day29

Day 29 - Search and Sort Algorithms Part B: The Binary Search

Binary Search is searching technique which works on Divide and Conquer approach. Indeed an efficient searching algorithm with a runtime of O(log(n)), but it works only on sorted arrays.

Question

Given a sorted array, and a number n, write a program to implement binary search and finid the index of the given number.

Try to do it using recursion as well as iteration.

Example

input: arr = [1, 2, 3, 4, 5, 8, 9], num = 8
output: 5 (index of found element)

input: arr = [1, 2, 3, 4, 5, 8, 9], num = 7
output: undefined

Solution

JavaScript Implementation

function binary (arr, n) {
    let left = 0,
        right = arr.length - 1, mid;

    while (left <= right) {
        mid = Math.floor((left + right)/2);
        if (arr[mid] === n)  return mid;
        else if (arr[mid] < n)  left = mid+1;
        else right = mid-1;
    }

    return -1;
}

console.log (binarySearch ([1, 2, 3, 4, 5, 8, 9], 8));
console.log (binarySearch ([1, 2, 3, 4, 5, 8, 9], 7));

C Implementation

Solution

/*
 * @author : ashwek
 * @date : 29/01/2019
 */

#include <stdio.h>
#include <stdlib.h>

int binary_search(int *arr, int n, int search){

    int low = 0, high = n-1;
    int mid;

    while( low <= high ){

        mid  = (high - low) / 2 + low;

        if( arr[mid] == search ){
            return mid;
        }
        else if( search < arr[mid] ){
            high--;
        }
        else{
            low++;
        }

    }

    return -1;
}

void main(){

    int *arr, n, search, i;

    printf("Enter size of array = ");
    scanf("%d", &n);

    arr = (int *)malloc(sizeof(int)*n);

    printf("Enter %d elements (in sorted order) : ", n);
    for(i=0; i<n; i++){
        scanf("%d", &arr[i]);
    }

    printf("Enter a value to search = ");
    scanf("%d", &search);

    i = binary_search(arr, n, search);

    if( i == -1 ){
        printf("%d not found in array", search);
    }
    else{
        printf("%d found at %d index", search, i);
    }
}

Java Implementation

Solution

/**
 * @date 29/01/19
 * @author SPREEHA DUTTA
 */
import java.util.*;
public class binarySearch {
    public static int search(int []arr,int p,int l,int r)
    {
        int mid=0;
        if(r>=l)
        {
          mid=(l+r)/2;  
        if(p==arr[mid])
            return mid;
        else if(p>arr[mid])
            return search(arr,p,mid+1,r);
        else
            return search(arr,p,l,mid-1);
        }
        return -1;
    }
    public static void main(String []args)
    {
        Scanner sc=new Scanner(System.in);
        int n,i;int p,f;
        n=sc.nextInt();
        int arr[]=new int[n];
        for(i=0;i<n;i++)
            arr[i]=sc.nextInt();
        System.out.println("Enter element to be searched");
        p=sc.nextInt();
        f=search(arr,p,0,arr.length-1);
        if(f==-1)
            System.out.println("undefined");
        else
            System.out.println(f);
    }
}

C++ Implementation

Solution by @imkaka

/*
* @author : imkaka
* @date   : 31/1/2019
*/

#include<iostream>
#include<cstdio>

using namespace std;

int binary_search_itr(int arr[], int size, int val){
    int l = 0, r = size-1;
    int mid = (l + r) / 2;

    while(arr[mid] != val && l <= r){
        if(val < arr[mid]){
            r = mid - 1;
        }
        else{
            l = mid +1;
        }

        mid = (l + r) / 2;
    }

    if(arr[mid] == val)
        return mid;

    return -1;
}

int binary_search_rec(int arr[], int left, int right, int val){
    if(right >= left){

        int mid = left + (right - left) / 2;

        if(arr[mid] == val) return mid;

        if(arr[mid] > val)
            return binary_search_rec(arr, left, mid-1, val);
        return binary_search_rec(arr, mid+1, right, val);
    }

    return -1;
}

int main(){

    int arr[] = {1, 2, 3, 6, 9, 15, 16, 14};
    int val = 9;
    int size = sizeof(arr)/sizeof(arr[0]);
    cout << "Iterative: " << binary_search_itr(arr, size, val) << endl;
    cout << "Recursive: " << binary_search_rec(arr, 0, size-1, val) << endl;
    return 0;
}