  }
+}
diff --git a/valid-palindrome/ekgns33.java b/valid-palindrome/ekgns33.java
new file mode 100644
index 000000000..b65b10001
--- /dev/null
+++ b/valid-palindrome/ekgns33.java
@@ -0,0 +1,75 @@
+/*
+start : 7:18 ~
+input : String s
+output : return if given s is valid palindrome
+constraint:
+1) valid palindrome?
+    convert all uppercase letter to lowercase,
+    remove non-alphanumeric chars,
+    reads same forward and backward.
+    ex) abba, aba, a, aacbcaa
+2) length of the input string
+    [1, 2*10^5]
+3) does input string contains non-alphanumeric chars? such as whitespace
+    yes, but only ASCII chars
+
+-------
+
+brute force:
+read each character of input string.
+create a new string that only contains
+lowercase letters and numbers << O(n), when n is length of string s
+
+create string which one is reversed. < O(n)
+
+compare. < O(n)
+
+time : O(n)-3loops, space : O(n)
+
+-------
+better : two-pointer
+
+same as brute force. build new string.
+
+use two pointer left and right
+
+while left <= right
+    if s[left] != s[right] return false;
+
+return true;
+-------
+optimal :
+
+use two pointer left and right
+while left <= right
+    if s[left] is non-alpha left++
+    elif s[right] is non-alpha right--
+    elif s[left] != s[right] return false
+    else left++ right--
+
+return true
+time : O(n) space : O(1)
+ */
+class Solution {
+  public boolean isPalindrome(String s) {
+    int left = 0;
+    int right = s.length() - 1;
+    char leftC;
+    char rightC;
+    while(left <= right) {
+      leftC = s.charAt(left);
+      rightC = s.charAt(right);
+      if(!Character.isLetterOrDigit(leftC)) {
+        left++;
+      } else if (!Character.isLetterOrDigit(rightC)) {
+        right--;
+      } else if (Character.toLowerCase(leftC) != Character.toLowerCase(rightC)) {
+        return false;
+      } else {
+        left++;
+        right--;
+      }
+    }
+    return true;
+  }
+}