diff --git a/coin-change/forest000014.java b/coin-change/forest000014.java
new file mode 100644
index 000000000..45d4cff91
--- /dev/null
+++ b/coin-change/forest000014.java
@@ -0,0 +1,28 @@
+/* 
+Time Complexity: O(coins.length * amount)
+Space Complexity: O(amount)
+
+1 ~ i-1원까지의 최적해를 알고 있다면, i원의 최적해를 구할 때 coins 배열 iteration으로 구할 수 있음
+*/
+class Solution {
+    public int coinChange(int[] coins, int amount) {
+        int c = coins.length;
+
+        int[] dp = new int[amount + 1];
+        Arrays.fill(dp, 99999);
+        dp[0] = 0;
+
+        for (int i = 1; i <= amount; i++) {
+            for (int j = 0; j < c; j++) {
+                if (i - coins[j] < 0) {
+                    continue;
+                }
+                if (dp[i - coins[j]] >= 0 && dp[i - coins[j]] + 1 < dp[i]) {
+                    dp[i] = dp[i - coins[j]] + 1;
+                }
+            }
+        }
+
+        return dp[amount] == 99999 ? -1 : dp[amount];
+    }
+}
diff --git a/merge-two-sorted-lists/forest000014.java b/merge-two-sorted-lists/forest000014.java
new file mode 100644
index 000000000..4a31f1313
--- /dev/null
+++ b/merge-two-sorted-lists/forest000014.java
@@ -0,0 +1,35 @@
+/**
+
+ (n = list1.length + list2.length)
+ 시간 복잡도: O(n)
+ - 최악의 경우 전체 노드 순회
+ 공간 복잡도: O(n)
+ - 최악의 경우 전체 노드만큼 새 노드 생성
+
+ */
+class Solution {
+    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
+        ListNode head = new ListNode();
+        ListNode curr = head;
+
+        while (list1 != null && list2 != null) {
+            if (list1.val <= list2.val) {
+                curr.next = new ListNode(list1.val);
+                curr = curr.next;
+                list1 = list1.next;
+            } else {
+                curr.next = new ListNode(list2.val);
+                curr = curr.next;
+                list2 = list2.next;
+            }
+        }
+
+        if (list1 == null) {
+            curr.next = list2;
+        } else if (list2 == null) {
+            curr.next = list1;
+        }
+
+        return head.next;
+    }
+}
diff --git a/missing-number/forest000014.java b/missing-number/forest000014.java
new file mode 100644
index 000000000..ddac2e98f
--- /dev/null
+++ b/missing-number/forest000014.java
@@ -0,0 +1,16 @@
+/*
+시간 복잡도: O(n)
+공간 복잡도: O(1)
+
+1 ~ n의 합이 n * (n + 1) / 2 라는 점을 활용
+*/
+class Solution {
+    public int missingNumber(int[] nums) {
+        int n = nums.length;
+        int ans = n * (n + 1) / 2;
+        for (int i = 0; i < n; i++) {
+            ans -= nums[i];
+        }
+        return ans;
+    }
+}
diff --git a/palindromic-substrings/forest000014.java b/palindromic-substrings/forest000014.java
new file mode 100644
index 000000000..42e69387d
--- /dev/null
+++ b/palindromic-substrings/forest000014.java
@@ -0,0 +1,38 @@
+/*
+time complexity: O(n^2)
+space complexity: O(1)
+
+모든 가능한 조합(O(n^2))에 대해 palindrome 여부를 검사(O(n))하는 brute force는 O(n^3).
+i번째 문자에서 시작하여, 앞/뒤로 한 글자씩 늘려가면서 탐색하되, 한번이라도 앞/뒤 글자가 서로 다르다면 그 이후는 탐색하지 않을 수 있음. 이 경우는 O(n^2).
+*/
+class Solution {
+    public int countSubstrings(String s) {
+        int ans = 0;
+        for (int i = 0; i < s.length(); i++) {
+            int head = i, tail = i;
+            while (head >= 0 && tail < s.length()) {
+                if (s.charAt(head) == s.charAt(tail)) {
+                    ans++;
+                } else {
+                    break;
+                }
+                head--;
+                tail++;
+            }
+
+            head = i;
+            tail = i + 1;
+            while (head >= 0 && tail < s.length()) {
+                if (s.charAt(head) == s.charAt(tail)) {
+                    ans++;
+                } else {
+                    break;
+                }
+                head--;
+                tail++;
+            }
+        }
+
+        return ans;
+    }
+}
diff --git a/word-search/forest000014.java b/word-search/forest000014.java
new file mode 100644
index 000000000..f151e9fe8
--- /dev/null
+++ b/word-search/forest000014.java
@@ -0,0 +1,84 @@
+/*
+Time Complexity: O(m * n * 4^(word.length))
+Space Complexity: O(m * n)
+*/
+
+class Solution {
+    boolean[][] visited;
+    int m, n;
+    int len;
+    int[] dr = {-1, 0, 1, 0}; // clockwise traversal
+    int[] dc = {0, 1, 0, -1};
+    char board2[][];
+    String word2;
+
+    public boolean exist(char[][] board, String word) {
+        int[] cnt = new int[52];
+        board2 = board;
+        word2 = word;
+        m = board.length;
+        n = board[0].length;
+        visited = new boolean[m][n];
+        len = word.length();
+
+        // 1. for pruning, count characters in board and word respectively
+        for (int i = 0; i < m; i++) {
+            for (int j = 0; j < n; j++) {
+                cnt[charToInt(board[i][j])]++;
+            }
+        }
+        for (int i = 0; i < len; i++) {
+            int idx = charToInt(word.charAt(i));
+            if (--cnt[idx] < 0) {
+                return false;
+            }
+        }
+
+        // 2. DFS
+        for (int i = 0; i < m; i++) {
+            for (int j = 0; j < n; j++) {
+                if (board2[i][j] != word.charAt(0)) {
+                    continue;
+                }
+                if (dfs(i, j, 1)) {
+                    return true;
+                }
+            }
+        }
+        return false;
+    }
+
+    private boolean dfs(int row, int col, int idx) {
+        if (idx == len) { // end of word
+            return true;
+        }
+        visited[row][col] = true;
+
+        for (int i = 0; i < 4; i++) {
+            int nr = row + dr[i];
+            int nc = col + dc[i];
+            if (nr < 0 || nr >= m || nc < 0 || nc >= n) { // check boundary of the board
+                continue;
+            }
+            if (visited[nr][nc]) { // check visited
+                continue;
+            }
+            if (board2[nr][nc] == word2.charAt(idx)) {
+                if (dfs(nr, nc, idx + 1)) {
+                    return true;
+                }
+            }
+        }
+
+        visited[row][col] = false;
+        return false;
+    }
+
+    private int charToInt(char ch) {
+        if (ch <= 'Z') {
+            return ch - 'A';
+        } else {
+            return ch - 'a' + 26;
+        }
+    }
+}