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BinarySearch.java

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+// { Driver Code Starts
+//Initial Template for Java
+
+import java.io.*;
+import java.util.*;
+
+public class GFG {
+    public static void main (String[] args) throws IOException {
+        Scanner sc = new Scanner(System.in);
+        int T = sc.nextInt();
+        while(T>0)
+        {
+            int n = sc.nextInt();
+            int arr[] = new int[n];
+            for(int i=0;i<n;i++)
+            {
+                arr[i]=sc.nextInt();
+            }
+
+            int key =sc.nextInt();
+            Solution g = new Solution();
+            System.out.println(g.binarysearch(arr,n,key));
+            T--;
+        }
+    }
+}
+
+// } Driver Code Ends
+
+
+//User function Template for Java
+
+class Solution {
+
+    static int bs(int a[],int l,int h,int k){
+       if(l<=h){
+        int m=(l+h)/2;
+        if(a[m]==k)
+        return m;
+        else if(a[m]<k)
+            return bs(a,m+1,h,k);
+        else
+            return bs(a,l,m-1,k);
+        }else
+            return -1;
+   }
+
+
+    int binarysearch(int arr[], int n, int k){
+        // code here
+        int ans=bs(arr,0,n-1,k);
+        return ans;
+    }
+}

Factorials of large numbers.java

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+// { Driver Code Starts
+//Initial Template for Java
+
+import java.io.*;
+import java.util.*;
+
+class Hacktober
+{
+    public static void main(String args[])
+    {
+        Scanner sc = new Scanner(System.in);
+        int t = sc.nextInt();
+        while(t-->0)
+        {
+            int N = sc.nextInt();
+            Solution ob = new Solution();
+            ArrayList<Integer> ans = ob.factorial(N);
+            for (Integer val: ans) 
+                System.out.print(val); 
+            System.out.println();
+        }   
+    }
+}// } Driver Code Ends
+
+
+//User function Template for Java
+
+class Solution {
+    static ArrayList<Integer> factorial(int N){
+        //code here
+        ArrayList<Integer> a=new ArrayList<>();
+        a.add(1);
+        int i,;
+        for(i=2;i<=N;i++){
+            mul(i,a);
+        }
+        //System.out.println(a);
+    Collections.reverse(a);   
+    return  a;  
+    }
+    static void mul(int x, ArrayList<Integer> l){
+        int i, prod,carry=0;
+
+        for(i=0;i<l.size();i++){
+            prod=l.get(i) * x  +carry;
+            l.set(i,prod%10);
+            carry=prod/10;
+        }
+        while (carry!=0)
+        {
+            l.add(carry % 10);
+            carry = carry / 10;
+        }
+
+    }
+}

Middle Element LinkedList.java

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+// Given a singly linked list of N nodes.
+// The task is to find the middle of the linked list. For example, if the linked list is
+// 1-> 2->3->4->5, then the middle node of the list is 3.
+// If there are two middle nodes(in case, when N is even), print the second middle element.
+// For example, if the linked list given is 1->2->3->4->5->6, then the middle node of the list is 4.
+
+
+
+import java.util.*;
+import java.io.*;
+
+class Node{
+    int data;
+    Node next;
+
+    Node(int x){
+        data = x;
+        next = null;
+    }
+
+}
+class Middle{
+	static void printList(Node node) 
+	{ 
+		while (node != null) 
+		{ 
+			System.out.print(node.data + " "); 
+			node = node.next; 
+		} 
+		System.out.println(); 
+	}
+    public static void main(String args[]) throws IOException { 
+        Scanner sc = new Scanner(System.in);
+        int t = sc.nextInt();
+        while(t > 0){
+        	int n = sc.nextInt();
+            Node head = new Node(sc.nextInt());
+            Node tail = head;
+            for(int i=0; i<n-1; i++)
+            {
+                tail.next = new Node(sc.nextInt());
+                tail = tail.next;
+            }
+            Solution g = new Solution();
+            int ans = g.getMiddle(head);
+            System.out.println(ans);
+            //printList(head); 
+            t--;
+        }
+    } 
+} 
+
+class Solution
+{
+    int getMiddle(Node head)
+    {
+         // Your code here.
+         Node n=head;
+         int c=0;
+         while(n.next!=null){
+             c++;
+             n=n.next;
+         }
+         c=((c & 1)==0)?c/2:(c/2)+1;
+         n=head;
+         for(int i=0;i<c;i++)
+         n=n.next;
+         return(n.data);
+    }
+}