# Search an element in a sorted and rotated array

#### Bygeekycodesco

Oct 27, 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.

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

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:

```/* Program to search an element in
a sorted and pivoted array*/
#include <stdio.h>

int findPivot(int[], int, int);
int binarySearch(int[], int, int, int);

/* Searches an element key in a pivoted
sorted array arrp[] 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);
}

/* 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);
}

/* 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);
return binarySearch(arr, low, (mid - 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;
printf("Index of the element is : %d",
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.
##### 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:

```// Search an element in sorted and rotated
// array using single pass of Binary Search
#include <bits/stdc++.h>
using namespace std;

// Returns index of key in arr[l..h] if
// key is present, otherwise returns -1
int search(int arr[], int l, int h, int key)
{
if (l > h)
return -1;

int 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] && key <= arr[mid])
return search(arr, l, mid - 1, key);
/*If key not lies in first half subarray,
Divide other half into two subarrays,
such that we can quickly check if key lies
in other half */
return search(arr, mid + 1, h, key);
}

/* If arr[l..mid] first subarray is not sorted, then arr[mid... h]
must be sorted subarray */
if (key >= arr[mid] && key <= arr[h])
return search(arr, mid + 1, h, key);

return search(arr, l, mid - 1, key);
}

// Driver program
int main()
{
int arr[] = { 4, 5, 6, 7, 8, 9, 1, 2, 3 };
int n = sizeof(arr) / sizeof(arr[0]);
int key = 6;
int i = search(arr, 0, n - 1, key);

if (i != -1)
cout << "Index: " << i << endl;
else
}
```

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.

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.

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