Find an element in a sorted and rotated array Basic solution

Software Development
#1

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:

  • C++

/* C++ Program to search an element

`` in a sorted and pivoted array*/

#include <bits/stdc++.h>

using namespace std;

/* Standard Binary Search function*/

int binarySearch( int arr[], int low,

`` int high, int key)

{

`` if (high < low)

`` return -1;

`` int mid = (low + high) / 2; /*low + (high - low)/2;*/

`` if (key == arr[mid])

`` return mid;

`` if (key > arr[mid])

`` return binarySearch(arr, (mid + 1), high, key);

`` // else

`` return binarySearch(arr, low, (mid - 1), key);

}

/* Function to get pivot. For array 3, 4, 5, 6, 1, 2

`` it returns 3 (index of 6) */

int findPivot( int arr[], int low, int high)

{

`` // base cases

`` if (high < low)

`` return -1;

`` if (high == low)

`` return low;

`` int mid = (low + high) / 2; /*low + (high - low)/2;*/

`` if (mid < high && arr[mid] > arr[mid + 1])

`` return mid;

`` if (mid > low && 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);

}

/* Searches an element key in a pivoted

`` sorted array arr[] of size n */

int pivotedBinarySearch( int arr[], int n, int key)

{

`` int 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);

}

/* Driver program to check above functions */

int main()

{

`` // Let us search 3 in below array

`` int arr1[] = { 5, 6, 7, 8, 9, 10, 1, 2, 3 };

`` int n = sizeof (arr1) / sizeof (arr1[0]);

`` int key = 3;

`` // Function calling

`` cout << "Index of the element is : "

`` << pivotedBinarySearch(arr1, n, key);

`` return 0;

}

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.