diff --git a/combination-sum/ekgns33.java b/combination-sum/ekgns33.java
new file mode 100644
index 000000000..1e037722b
--- /dev/null
+++ b/combination-sum/ekgns33.java
@@ -0,0 +1,42 @@
+import java.util.ArrayList;
+import java.util.List;
+
+/**
+ input : array of distinct integers, single integer target
+ output : all unique combinations of candidates where chosen ones sum is target
+ constraints:
+ 1) can we use same integer multiple times?
+ yes
+ 2) input array can be empty?
+ no. [1, 30]
+
+
+ solution 1)
+ combination >> back-tracking
+ O(2^n) at most 2^30 << (2^10)^3 ~= 10^9
+
+ tc : O(2^n) sc : O(n) call stack
+ */
+class Solution {
+  public List<List<Integer>> combinationSum(int[] candidates, int target) {
+    List<List<Integer>> answer = new ArrayList<>();
+    List<Integer> prev = new ArrayList<>();
+    backTrackingHelper(answer, prev, candidates, target, 0, 0);
+    return answer;
+  }
+  private void backTrackingHelper(List<List<Integer>> ans, List<Integer> prev, int[] cands, int target, int curSum, int p) {
+    if(curSum == target) {
+      ans.add(new ArrayList(prev));
+      return;
+    }
+    if(p >= cands.length) return;
+
+    for(int i = p; i< cands.length; i++) {
+      if((curSum + cands[i]) <= target) {
+        prev.add(cands[i]);
+        backTrackingHelper(ans, prev, cands, target, curSum + cands[i],i);
+        prev.remove(prev.size() - 1);
+      }
+    }
+  }
+}
diff --git a/maximum-subarray/ekgns33.java b/maximum-subarray/ekgns33.java
new file mode 100644
index 000000000..f87957c53
--- /dev/null
+++ b/maximum-subarray/ekgns33.java
@@ -0,0 +1,41 @@
+/*
+input : array of integer
+output : largest sum of subarray
+constraints : 
+1) is the input array not empty?
+yes. at least one el
+2) range of integers
+[-10^4, 10^4]
+3) maximum lenght of input
+[10^5]
+>> maximum sum = 10^5 * 10^4 = 10 ^ 9 < INTEGER
+
+sol1) brute force
+nested for loop : O(n^2)
+tc : O(n^2), sc : O(1)
+
+sol2) dp?
+Subarray elements are continuous in the original array, so we can use dp.
+let dp[i] represent the largest sum of a subarray where the ith element is the last element of the subarray.
+
+if dp[i-1] + curval  < cur val : take curval
+if dp[i-1] + cur val >= curval :  take dp[i-1] + curval
+tc : O(n) sc : O(n)
+ */
+class Solution {
+  public int maxSubArray(int[] nums) {
+    int n = nums.length;
+    int[] dp = new int[n];
+    int maxSum = nums[0];
+    dp[0] = nums[0];
+    for(int i = 1; i < n; i++) {
+      if(dp[i-1] + nums[i] < nums[i]) {
+        dp[i] = nums[i];
+      } else {
+        dp[i] = nums[i] + dp[i-1];
+      }
+      maxSum = Math.max(maxSum, dp[i]);
+    }
+    return maxSum;
+  }
+}
diff --git a/product-of-array-except-self/ekgns33.java b/product-of-array-except-self/ekgns33.java
new file mode 100644
index 000000000..14e313f2a
--- /dev/null
+++ b/product-of-array-except-self/ekgns33.java
@@ -0,0 +1,72 @@
+/*
+input : array of integer
+output : array of integer that each element
+     is product of all the elements except itself
+constraint
+1) is the result of product also in range of integer?
+yes product of nums is guaranteed to fit 32-bit int
+2) how about zero?
+doesn't matter if the prduct result is zero
+3) is input array non-empty?
+yes. length of array is in range [2, 10^5]
+
+solution1) brute force
+
+calc all the product except current index element
+
+tc : O(n^2) sc : O(n) << for the result. when n is the length of input array
+
+solution 2) better?
+ds : array
+algo : hmmm we can reuse the product using prefix sum
+
+1) get prefix sum from left to right and vice versa : 2-O(n) loop + 2-O(n) space
+2) for i = 0 to n-1 when n is the length of input
+    get product of leftPrfex[i-1] * rightPrefix[i+1]
+       // edge : i == 0, i == n-1
+
+tc : O(n) sc : O(n)
+
+solution 3) optimal?
+can we reduce space?
+1) product of all elements. divide by current element.
+
+    > edge : what if current element is zero?
+    2) if there exists only one zero:
+        all the elements except zero index will be zero
+    3) if there exist multiple zeros:
+        all the elements are zero
+    4) if there is no zero
+        do 1)
+ */
+class Solution {
+  public int[] productExceptSelf(int[] nums) {
+    int n = nums.length, product = 1, zeroCount = 0;
+    for(int num : nums) {
+      if(num == 0) {
+        zeroCount ++;
+        if(zeroCount > 1) break;
+      } else {
+        product *= num;
+      }
+    }
+
+    int[] answer = new int[n];
+    if(zeroCount > 1) {
+      return answer;
+    } else if (zeroCount == 1) {
+      for(int i = 0; i < n; i++) {
+        if(nums[i] != 0) {
+          answer[i] = 0;
+          continue;
+        }
+        answer[i] = product;
+      }
+    } else {
+      for(int i = 0; i < n; i++) {
+        answer[i] = product / nums[i];
+      }
+    }
+    return answer;
+  }
+}
diff --git a/reverse-bits/ekgns33.java b/reverse-bits/ekgns33.java
new file mode 100644
index 000000000..b78c5b73e
--- /dev/null
+++ b/reverse-bits/ekgns33.java
@@ -0,0 +1,39 @@
+/*
+input : 32 bit unsigned integer n
+output : unsigned integer representation of reversed bit
+constraint :
+1) input is always 32 bit unsigned integer
+2) implementation should not be affected by programming language
+
+solution 1)
+
+get unsigned integer bit representation
+
+build string s O(n)
+reverse O(n)
+get integer O(n)
+
+tc : O(n) sc : O(n)
+
+solution 2) one loop
+
+nth bit indicates (1<<n) if reverse (1<< (32 - n))
+add up in one loop
+return 
+
+tc : O(n), sc:  O(1)
+
+ */
+public class Solution {
+  // you need treat n as an unsigned value
+  static final int LENGTH = 32;
+  public int reverseBits(int n) {
+    int answer = 0;
+    for(int i = 0; i < LENGTH; i++) {
+      if(((1<<i) & n) != 0) {
+        answer += (1 << (LENGTH -  i - 1));
+      }
+    }
+    return answer;
+  }
+}
diff --git a/two-sum/ekgns33.java b/two-sum/ekgns33.java
new file mode 100644
index 000000000..097903bb7
--- /dev/null
+++ b/two-sum/ekgns33.java
@@ -0,0 +1,39 @@
+import java.util.HashMap;
+import java.util.Map;
+
+/*
+input : array of integers, single integer target
+output : indices of the two numbers that they add up to target
+constraint:
+1) is integer positive? 
+no [-10^9, 10^9]
+2) is there any duplicates?
+yes. but only one valid answer exists
+3) can i reuse elem?
+no
+
+sol1) brute force
+nested for loop. tc: O(n^2), sc: O(1) when n is the length of input
+
+sol2) better solution with hash map
+iterate through the array
+  check if target - current elem exists
+    if return pair of indices
+    else save current elem and continue
+
+tc : O(n), sc: O(n) when n is the length of input
+
+ */
+class Solution {
+  public int[] twoSum(int[] nums, int target) {
+    Map<Integer, Integer> prev = new HashMap<>();
+    for(int i = 0; i < nums.length; i++) {
+      int key = target - nums[i];
+      if(prev.containsKey(key)) {
+        return new int[] {prev.get(key), i};
+      }
+      prev.put(nums[i], i);
+    }
+    return null;
+  }
+}