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 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 :
The 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.
- Compare x with the middle element.
- If x matches with middle element, we return the mid index.
- Else If x is greater than the mid element, then x can only lie in right half subarray after the mid element. So we recur for right half.
- 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 -1int 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 -1int 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 Searchclass 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 -1def 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 Codearr = [ 2, 3, 4, 10, 40 ]x = 10 # Function callresult = 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 Searchusing 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 -1function 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, $x);} // We reach here when element // is not present in arrayreturn -1;} // Driver Code$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";elseecho "Element is present at index ", $result; // This code is contributed by anuj_67.?> |
Output :
Element is present at index 3
Iterative implementation of Binary Search
C++
// C++ program to implement recursive Binary Search#include <bits/stdc++.h>using namespace std; // A iterative binary search function. It returns// location of x in given array arr[l..r] if present,// otherwise -1int binarySearch(int arr[], int l, int r, int x){ 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;} 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 iterative Binary Search#include <stdio.h> // A iterative binary search function. It returns// location of x in given array arr[l..r] if present,// otherwise -1int binarySearch(int arr[], int l, int r, int x){ 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;} 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 iterative Binary Searchclass 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); }} |
Python3
# Python3 code to implement iterative Binary # Search. # It returns location of x in given array arr # if present, else returns -1def 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 # Driver Codearr = [ 2, 3, 4, 10, 40 ]x = 10 # Function callresult = 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 iterative Binary Searchusing System; class GFG { // Returns index of x if it is present in arr[], // else return -1 static 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() { int[] arr = { 2, 3, 4, 10, 40 }; int n = arr.Length; int x = 10; int result = binarySearch(arr, 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// iterative Binary Search // A iterative binary search // function. It returns location // of x in given array arr[l..r] // if present, otherwise -1function binarySearch($arr, $l, $r, $x){ while ($l <= $r) { $m = $l + ($r - $l) / 2; // Check if x is present at mid if ($arr[$m] == $x) return floor($m); // If x greater, ignore // left half if ($arr[$m] < $x) $l = $m + 1; // If x is smaller, // ignore right half else $r = $m - 1; } // if we reach here, then // element was not present return -1;} // Driver Code$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";elseecho "Element is present at index ", $result; // This code is contributed by anuj_67.?> |
Output :
Element is present at index 3
Time Complexity:
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 
.
Auxiliary Space: O(1) in case of iterative implementation. In case of recursive implementation, O(Logn) recursion call stack space.
Algorithmic Paradigm: Decrease and Conquer.
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Coding Practice Questions on Binary Search
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