Answers for "binary search algorithm."

8

binary search

#binary search python
def binaryy(ar, ele):
    low = 0 
    high = len(ar)-1
    if ele not in ar:
        return "Not Found"
    while low <= high:
        mid = (low + high) // 2
        if ar[mid] < ele:
            low = mid + 1
        elif ar[mid] > ele:
            high = mid - 1
        else:
            return mid


ar = [10, 20, 30, 40, 50]
ele = 55
print(binaryy(ar, ele))
Posted by: Guest on July-24-2021
0

binary search

#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;
}
Posted by: Guest on July-16-2021
1

binary search

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");
		}
		

	}

}
Posted by: Guest on August-03-2020
0

Binary Search

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")
Posted by: Guest on July-23-2021
1

binary search iterative

// Binary Search using Iterative Approach

import java.io.*;
class Binary_Search
{
	public static void main(String[] args) throws Exception
	{
		Binary_Search obj = new Binary_Search();
		InputStreamReader isr = new InputStreamReader(System.in);
		BufferedReader br = new BufferedReader(isr);
		System.out.println("Insert the length of the Array : ");
		int n = Integer.parseInt(br.readLine());
		int arr[] = new int[n];
		System.out.println("Insert elements into the array : ");
		for(int i=0;i<n;i++)
		{
			arr[i] = Integer.parseInt(br.readLine());
		}
		System.out.println("Enter the num which you want to Search : ");
		int num = Integer.parseInt(br.readLine());
		obj.logic(arr,num);
	}
	void logic(int arr[],int num)
	{
		int r = arr.length - 1;
		int l = 0;
		int mid;
		while(l<=r)
		{
			mid = l + (r-l)/2;
			if(arr[mid] == num)
			{
				System.out.println("Number found at "+mid+"th index");
				break;
			}
			else if(arr[mid]>num)
			{
				r = mid - 1;
			}
			else
			{
				l = mid + 1;
			}
		}
	}
}
Posted by: Guest on June-23-2020
0

binary search algorithm

#include <bits/stdc++.h>

using namespace std;

int binarySearch(int arr[], int l, int h, int key){
    if(l<=h){
        int mid = l + (h-l)/2;

        if(arr[mid] == key){
            return mid;
        }

        else if(arr[mid] > key){
            return binarySearch(arr, l, mid-1, key);
        }

        else if(arr[mid] < key){
            return binarySearch(arr,mid+1, h, key);
        }
    }       

    return -1;
}

int main(){
    int arr[] = {1,2,3,4,5,6,7,8,9,10};
    int n = sizeof(arr)/sizeof(arr[0]);
    int key = 7;

    int result = binarySearch(arr,0,n-1,key);

    (result==-1)
        ? cout << "Element is not found in the array" << endl
        : cout << "Element is found at index " << result;

    return 0;

}
Posted by: Guest on January-16-2021

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