Linear Search & Binary Search- 2 Notes | EduRev

Algorithms

Computer Science Engineering (CSE) : Linear Search & Binary Search- 2 Notes | EduRev

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Binary Search

Given a sorted array arr[] of n elements, write a function to search a given element x in arr[].
A simple approach is to do a linear search. The time complexity of the above algorithm is O(n). Another approach to perform the same task is using Binary Search. 

Binary Search: Search a sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. Otherwise, narrow it to the upper half. Repeatedly check until the value is found or the interval is empty.
Example :
Linear Search & Binary Search- 2 Notes | EduRevThe idea of binary search is to use the information that the array is sorted and reduce the time complexity to O(Log n). 

We basically ignore half of the elements just after one comparison.

  1. Compare x with the middle element.
  2. If x matches with the middle element, we return the mid index.
  3. Else If x is greater than the mid element, then x can only lie in the right half subarray after the mid element. So we recur for the right half.
  4. Else (x is smaller) recur for the left half.

Recursive implementation of Binary Search

  • C++
    // C++ program to implement recursive Binary Search
    #include <bits/stdc++.h>
    using namespace std;
    // A recursive binary search function. It returns
    // location of x in given array arr[l..r] is present,
    // otherwise -1
    int binarySearch(int arr[], int l, int r, int x)
    {
        if (r >= l) {
            int mid = l + (r - l) / 2;
            // If the element is present at the middle
            // itself
            if (arr[mid] == x)
                return mid;
            // If element is smaller than mid, then
            // it can only be present in left subarray
            if (arr[mid] > x)
                return binarySearch(arr, l, mid - 1, x);
            // Else the element can only be present
            // in right subarray
            return binarySearch(arr, mid + 1, r, x);
        }
        // We reach here when element is not
        // present in array
        return -1;
    }
    int main(void)
    {
        int arr[] = { 2, 3, 4, 10, 40 };
        int x = 10;
        int n = sizeof(arr) / sizeof(arr[0]);
        int result = binarySearch(arr, 0, n - 1, x);
        (result == -1) ? cout << "Element is not present in array"
                       : cout << "Element is present at index " << result;
        return 0;
    }
  • C
    // C program to implement recursive Binary Search
    #include <stdio.h>
    // A recursive binary search function. It returns
    // location of x in given array arr[l..r] is present,
    // otherwise -1
    int binarySearch(int arr[], int l, int r, int x)
    {
        if (r >= l) {
            int mid = l + (r - l) / 2;
            // If the element is present at the middle
            // itself
            if (arr[mid] == x)
                return mid;
            // If element is smaller than mid, then
            // it can only be present in left subarray
            if (arr[mid] > x)
                return binarySearch(arr, l, mid - 1, x);
            // Else the element can only be present
            // in right subarray
            return binarySearch(arr, mid + 1, r, x);
        }
        // We reach here when element is not
        // present in array
        return -1;
    }
    int main(void)
    {
        int arr[] = { 2, 3, 4, 10, 40 };
        int n = sizeof(arr) / sizeof(arr[0]);
        int x = 10;
        int result = binarySearch(arr, 0, n - 1, x);
        (result == -1) ? printf("Element is not present in array")
                       : printf("Element is present at index %d",
                                result);
        return 0;
    }
  • Java
    // Java implementation of recursive Binary Search
    class BinarySearch {
        // Returns index of x if it is present in arr[l..
        // r], else return -1
        int binarySearch(int arr[], int l, int r, int x)
        {
            if (r >= l) {
                int mid = l + (r - l) / 2;
                // If the element is present at the
                // middle itself
                if (arr[mid] == x)
                    return mid;
                // If element is smaller than mid, then
                // it can only be present in left subarray
                if (arr[mid] > x)
                    return binarySearch(arr, l, mid - 1, x);
                // Else the element can only be present
                // in right subarray
                return binarySearch(arr, mid + 1, r, x);
            }
            // We reach here when element is not present
            // in array
            return -1;
        }
        // Driver method to test above
        public static void main(String args[])
        {
            BinarySearch ob = new BinarySearch();
            int arr[] = { 2, 3, 4, 10, 40 };
            int n = arr.length;
            int x = 10;
            int result = ob.binarySearch(arr, 0, n - 1, x);
            if (result == -1)
                System.out.println("Element not present");
            else
                System.out.println("Element found at index " + result);
        }
    }
    /* This code is contributed by Rajat Mishra */
  • Python3
    # Python3 Program for recursive binary search.
    # Returns index of x in arr if present, else -1
    def binarySearch (arr, l, r, x):
        # Check base case
        if r >= l:
            mid = l + (r - l) // 2
            # If element is present at the middle itself
            if arr[mid] == x:
                return mid
            # If element is smaller than mid, then it
            # can only be present in left subarray
            elif arr[mid] > x:
                return binarySearch(arr, l, mid-1, x)
            # Else the element can only be present
            # in right subarray
            else:
                return binarySearch(arr, mid + 1, r, x)
        else:
            # Element is not present in the array
            return -1
    # Driver Code
    arr = [ 2, 3, 4, 10, 40 ]
    x = 10
    # Function call
    result = binarySearch(arr, 0, len(arr)-1, x)
    if result != -1:
        print ("Element is present at index % d" % result)
    else:
        print ("Element is not present in array")
  • C#
    // C# implementation of recursive Binary Search
    using System;
    class GFG {
        // Returns index of x if it is present in
        // arr[l..r], else return -1
        static int binarySearch(int[] arr, int l,
                                int r, int x)
        {
            if (r >= l) {
                int mid = l + (r - l) / 2;
                // If the element is present at the
                // middle itself
                if (arr[mid] == x)
                    return mid;
                // If element is smaller than mid, then
                // it can only be present in left subarray
                if (arr[mid] > x)
                    return binarySearch(arr, l, mid - 1, x);
                // Else the element can only be present
                // in right subarray
                return binarySearch(arr, mid + 1, r, x);
            }
            // We reach here when element is not present
            // in array
            return -1;
        }
        // Driver method to test above
        public static void Main()
        {
            int[] arr = { 2, 3, 4, 10, 40 };
            int n = arr.Length;
            int x = 10;
            int result = binarySearch(arr, 0, n - 1, x);
            if (result == -1)
                Console.WriteLine("Element not present");
            else
                Console.WriteLine("Element found at index "
                                  + result);
        }
    }
    // This code is contributed by Sam007.
  • PHP
    <?php
    // PHP program to implement
    // recursive Binary Search
    // A recursive binary search
    // function. It returns location
    // of x in given array arr[l..r]
    // is present, otherwise -1
    function binarySearch($arr, $l, $r, $x)
    {
    if ($r >= $l)
    {
            $mid = ceil($l + ($r - $l) / 2);
            // If the element is present
            // at the middle itself
            if ($arr[$mid] == $x)
                return floor($mid);
            // If element is smaller than
            // mid, then it can only be
            // present in left subarray
            if ($arr[$mid] > $x)
                return binarySearch($arr, $l,
                                    $mid - 1, $x);
            // Else the element can only
            // be present in right subarray
            return binarySearch($arr, $mid + 1,
                                $r, $x);
    }
    // We reach here when element
    // is not present in array
    return -1;
    }
    // Driver Code
    $arr = array(2, 3, 4, 10, 40);
    $n = count($arr);
    $x = 10;
    $result = binarySearch($arr, 0, $n - 1, $x);
    if(($result == -1))
    echo "Element is not present in array";
    else
    echo "Element is present at index ",
                                $result;
    // This code is contributed by anuj_67.
    ?>
  • Javascript
    <script>
    // JavaScript program to implement recursive Binary Search
    // A recursive binary search function. It returns
    // location of x in given array arr[l..r] is present,
    // otherwise -1
    function binarySearch(arr, l, r, x){
        if (r >= l) {
            let mid = l + Math.floor((r - l) / 2);
            // If the element is present at the middle
            // itself
            if (arr[mid] == x)
                return mid;
            // If element is smaller than mid, then
            // it can only be present in left subarray
            if (arr[mid] > x)
                return binarySearch(arr, l, mid - 1, x);
            // Else the element can only be present
            // in right subarray
            return binarySearch(arr, mid + 1, r, x);
        }
        // We reach here when element is not
        // present in array
        return -1;
    }
    let arr = [ 2, 3, 4, 10, 40 ];
    let x = 10;
    let n = arr.length
    let result = binarySearch(arr, 0, n - 1, x);
    (result == -1) ? document.write( "Element is not present in array")
                       : document.write("Element is present at index " +result);
    </script>

Output: Element is present at index 3
Here you can create a check function for easier implementation.
Here is recursive implementation with check function which I feel is a much easier 

Implementation:

  • C++
    C++
    #include <bits/stdc++.h>
    using namespace std;
    //define array globally
    const int N = 1e6 +4;
    int a[N];
    int n;//array size
    //elememt to be searched in array
       int k;
    bool check(int dig)
    {
        //element at dig position in array
        int ele=a[dig];
        //if k is less than
        //element at dig position
        //then we need to bring our higher ending to dig
        //and then continue further
        if(k<=ele)
        {
            return 1;
        }
        else
        {
        return 0;
        }
    }
    void binsrch(int lo,int hi)
    {
        while(lo<hi)
        {
            int mid=(lo+hi)/2;
            if(check(mid))
            {
                hi=mid;
            }
            else
            {
              lo=mid+1;
            }
        }
        //if a[lo] is k
        if(a[lo]==k)
            cout<<"Element found at index "<<lo;// 0 based indexing
        else
            cout<<"Element doesnt exist in array";//element was not in our array
    }
    int main()
    {  
        cin>>n;
       for(int i=0; i<n; i++)
       {
        cin>>a[i];
       }
       cin>>k;
       //it is being given array is sorted
       //if not then we have to sort it
       //minimum possible point where our k can be is starting index
       //so lo=0
       //also k cannot be outside of array so end point
       //hi=n
       binsrch(0,n);
        return 0;
    }
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