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Python3 Program for Search an element in a sorted and rotated array

  • Last Updated : 11 Dec, 2021

An element in a sorted array can be found in O(log n) time via binary search. But suppose we rotate an ascending order sorted array at some pivot unknown to you beforehand. So for instance, 1 2 3 4 5 might become 3 4 5 1 2. Devise a way to find an element in the rotated array in O(log n) time.
 

sortedPivotedArray

Example: 
 

Input  : arr[] = {5, 6, 7, 8, 9, 10, 1, 2, 3};
         key = 3
Output : Found at index 8

Input  : arr[] = {5, 6, 7, 8, 9, 10, 1, 2, 3};
         key = 30
Output : Not found

Input : arr[] = {30, 40, 50, 10, 20}
        key = 10   
Output : Found at index 3

 

 

 

All solutions provided here assume that all elements in the array are distinct.
Basic Solution: 
Approach: 
 

  1. The idea is to find the pivot point, divide the array in two sub-arrays and perform binary search.
  2. The main idea for finding pivot is – for a sorted (in increasing order) and pivoted array, pivot element is the only element for which next element to it is smaller than it.
  3. Using the above statement and binary search pivot can be found.
  4. After the pivot is found out divide the array in two sub-arrays.
  5. Now the individual sub – arrays are sorted so the element can be searched using Binary Search.

Implementation: 
 

Input arr[] = {3, 4, 5, 1, 2}
Element to Search = 1
  1) Find out pivot point and divide the array in two
      sub-arrays. (pivot = 2) /*Index of 5*/
  2) Now call binary search for one of the two sub-arrays.
      (a) If element is greater than 0th element then
             search in left array
      (b) Else Search in right array
          (1 will go in else as 1 < 0th element(3))
  3) If element is found in selected sub-array then return index
     Else return -1.

Below is the implementation of the above approach: 
 

Python3




# Python Program to search an element
# in a sorted and pivoted array
  
# Searches an element key in a pivoted
# sorted array arrp[] of size n 
def pivotedBinarySearch(arr, n, key):
  
    pivot = findPivot(arr, 0, n-1);
  
    # If we didn't find a pivot, 
    # then array is not rotated at all
    if pivot == -1:
        return binarySearch(arr, 0, n-1, key);
  
    # If we found a pivot, then first
    # compare with pivot and then
    # search in two subarrays around pivot
    if arr[pivot] == key:
        return pivot
    if arr[0] <= key:
        return binarySearch(arr, 0, pivot-1, key);
    return binarySearch(arr, pivot + 1, n-1, key);
  
  
# Function to get pivot. For array 
# 3, 4, 5, 6, 1, 2 it returns 3 
# (index of 6) 
def findPivot(arr, low, high):
      
    # base cases
    if high < low:
        return -1
    if high == low:
        return low
      
    # low + (high - low)/2;
    mid = int((low + high)/2)
      
    if mid < high and arr[mid] > arr[mid + 1]:
        return mid
    if mid > low and arr[mid] < arr[mid - 1]:
        return (mid-1)
    if arr[low] >= arr[mid]:
        return findPivot(arr, low, mid-1)
    return findPivot(arr, mid + 1, high)
  
# Standard Binary Search function*/
def binarySearch(arr, low, high, key):
  
    if high < low:
        return -1
          
    # low + (high - low)/2;    
    mid = int((low + high)/2)
      
    if key == arr[mid]:
        return mid
    if key > arr[mid]:
        return binarySearch(arr, (mid + 1), high,
                                            key);
    return binarySearch(arr, low, (mid -1), key);
  
  
# Driver program to check above functions */
# Let us search 3 in below array
arr1 = [5, 6, 7, 8, 9, 10, 1, 2, 3]
n = len(arr1)
key = 3
print("Index of the element is : "
      pivotedBinarySearch(arr1, n, key))
        
# This is contributed by Smitha Dinesh Semwal

Output: 
 

Index of the element is : 8

Complexity Analysis: 
 

  • Time Complexity: O(log n). 
    Binary Search requires log n comparisons to find the element. So time complexity is O(log n).
  • Space Complexity:O(1), No extra space is required.

Thanks to Ajay Mishra for initial solution.
Improved Solution: 
Approach: Instead of two or more pass of binary search the result can be found in one pass of binary search. The binary search needs to be modified to perform the search. The idea is to create a recursive function that takes l and r as range in input and the key.
 

1) Find middle point mid = (l + h)/2
2) If key is present at middle point, return mid.
3) Else If arr[l..mid] is sorted
    a) If key to be searched lies in range from arr[l]
       to arr[mid], recur for arr[l..mid].
    b) Else recur for arr[mid+1..h]
4) Else (arr[mid+1..h] must be sorted)
    a) If key to be searched lies in range from arr[mid+1]
       to arr[h], recur for arr[mid+1..h].
    b) Else recur for arr[l..mid] 

Below is the implementation of above idea: 
 

Python3




# Search an element in sorted and rotated array using
# single pass of Binary Search
  
# Returns index of key in arr[l..h] if key is present,
# otherwise returns -1
def search (arr, l, h, key):
    if l > h:
        return -1
      
    mid = (l + h) // 2
    if arr[mid] == key:
        return mid
  
    # If arr[l...mid] is sorted 
    if arr[l] <= arr[mid]:
  
        # As this subarray is sorted, we can quickly
        # check if key lies in half or other half 
        if key >= arr[l] and key <= arr[mid]:
            return search(arr, l, mid-1, key)
        return search(arr, mid + 1, h, key)
  
    # If arr[l..mid] is not sorted, then arr[mid... r]
    # must be sorted
    if key >= arr[mid] and key <= arr[h]:
        return search(a, mid + 1, h, key)
    return search(arr, l, mid-1, key)
  
# Driver program
arr = [4, 5, 6, 7, 8, 9, 1, 2, 3]
key = 6
i = search(arr, 0, len(arr)-1, key)
if i != -1:
    print ("Index: % d"% i)
else:
    print ("Key not found")
  
# This code is contributed by Shreyanshi Arun

Output: 

Index: 2

Complexity Analysis: 
 

  • Time Complexity: O(log n). 
    Binary Search requires log n comparisons to find the element. So time complexity is O(log n).
  • Space Complexity: O(1). 
    As no extra space is required.

Thanks to Gaurav Ahirwar for suggesting above solution. 
How to handle duplicates? 
It doesn’t look possible to search in O(Logn) time in all cases when duplicates are allowed. For example consider searching 0 in {2, 2, 2, 2, 2, 2, 2, 2, 0, 2} and {2, 0, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2}. 
It doesn’t look possible to decide whether to recur for the left half or right half by doing a constant number of comparisons at the middle.
 

Similar Articles: 
 

Please write comments if you find any bug in the above codes/algorithms, or find other ways to solve the same problem.
 

Please refer complete article on Search an element in a sorted and rotated array for more details!


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