Binary Search Tree Coding

Binary Search

Binary Search - search and sorting - Search a sorted array by dividing the search interval in half. Begin with an interval covering the whole array.

 

 

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 linear search.The time complexity of 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:

Binary Search

c and c++

c and c++
#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;
}

Output

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Element is present at index 3

Python

python
# Iterative Binary Search Function
# It returns location of x in given array arr if present,
# else returns -1
def binarySearch(arr, l, r, x):
 
    while l <= r:
 
        mid = l + (r - l)/2;
         
        # Check if x is present at mid
        if arr[mid] == x:
            return mid
 
        # If x is greater, ignore left half
        elif arr[mid] < x:
            l = mid + 1
 
        # If x is smaller, ignore right half
        else:
            r = mid - 1
     
    # If we reach here, then the element was not present
    return -1
 
 
# Test array
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"

Output

Element is present at index 3

Java

java
// Java implementation of iterative Binary Search
class BinarySearch
{
    // Returns index of x if it is present in arr[], else
    // return -1
    int binarySearch(int arr[], int x)
    {
        int l = 0, r = arr.length - 1;
        while (l <= r)
        {
            int m = l + (r-l)/2;
 
            // Check if x is present at mid
            if (arr[m] == x)
                return m;
 
            // If x greater, ignore left half
            if (arr[m] < x)
                l = m + 1;
 
            // If x is smaller, ignore right half
            else
                r = m - 1;
        }
 
        // if we reach here, then element was not present
        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, x);
        if (result == -1)
            System.out.println("Element not present");
        else
            System.out.println("Element found at index "+result);
    }
}

Output

Element is present at index 3

Time Complexity:

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The time complexity of Binary Search can be written as

T(n) = T(n/2) + c

The above recurrence can be solved either using Recurrence T ree method or Master method. It falls in case II of Master Method and solution of the recurrence is \Theta(Logn).

Auxiliary Space: O(1) in case of iterative implementation. In case of recursive implementation, O(Logn) recursion call stack space.

Algorithmic Paradigm: Divide and Conquer.

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Venkatesan Prabu

Venkatesan Prabu

Wikitechy Founder, Author, International Speaker, and Job Consultant. My role as the CEO of Wikitechy, I help businesses build their next generation digital platforms and help with their product innovation and growth strategy. I'm a frequent speaker at tech conferences and events.

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