diff --git a/coin-change/minji-go.java b/coin-change/minji-go.java
new file mode 100644
index 000000000..527a23d8e
--- /dev/null
+++ b/coin-change/minji-go.java
@@ -0,0 +1,23 @@
+/*
+    Problem: https://leetcode.com/problems/coin-change/
+    Description: return the fewest number of coins that you need to make up that amount
+    Concept: Array, Dynamic Programming, Breadth-First Search
+    Time Complexity: O(NM), Runtime 15ms   - M is the amount
+    Space Complexity: O(M), Memory 44.28MB
+*/
+class Solution {
+    public int coinChange(int[] coins, int amount) {
+        int[] dp = new int[amount+1];
+        Arrays.fill(dp, amount+1);
+        dp[0]=0;
+
+        for(int i=1; i<=amount; i++){
+            for(int coin : coins){
+                if(i >= coin) {
+                    dp[i] = Math.min(dp[i], dp[i-coin] +1);
+                }
+            }
+        }
+        return dp[amount]>amount? -1: dp[amount];
+    }
+}
diff --git a/combination-sum/minji-go.java b/combination-sum/minji-go.java
index ab895923a..fac01b813 100644
--- a/combination-sum/minji-go.java
+++ b/combination-sum/minji-go.java
@@ -2,10 +2,8 @@
     Problem: https://leetcode.com/problems/combination-sum/
     Description: return a list of all unique combinations of candidates where the chosen numbers sum to target
     Concept: Array, Backtracking
-    Time Complexity: O(Nⁿ), Runtime 2ms
-    Space Complexity: O(N), Memory 44.88MB
-
-    - Time Complexity, Space Complexity를 어떻게 계산해야할지 어렵네요 :(
+    Time Complexity: O(Nᵀ), Runtime 2ms 
+    Space Complexity: O(T), Memory 44.88MB
 */
 class Solution {
     public List<List<Integer>> answer = new ArrayList<>();
diff --git a/merge-two-sorted-lists/minji-go.java b/merge-two-sorted-lists/minji-go.java
new file mode 100644
index 000000000..d70eac76f
--- /dev/null
+++ b/merge-two-sorted-lists/minji-go.java
@@ -0,0 +1,36 @@
+/*
+    Problem: https://leetcode.com/problems/merge-two-sorted-lists/
+    Description: return the head of the merged linked list of two sorted linked lists
+    Concept: Linked List, Recursion
+    Time Complexity: O(N+M), Runtime 0ms
+    Space Complexity: O(N+M), Memory 42.74MB
+*/
+
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode() {}
+ *     ListNode(int val) { this.val = val; }
+ *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+ * }
+ */
+class Solution {
+    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
+
+        ListNode head = new ListNode(0);
+        ListNode tail = head;
+
+        while(list1 != null || list2 != null) {
+            if (list2 == null || (list1 != null && list1.val <= list2.val)) {
+                tail = tail.next = new ListNode(list1.val);
+                list1 = list1.next;
+            } else {
+                tail = tail.next = new ListNode(list2.val);
+                list2 = list2.next;
+            }
+        }
+        return head.next;
+    }
+}
diff --git a/missing-number/minji-go.java b/missing-number/minji-go.java
new file mode 100644
index 000000000..61cc2894f
--- /dev/null
+++ b/missing-number/minji-go.java
@@ -0,0 +1,19 @@
+/*
+    Problem: https://leetcode.com/problems/missing-number/
+    Description: return the only number in the range that is missing from the array.
+    Concept: Array, Hash Table, Math, Binary Search, Bit Manipulation, Sorting
+    Time Complexity: O(N), Runtime 0ms
+    Space Complexity: O(1), Memory 45.71MB
+*/
+class Solution {
+    public int missingNumber(int[] nums) {
+        int n = nums.length;
+        int missing = n;
+
+        for(int i=0; i<n; i++){
+            missing ^= i;
+            missing ^= nums[i];
+        }
+        return missing;
+    }
+}
diff --git a/palindromic-substrings/minji-go.java b/palindromic-substrings/minji-go.java
new file mode 100644
index 000000000..cecce3457
--- /dev/null
+++ b/palindromic-substrings/minji-go.java
@@ -0,0 +1,27 @@
+/*
+    Problem: https://leetcode.com/problems/palindromic-substrings/
+    Description: return the number of palindromic substrings in it.
+    Concept: Two Pointers, String, Dynamic Programming
+    Time Complexity: O(N²), Runtime 6ms
+    Space Complexity: O(1), Memory 41.62MB
+*/
+class Solution {
+    public int countSubstrings(String s) {
+        int totalCount = 0;
+        for(int i=0; i<s.length(); i++){
+            totalCount+= countPalindromes(s, i, i);
+            totalCount+= countPalindromes(s, i, i+1);
+        }
+        return totalCount;
+    }
+
+    public int countPalindromes(String s, int left, int right){
+        int count = 0;
+        while(left>=0 && right<s.length() && s.charAt(left)==s.charAt(right)) {
+            count++;
+            right++;
+            left--;
+        }
+        return count;
+    }
+}
diff --git a/word-search/minji-go.java b/word-search/minji-go.java
new file mode 100644
index 000000000..51ca4eed7
--- /dev/null
+++ b/word-search/minji-go.java
@@ -0,0 +1,47 @@
+/*
+    Problem: https://leetcode.com/problems/word-search/
+    Description: return true if word exists in the grid
+    Concept: Array, String, Backtracking, Matrix
+    Time Complexity: O(MN4ᵀ), Runtime 147ms
+    Space Complexity: O(MN), Memory 42.11MB
+*/
+class Solution {
+    public char[][] board;
+    public String word;
+    public boolean[][] visited;
+    public int n, m;
+
+    public boolean exist(char[][] board, String word) {
+        this.board = board;
+        this.word = word;
+        this.m = board.length;
+        this.n = board[0].length;
+        this.visited = new boolean[m][n];
+
+        for(int i=0; i<m; i++){
+            for(int j=0; j<n; j++){
+                if(word.charAt(0) != board[i][j]) continue;
+                if(wordExists(i, j, 1)) return true;
+            }
+        }
+        return false;
+    }
+
+    public int[] dr = new int[]{-1,1,0,0};
+    public int[] dc = new int[]{0,0,-1,1};
+    public boolean wordExists(int cr, int cc, int i){
+        if(i==word.length()) return true;
+
+        visited[cr][cc] = true;
+        for(int k=0; k<4; k++){
+            int nr = cr+dr[k];
+            int nc = cc+dc[k];
+            if(nr<0||nc<0||nr>m-1||nc>n-1||visited[nr][nc]) continue;
+            if(board[nr][nc]!=word.charAt(i)) continue;
+            if(wordExists(nr, nc, i+1)) return true;
+        }
+        visited[cr][cc] = false;
+
+        return false;
+    }
+}