diff --git a/Topic1_Arrays/Day2207/excercise_3.cpp b/Topic1_Arrays/Day2207/excercise_3.cpp
new file mode 100644
index 0000000..6dc1b38
--- /dev/null
+++ b/Topic1_Arrays/Day2207/excercise_3.cpp
@@ -0,0 +1,31 @@
+// Use the two pointers technique to solve the problem
+// Time complexity : O(m + n)
+// Space complexity: O(m + n) (due to creating a new vector to store the output)
+
+class Solution {
+public:
+    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
+        vector<int> tmp;
+        tmp = nums1;
+        int pointer1 = 0, pointer2 = 0, index = 0;
+        while (pointer1 < m && pointer2 < n) {
+            if (tmp[pointer1] < nums2[pointer2]) {
+                nums1[index] = tmp[pointer1];
+                pointer1++; index++;
+            } else {
+                nums1[index] = nums2[pointer2];
+                pointer2++; index++;
+            }
+        }
+        while (pointer1 < m) {
+            nums1[index] = tmp[pointer1];
+            pointer1++;
+            index++;
+        }
+        while (pointer2 < n) {
+            nums1[index] = nums2[pointer2];
+            pointer2++;
+            index++;
+        }
+    }
+};
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diff --git a/Topic1_Arrays/Day2207/exercise_1.cpp b/Topic1_Arrays/Day2207/exercise_1.cpp
new file mode 100644
index 0000000..42e7f45
--- /dev/null
+++ b/Topic1_Arrays/Day2207/exercise_1.cpp
@@ -0,0 +1,18 @@
+// Use nested loop to calculate all the possible cases and find the answer
+// Time complexity : O(n^2)
+// Space complexity: O(1) (because no extra spaces are needed)
+#include <bits/stdc++.h>
+class Solution {
+public:
+    void twoSum(vector<int>& nums, int target) {
+        for (int i = 0; i < nums.size(); i++) {
+            for (int j = i + 1; j < nums.size(); j++) {
+                if (nums[i] + nums[j] == target) {
+                    //return {i, j};
+                }
+            }
+        }
+        //return {-1, -1};
+    }
+};
+
diff --git a/Topic1_Arrays/Day2207/exercise_2.cpp b/Topic1_Arrays/Day2207/exercise_2.cpp
new file mode 100644
index 0000000..75f5553
--- /dev/null
+++ b/Topic1_Arrays/Day2207/exercise_2.cpp
@@ -0,0 +1,30 @@
+// Use DFS to count the number of islands
+// Time complexity : O(n^2)
+// Space complexity : O(n) (The number of recusive function calls in the call stack)
+class Solution {
+public:
+    int dx[4] = {-1, 0, 0, 1};
+    int dy[4] = {0, -1, 1, 0};
+    void loang(int i, int j, vector<vector<char>>& grid) {
+        grid[i][j] = '0';
+        for (int k = 0; k < 4; k++) {
+            int i1 = i + dx[k], j1 = j + dy[k];
+            if (i1 >= 0 && i1 < grid.size() && j1 >= 0 && j1 < grid[i].size() &&
+                grid[i1][j1] == '1') {
+                loang(i1, j1, grid);
+            }
+        }
+    }
+    int numIslands(vector<vector<char>>& grid) {
+        int res = 0;
+        for (int i = 0; i < grid.size(); i++) {
+            for (int j = 0; j < grid[i].size(); j++) {
+                if (grid[i][j] == '1') {
+                    loang(i, j, grid);
+                    res++;
+                }
+            }
+        }
+        return res;
+    }
+};
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