Added codes for 20 April

GeeksForGeeks/April/20-4-24/GFG.java

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+//{ Driver Code Starts
+//Initial Template for Java
+
+
+import java.util.*;
+import java.io.*;
+import java.lang.*;
+
+class Main
+{
+    public static void main(String args[])throws IOException
+    {
+        BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
+        int t  =Integer.parseInt(read.readLine());
+
+        while(t-- > 0)
+        {
+            String st[] = read.readLine().trim().split(" ");
+            int N = Integer.parseInt(st[0]);
+            int M = Integer.parseInt(st[1]);
+
+            int arr1[] = new int[N];
+            int arr2[] = new int[M];
+
+            st = read.readLine().trim().split(" ");
+            for(int i = 0; i < N; i++)
+              arr1[i] = Integer.parseInt(st[i]);
+
+            st = read.readLine().trim().split(" ");  
+            for(int i = 0; i< M; i++)
+              arr2[i] = Integer.parseInt(st[i]);
+
+            Solution obj = new Solution();
+            ArrayList<Integer> res = new ArrayList<Integer>();
+            res = obj.findUnion(arr1, arr2, N, M);
+            for(int i = 0;i<res.size();i++)
+                System.out.print(res.get(i) + " ");
+            System.out.println(); 
+        }
+    }
+}
+
+
+
+// } Driver Code Ends
+
+
+//User function Template for Java
+
+//arr1,arr2 : the arrays
+// n, m: size of arrays
+class Solution
+{
+    //Function to return a list containing the union of the two arrays.
+    public static ArrayList<Integer> findUnion(int arr1[], int arr2[], int n, int m)
+    {
+        // add your code here
+        ArrayList<Integer> ans= new ArrayList<Integer>();
+
+        Set<Integer> set= new HashSet<>();
+
+        for(int a:arr1)
+            set.add(a);
+
+        for(int b:arr2)
+            set.add(b);
+
+        for(int i:set)
+            ans.add(i);
+
+        Collections.sort(ans);
+        return ans;
+    }
+}

GeeksForGeeks/April/20-4-24/README.md

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+Time complexity - O(n+m)
+Space complexity - O(n+m)

LeetCode/April/20-4-24/README.md

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+Time complexity - O(n*m)
+Space complexity - O(n*m)

LeetCode/April/20-4-24/Solution.java

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+class Solution 
+{
+    public int[][] findFarmland(int[][] land) 
+    {
+        List<int[]> ans = new ArrayList<>();
+
+        for (int i = 0; i < land.length; ++i)
+            for (int j = 0; j < land[0].length; ++j)
+                if (land[i][j] == 1) 
+                {
+                    int[] cell = new int[] {i, j};
+                    dfs(land, i, j, cell);
+                    ans.add(new int[] {i, j, cell[0], cell[1]});
+                }
+
+        return ans.stream().toArray(int[][] ::new);
+    }
+
+    private void dfs(int[][] land, int i, int j, int[] cell) 
+    {
+        if (i < 0 || i == land.length || j < 0 || j == land[0].length)
+            return;
+        if (land[i][j] != 1)
+            return;
+
+        land[i][j] = 2;
+        cell[0] = Math.max(cell[0], i);
+        cell[1] = Math.max(cell[1], j);
+        dfs(land, i + 1, j, cell);
+        dfs(land, i, j + 1, cell);
+    }
+}