diff --git a/course-schedule/Geegong.java b/course-schedule/Geegong.java
new file mode 100644
index 0000000000..a5d0373f9d
--- /dev/null
+++ b/course-schedule/Geegong.java
@@ -0,0 +1,64 @@
+import java.util.*;
+
+public class Geegong {
+
+    /**
+     * in-degree array, graph, queue 이용
+     * directed acyclic graph (DAG) 문제..
+     * in-degree array : 타 노드에서 유입이 되는 갯수
+     * graph : [a,b] 이렇게 있으면 b -> a 로 가는 방향 edge 를 표현
+     * queue : in-degree array 에서 0 인 노드들. 즉 바로 수강할 수 있는 녀석들을 차례차례 넣어 카운팅할 예정
+     * (예전에도 동일하게 풀어서 .. 한번 더 복습하는 느낌으로 풀었습니다)
+     * @param numCourses
+     * @param prerequisites
+     * @return
+     */
+    public boolean canFinish(int numCourses, int[][] prerequisites) {
+
+        int[] indegreeCnt = new int[numCourses];
+        List<List<Integer>> outdegree = new ArrayList<>();
+        // set outdegree
+        for (int idx=0; idx<numCourses; idx++) {
+            outdegree.add(new ArrayList<>());
+        }
+
+        // set indgree, outdegree, candidateNoIndgree
+        for (int[] pre : prerequisites) {
+            int course = pre[0];
+            int prerequisite = pre[1];
+
+            if (course == prerequisite) {
+                return false;
+            }
+
+            outdegree.get(prerequisite).add(course);
+            indegreeCnt[course]++;
+        }
+
+        // find indgree zero
+        Queue<Integer> queue = new ArrayDeque<>();
+        for (int idx=0; idx<numCourses; idx++) {
+            if (indegreeCnt[idx] == 0) {
+                queue.add(idx);
+            }
+        }
+
+
+        int currentTakenCnt = 0;
+        while(!queue.isEmpty()) {
+            int indegreeZero = queue.poll();
+            currentTakenCnt++;
+
+            List<Integer> targets = outdegree.get(indegreeZero);
+            for (int course : targets) {
+                indegreeCnt[course]--;
+                if (indegreeCnt[course] == 0) {
+                    queue.add(course);
+                }
+            }
+        }
+
+        return currentTakenCnt == numCourses;
+    }
+
+}
diff --git a/design-add-and-search-words-data-structure/Geegong.java b/design-add-and-search-words-data-structure/Geegong.java
index 08dfc87e27..c3774e3c00 100644
--- a/design-add-and-search-words-data-structure/Geegong.java
+++ b/design-add-and-search-words-data-structure/Geegong.java
@@ -1,94 +1,184 @@
-import java.util.ArrayList;
-import java.util.HashMap;
-import java.util.List;
-import java.util.Map;
+import java.util.*;
 
 public class Geegong {
 
     /**
-     * case 1. Letter 라는 클래스 안에 각 char 를 가지고 있고 하나씩 순회하면서 search 를 하니 응답속도가 너무 느림..
+     * 이전 기수와는 다르게 배열을 활용해서 문제 풀이
+     * time complexity : O(26^N) (worst case)
+     * space complexity : O(26 * N) => O(N)
      */
-    class WordDictionary {
-        public Map<Integer, List<Letter>> wordsLengthMap;
+    public static class Node {
+        private Node[] next = new Node[26];
+        private boolean end;
+
+        public void setNext(char ch) {
+            int idx = ch - 'a';
+            if (next[idx] == null) {
+                next[idx] = new Node();
+            }
+        }
 
-        public WordDictionary() {
-            this.wordsLengthMap = new HashMap<>();
+        public Node getSpecificNext(char ch) {
+            int idx = ch - 'a';
+            return this.next[idx];
         }
 
-        public void addWord(String word) {
-            char[] wordChars = word.toCharArray();
+        public void setEnd(boolean isEnd) {
+            this.end = isEnd;
+        }
+    }
 
-            Letter prev = new Letter(null);
-            Letter root = new Letter(prev);
-            for (char wordChar : wordChars) {
-                Letter curr = new Letter(wordChar, false);
 
-                prev.nextLetter = curr;
 
-                prev = curr;
-            }
 
-            prev.isEnd = true;
 
-            wordsLengthMap.computeIfAbsent(wordChars.length, ArrayList::new).add(root.nextLetter);
+    /**
+     * case 1. Letter 라는 클래스 안에 각 char 를 가지고 있고 하나씩 순회하면서 search 를 하니 응답속도가 너무 느림..
+     */
+    public class WordDictionary {
+        private Node entry;
+
+        public WordDictionary(Node entry) {
+            this.entry = entry;
         }
 
-        public boolean search(String word) {
-            char[] wordChars = word.toCharArray();
-            int length = wordChars.length;
+        public WordDictionary() { }
 
-            if (!wordsLengthMap.containsKey(length)) {
-                return false;
-            }
+        public void addWord(String word) {
+            char[] chArray = word.toCharArray();
 
-            List<Letter> letters = wordsLengthMap.get(length);
-            for (Letter letter : letters) {
-                if (letter.isMatched(wordChars)) {
-                    return true;
-                }
+            if (entry == null) {
+                entry = new Node();
             }
-
-            return false;
+            Node curr = entry;
+            for (char ch : chArray) {
+                curr.setNext(ch);
+                curr = curr.getSpecificNext(ch);
+            }
+            curr.setEnd(true);
         }
-    }
 
-    public static class Letter {
-        public char currentLetterOfAsciiCode; // '?'-'a'
-        public Letter nextLetter;
-        public boolean isEnd; // 마지막 글자 인지 여부
+        public boolean search(String word) {
 
-        public Letter(Letter next) {
-            this.nextLetter = next;
-        }
+            char[] chArray = word.toCharArray();
 
-        public Letter(char wordChar, boolean isEnd) {
-            this.currentLetterOfAsciiCode = wordChar;
-            this.isEnd = isEnd;
-        }
+            Node curr = entry;
+            for (int idx = 0; idx<chArray.length; idx++) {
+                char ch = chArray[idx];
 
-        public boolean isMatched(char[] wordChars) {
+                if (ch == '.') {
+                    // dfs, back tracking,
+                    // . 이 있는 경우에는 그 다음 문자열부터 찾음
+                    String temp = word.substring(idx + 1, chArray.length);
 
-            Letter current = this.nextLetter;
-            // 본인 시점으로 부터 wordChars 마지막까지 체크
-            for (char wordChar : wordChars) {
-
-                // . 이면 그냥 패스
-                boolean isMatched = wordChar == '.' || wordChar == current.currentLetterOfAsciiCode;
-                if (isMatched) {
-                    if (current.isEnd) {
-                        return true;
-                    } else {
-                        current = current.nextLetter;
+                    for (Node next : curr.next) {
+                        WordDictionary newWordDict = new WordDictionary(next);
+                        if (newWordDict.search(temp)) {
+                            return true;
+                        }
                     }
+                    return false;
+                }
 
-                } else {
+                if (curr == null || curr.getSpecificNext(ch) == null) {
                     return false;
                 }
-            }
 
-            return false;
+                curr = curr.getSpecificNext(ch);
+            }
 
+            return curr.end;
         }
-    }
+
+
+
+
+
+
+
+
+
+// 이전 기수에서 풀었던 방법, 전에는 List, Map을 이용하여 코드 풀이 하였음
+//        public Map<Integer, List<Letter>> wordsLengthMap;
+//
+//        public WordDictionary() {
+//            this.wordsLengthMap = new HashMap<>();
+//        }
+//
+//        public void addWord(String word) {
+//            char[] wordChars = word.toCharArray();
+//
+//            Letter prev = new Letter(null);
+//            Letter root = new Letter(prev);
+//            for (char wordChar : wordChars) {
+//                Letter curr = new Letter(wordChar, false);
+//
+//                prev.nextLetter = curr;
+//
+//                prev = curr;
+//            }
+//
+//            prev.isEnd = true;
+//
+//            wordsLengthMap.computeIfAbsent(wordChars.length, ArrayList::new).add(root.nextLetter);
+//        }
+//
+//        public boolean search(String word) {
+//            char[] wordChars = word.toCharArray();
+//            int length = wordChars.length;
+//
+//            if (!wordsLengthMap.containsKey(length)) {
+//                return false;
+//            }
+//
+//            List<Letter> letters = wordsLengthMap.get(length);
+//            for (Letter letter : letters) {
+//                if (letter.isMatched(wordChars)) {
+//                    return true;
+//                }
+//            }
+//
+//            return false;
+//        }
+//    }
+//
+//    public static class Letter {
+//        public char currentLetterOfAsciiCode; // '?'-'a'
+//        public Letter nextLetter;
+//        public boolean isEnd; // 마지막 글자 인지 여부
+//
+//        public Letter(Letter next) {
+//            this.nextLetter = next;
+//        }
+//
+//        public Letter(char wordChar, boolean isEnd) {
+//            this.currentLetterOfAsciiCode = wordChar;
+//            this.isEnd = isEnd;
+//        }
+//
+//        public boolean isMatched(char[] wordChars) {
+//
+//            Letter current = this.nextLetter;
+//            // 본인 시점으로 부터 wordChars 마지막까지 체크
+//            for (char wordChar : wordChars) {
+//
+//                // . 이면 그냥 패스
+//                boolean isMatched = wordChar == '.' || wordChar == current.currentLetterOfAsciiCode;
+//                if (isMatched) {
+//                    if (current.isEnd) {
+//                        return true;
+//                    } else {
+//                        current = current.nextLetter;
+//                    }
+//
+//                } else {
+//                    return false;
+//                }
+//            }
+//
+//            return false;
+//
+//        }
+//    }
 }
 
diff --git a/maximum-product-subarray/Geegong.java b/maximum-product-subarray/Geegong.java
new file mode 100644
index 0000000000..69b83d6ca0
--- /dev/null
+++ b/maximum-product-subarray/Geegong.java
@@ -0,0 +1,29 @@
+public class Geegong {
+
+    /**
+     * kadane's algorithm 으로 풀이
+     * 위 알고리즘은 contiguous array 에 대한 문제를 풀이할 때 자주 쓰임
+     * time complexity : O(N)
+     * space complexity : O(1)
+     * @param nums
+     * @return
+     */
+    public int maxProduct(int[] nums) {
+        int currMaxProd = nums[0];
+        int currMinProd = nums[0];
+        int max = nums[0];
+
+        for (int idx=1; idx<nums.length; idx++) {
+            int num = nums[idx];
+
+            // currMaxProd 가 변경되고 난 후에 currMinProd 를 구하기 때문에 currMinprod 를 구하기 위해 임시 저장
+            int tempMaxProd = currMaxProd;
+            currMaxProd = Math.max(num, Math.max(currMaxProd * num, currMinProd * num));
+            currMinProd = Math.min(num, Math.min(currMinProd * num, tempMaxProd * num));
+
+            max = Math.max(currMaxProd, max);
+        }
+
+        return max;
+    }
+}
diff --git a/merge-intervals/Geegong.java b/merge-intervals/Geegong.java
new file mode 100644
index 0000000000..0ff9669211
--- /dev/null
+++ b/merge-intervals/Geegong.java
@@ -0,0 +1,36 @@
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.Comparator;
+import java.util.List;
+
+public class Geegong {
+
+    /**
+     * inerval 들을 오름차순으로 sorting 한후 각각의 start 와 end 를 비교하면서 머지할 수 경우를 고려하여 머지한다
+     * time complexity : O(N) + O(N Long N) => O(N log N)
+     * space complexity : O(N)
+     * @param intervals
+     * @return
+     */
+    public int[][] merge(int[][] intervals) {
+        // 1. sort intervals
+        Arrays.sort(intervals, Comparator.comparingInt(o -> o[0]));
+        List<int[]> result = new ArrayList<>();
+
+        result.add(intervals[0]);
+        for (int i = 1; i < intervals.length; i++) {
+            int[] last = result.get(result.size() - 1);
+            int[] cur = intervals[i];
+
+            if (last[1] >= cur[0]) {
+                // merge (update last end)
+                last[1] = Math.max(last[1], cur[1]);
+            } else {
+                // no overlap
+                result.add(cur);
+            }
+        }
+
+        return result.toArray(t -> new int[result.size()][]);
+    }
+}