diff --git a/C++/Max Area of Island.cpp b/C++/Max Area of Island.cpp
new file mode 100644
index 0000000..98f1a9d
--- /dev/null
+++ b/C++/Max Area of Island.cpp	
@@ -0,0 +1,87 @@
+/*
+
+Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.
+
+Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)
+
+Example 1:
+
+[[0,0,1,0,0,0,0,1,0,0,0,0,0],
+
+[0,0,0,0,0,0,0,1,1,1,0,0,0],
+
+[0,1,1,0,1,0,0,0,0,0,0,0,0],
+
+[0,1,0,0,1,1,0,0,1,0,1,0,0],
+
+[0,1,0,0,1,1,0,0,1,1,1,0,0],
+
+[0,0,0,0,0,0,0,0,0,0,1,0,0],
+
+[0,0,0,0,0,0,0,1,1,1,0,0,0],
+
+[0,0,0,0,0,0,0,1,1,0,0,0,0]]
+
+Given the above grid, return 6. Note the answer is not 11, because the island must be connected 4-directionally.
+
+Example 2:
+
+[[0,0,0,0,0,0,0,0]]
+
+Given the above grid, return 0.
+
+Note: The length of each dimension in the given grid does not exceed 50.
+
+*/
+
+// DFS solution
+
+
+  
+
+int dfs(int x,int y, vector<vector<int>>&grid, vector<vector<bool>>&v)
+    {
+    
+    
+        if(x<0 || y<0 || x>=grid.size()|| y>=grid[0].size() || v[x][y] == true || grid[x][y] == 0)
+          {
+              return 0;
+          }
+        
+        else
+          {
+              v[x][y] = true;
+
+               int    top = dfs(x-1,y,grid,v);
+
+               int    left = dfs(x,y-1,grid,v);
+
+               int    right = dfs(x,y+1,grid,v);
+
+               int   down = dfs(x+1,y,grid,v);
+
+
+              return (1+(left+right+down+top));
+
+          }
+      }
+      
+ int maxAreaOfIsland(vector<vector<int>>& grid) {
+        
+        int ans = 0;
+        
+        vector<vector<bool>> visited (grid.size(), vector<bool>(grid[0].size(), false));
+        
+        int n = grid.size();
+        int m = grid[0].size();
+    
+            for(int i=0;i<n;i++)
+            {
+                for(int j = 0;j<m;j++)
+                {
+                    ans = max(ans,dfs(i,j,grid,visited));
+                }
+            }
+        
+        return ans;
+    }