Element in a sorted and rotated array

Software Development
#1

A sorted array is rotated at some unknown point, find the minimum element in it.
The following solution assumes that all elements are distinct.

Examples:

Input: {5, 6, 1, 2, 3, 4} Output: 1 Input: {1, 2, 3, 4} Output: 1 Input: {2, 1} Output: 1

A simple solution is to traverse the complete array and find a minimum. This solution requires O(n) time.
We can do it in O(Logn) using Binary Search. If we take a closer look at the above examples, we can easily figure out the following pattern:

  • The minimum element is the only element whose previous is greater than it. If there is no previous element, then there is no rotation (the first element is minimum). We check this condition for the middle element by comparing it with (mid-1)’th and (mid+1)’th elements.
  • If the minimum element is not at the middle (neither mid nor mid + 1), then the minimum element lies in either the left half or right half.
    1. If the middle element is smaller than the last element, then the minimum element lies in the left half
    2. Else minimum element lies in the right half.

We strongly recommend you to try it yourself before seeing the following implementation.

  • C
  • C++

// C program to find minimum element in a sorted and rotated array

#include <stdio.h>

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

{

`` // This condition is needed to handle the case when array is not

`` // rotated at all

`` if (high < low) return arr[0];

`` // If there is only one element left

`` if (high == low) return arr[low];

`` // Find mid

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

`` // Check if element (mid+1) is minimum element. Consider

`` // the cases like {3, 4, 5, 1, 2}

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

`` return arr[mid+1];

`` // Check if mid itself is minimum element

`` if (mid > low && arr[mid] < arr[mid - 1])

`` return arr[mid];

`` // Decide whether we need to go to left half or right half

`` if (arr[high] > arr[mid])

`` return findMin(arr, low, mid-1);

`` return findMin(arr, mid+1, high);

}

// Driver program to test above functions

int main()

{

`` int arr1[] = {5, 6, 1, 2, 3, 4};

`` int n1 = sizeof (arr1)/ sizeof (arr1[0]);

`` printf ( "The minimum element is %d\n" , findMin(arr1, 0, n1-1));

`` int arr2[] = {1, 2, 3, 4};

`` int n2 = sizeof (arr2)/ sizeof (arr2[0]);

`` printf ( "The minimum element is %d\n" , findMin(arr2, 0, n2-1));

`` int arr3[] = {1};

`` int n3 = sizeof (arr3)/ sizeof (arr3[0]);

`` printf ( "The minimum element is %d\n" , findMin(arr3, 0, n3-1));

`` int arr4[] = {1, 2};

`` int n4 = sizeof (arr4)/ sizeof (arr4[0]);

`` printf ( "The minimum element is %d\n" , findMin(arr4, 0, n4-1));

`` int arr5[] = {2, 1};

`` int n5 = sizeof (arr5)/ sizeof (arr5[0]);

`` printf ( "The minimum element is %d\n" , findMin(arr5, 0, n5-1));

`` int arr6[] = {5, 6, 7, 1, 2, 3, 4};

`` int n6 = sizeof (arr6)/ sizeof (arr6[0]);

`` printf ( "The minimum element is %d\n" , findMin(arr6, 0, n6-1));

`` int arr7[] = {1, 2, 3, 4, 5, 6, 7};

`` int n7 = sizeof (arr7)/ sizeof (arr7[0]);

`` printf ( "The minimum element is %d\n" , findMin(arr7, 0, n7-1));

`` int arr8[] = {2, 3, 4, 5, 6, 7, 8, 1};

`` int n8 = sizeof (arr8)/ sizeof (arr8[0]);

`` printf ( "The minimum element is %d\n" , findMin(arr8, 0, n8-1));

`` int arr9[] = {3, 4, 5, 1, 2};

`` int n9 = sizeof (arr9)/ sizeof (arr9[0]);

`` printf ( "The minimum element is %d\n" , findMin(arr9, 0, n9-1));

`` return 0;

}

Output:

The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1

How to handle duplicates?

The above approach in the worst case(If all the elements are the same) takes O(N).
Below is the code to handle duplicates in O(log n) time.

  • C++

// C++ program to find minimum element in a sorted

// and rotated array contacting duplicate elements.

#include <bits/stdc++.h>

using namespace std;

// Function to find minimum element

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

{

`` while (low < high)

`` {

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

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

`` high--;

`` else if (arr[mid] > arr[high])

`` low = mid + 1;

`` else

`` high = mid;

`` }

`` return arr[high];

}

// Driver code

int main()

{

`` int arr1[] = {5, 6, 1, 2, 3, 4};

`` int n1 = sizeof (arr1)/ sizeof (arr1[0]);

`` cout << "The minimum element is " << findMin(arr1, 0, n1-1) << endl;

``

`` int arr2[] = {1, 2, 3, 4};

`` int n2 = sizeof (arr2)/ sizeof (arr2[0]);

`` cout << "The minimum element is " << findMin(arr2, 0, n2-1) << endl;

``

`` int arr3[] = {1};

`` int n3 = sizeof (arr3)/ sizeof (arr3[0]);

`` cout<< "The minimum element is " <<findMin(arr3, 0, n3-1)<<endl;

``

`` int arr4[] = {1, 2};

`` int n4 = sizeof (arr4)/ sizeof (arr4[0]);

`` cout<< "The minimum element is " <<findMin(arr4, 0, n4-1)<<endl;

``

`` int arr5[] = {2, 1};

`` int n5 = sizeof (arr5)/ sizeof (arr5[0]);

`` cout<< "The minimum element is " <<findMin(arr5, 0, n5-1)<<endl;

``

`` int arr6[] = {5, 6, 7, 1, 2, 3, 4};

`` int n6 = sizeof (arr6)/ sizeof (arr6[0]);

`` cout<< "The minimum element is " <<findMin(arr6, 0, n6-1)<<endl;

``

`` int arr7[] = {1, 2, 3, 4, 5, 6, 7};

`` int n7 = sizeof (arr7)/ sizeof (arr7[0]);

`` cout << "The minimum element is " << findMin(arr7, 0, n7-1) << endl;

``

`` int arr8[] = {2, 3, 4, 5, 6, 7, 8, 1};

`` int n8 = sizeof (arr8)/ sizeof (arr8[0]);

`` cout << "The minimum element is " << findMin(arr8, 0, n8-1) << endl;

``

`` int arr9[] = {3, 4, 5, 1, 2};

`` int n9 = sizeof (arr9)/ sizeof (arr9[0]);

`` cout << "The minimum element is " << findMin(arr9, 0, n9-1) << endl;

``

`` return 0;

}

Output:

The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1 The minimum element is 1