Python Program for Binary Search (Recursive and Iterative)
In a nutshell, this search algorithm takes advantage of a collection of elements that is already sorted by ignoring half of the elements after just one comparison.
- Compare x with the middle element.
- If x matches with the middle element, we return the mid index.
- Else if x is greater than the mid element, then x can only lie in the right (greater) half subarray after the mid element. Then we apply the algorithm again for the right half.
- Else if x is smaller, the target x must lie in the left (lower) half. So we apply the algorithm for the left half.
Recursive :
Python3
# Python 3 program for recursive binary search. # Modifications needed for the older Python 2 are found in comments. # Returns index of x in arr if present, else -1 def binary_search(arr, low, high, x): # Check base case if high > = low: mid = (high + low) / / 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 binary_search(arr, low, mid - 1 , x) # Else the element can only be present in right subarray else : return binary_search(arr, mid + 1 , high, x) else : # Element is not present in the array return - 1 # Test array arr = [ 2 , 3 , 4 , 10 , 40 ] x = 10 # Function call result = binary_search(arr, 0 , len (arr) - 1 , x) if result ! = - 1 : print ( "Element is present at index" , str (result)) else : print ( "Element is not present in array" ) |
Output:
Element is present at index 3
Time Complexity: O(log n)
Auxiliary Space: O(logn) [NOTE: Recursion creates Call Stack]
Iterative:
Python3
# Iterative Binary Search Function # It returns index of x in given array arr if present, # else returns -1 def binary_search(arr, x): low = 0 high = len (arr) - 1 mid = 0 while low < = high: mid = (high + low) / / 2 # If x is greater, ignore left half if arr[mid] < x: low = mid + 1 # If x is smaller, ignore right half elif arr[mid] > x: high = mid - 1 # means x is present at mid else : return mid # 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 = binary_search(arr, x) if result ! = - 1 : print ( "Element is present at index" , str (result)) else : print ( "Element is not present in array" ) |
Output:
Element is present at index 3
Time Complexity: O(log n)
Auxiliary Space: O(1)
Please refer to the article Binary Search for more details!
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