Binary Search

                                    //it is using divide and conquer method
#include <iostream>

using namespace std;
void binarysearch(int arr[],int start,int end,int val)
{
    if(start<end)
    {
        int mid=(start+end)/2;
        if(arr[mid]==val)
        {
            cout<<"value found:"<<endl;
        }
        else if(arr[mid]<val)
        {
            binarysearch(arr,mid+1,end,val);
        }
        else if(arr[mid]>val)
        {
            binarysearch(arr,start,mid-1,val);
        }
    }
    else
    {
        cout<<"not present:"<<endl;
    }
}

int main()
{
    int n;
    cout<<"enter the size of the array:"<<endl;
    cin>>n;
    int arr[n];
    cout<<"enter the elements of the array:"<<endl;
    for(int i=0;i<n;i++)
    {
        cin>>arr[i];
    }
    cout<<"enter the value you want to search:"<<endl;
    int val;
    cin>>val;
    binarysearch(arr,0,n-1,val);

    return 0;
}

                                    
import java.util.Scanner;

public class Binarysearch {

	public static void main(String[] args) {
		int[] x= {1,2,3,4,5,6,7,8,9,10,16,18,20,21};
		Scanner scan=new Scanner(System.in);
		System.out.println("enter the key:");
		int key=scan.nextInt();
		int flag=0;
		int low=0;
		int high=x.length-1;
		int mid=0;
		while(low<=high)
		{
			mid=(low+high)/2;
			if(key<x[mid])
			{
				high=mid-1;
			}
			else if(key>x[mid])
			{
				low=mid+1;
			}
			else if(key==x[mid])
			{
				flag++;
				System.out.println("found at index:"+mid);
				break;
			}
		}
		if(flag==0)
		{
			System.out.println("Not found");
		}
		

	}

}

                                    function binarySearchRicorsivo(array A, int p, int r, int v)
    if p > r
      return -1
     if v < A[p] or v > A[r]
       return -1
     q= (p+r)/2
     if A[q] == v
       return q
     else if A[q] > v
       return binarySearchRicorsivo(A,p,q-1,v)
     else
     

                                    //Binary search can apply to sorted data only.
//Time complexity of binary search is O(log n ).
//It always divide the whole data in parts and compare  a search key to middle element only.


import java.util.*;
public class BinarySearch {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int[] a = {10,20,50,30,40};
		int key=sc.nextInt();
		
		Arrays.sort(a);					// An method in java.util.Arrays package to sort an array element.
		
		int first=0,end=a.length-1,mid=0,flag=0;

		while(first<=end)
		{
			mid=(first+end)/2;
			if(key<a[mid])				// Move to left part if key is smaller than middle element.
			{
				end = mid-1;
			}
			else if(key>a[mid])		   // Move to right part if key is greater than middle element.
			{
				first = mid+1;
			}
			else
			{
				flag=1;
				break;
			}
		}
		if(flag==1)
		{
			System.out.println("Success! found");
		}
		else
		{
			System.out.println("Error! This key (" + key + ") does not exist in the array");
		}
		
	}

}

                                    #include <bits/stdc++.h>
using namespace std;

int main(){
    int n;
    cin>>n;
    vector<int>v(n);
    for(int i = 0; i<n; i++){
        cin>>v[i];
    }
    int to_find;
    cin>>to_find;
    int lo = 0 , hi = n-1 , mid ;
    while(hi - lo > 1){
        int mid = (hi + lo)/2;
        if(v[mid] < to_find){
            lo = mid + 1;
        }else{
            hi = mid;
        }
    }
    if(v[lo] == to_find){
        cout<<lo<<endl;
    }else if(v[hi] == to_find){
        cout<<hi<<endl;
    }else{
        cout<<"Not Found"; 
    }

    return 0;
}

                                    def binarySearch(arr, k, low, high):   
    while low <= high:
        mid = low + (high - low)//2
        if arr[mid] == k:
            return mid
        elif arr[mid] < k:

            low = mid + 1
        else:

            high = mid - 1
    return -1

arr = [1, 3, 5, 7, 9]

k = 5
result = binarySearch(arr, k, 0, len(arr)-1)

if result != -1:

    print("Element is present at index " + str(result))

else:

    print("Not found")

                                    10 101 61 126 217 2876 6127 39162 98126 712687 1000000000100 6127 1 61 200 -10000 1 217 10000 1000000000