Find the minimum element in a sorted and rotated array

A sorted array arr[] of size N is rotated at some unknown point, find the minimum element in it. 

Note: It is assumed that all elements are distinct.

Examples:

Input: arr[] = {5, 6, 1, 2, 3, 4}
Output: 1

Input: arr[] = {1, 2, 3, 4}
Output: 1

Input: arr[] = {2, 1}
Output: 1

Naive Approach:

A simple solution is to use linear search to traverse the complete array and find a minimum. 

Time Complexity: O(N)
Auxiliary Space: O(1)

Efficient Approach (Using Binary Search): This problem can be solved using binary search based on the following idea:

As the array is sorted and rotated, there are two segments that are themselves sorted but their meeting point is the only position where the smallest element is and that is not sorted. 

So we just need to find the position whose neighbours are greater than it and based on the extreme end values we can decide in which half we should search for that element.

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: 
  • If the middle element is smaller than the last element, then the minimum element lies in the left half
  • Else minimum element lies in the right half.

Follow the below illustration for a better understanding.

Illustration:

Let the array be arr[]={15, 18, 2, 3, 6, 12}
low = 0 , high = 5.
=>  mid = 2
=>  arr[mid]=2 , arr[mid-1] > arr[mid] , hence condition is matched
=>  The required index = mid = 2

So the element is  found at index 2 and arr[2] = 2

Below is the code implementation of the above approach:

C++

// C++ program to find minimum
// element in a sorted and rotated array
#include <bits/stdc++.h>
using namespace std;

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]);
    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;
}

// This is code is contributed by rathbhupendra

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;
}

Java

// Java program to find minimum element in a sorted and
// rotated array
import java.io.*;
import java.lang.*;
import java.util.*;

class Minimum {
    static 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
    public static void main(String[] args)
    {
        int arr1[] = { 5, 6, 1, 2, 3, 4 };
        int n1 = arr1.length;
        System.out.println("The minimum element is "
                           + findMin(arr1, 0, n1 - 1));

        int arr2[] = { 1, 2, 3, 4 };
        int n2 = arr2.length;
        System.out.println("The minimum element is "
                           + findMin(arr2, 0, n2 - 1));

        int arr3[] = { 1 };
        int n3 = arr3.length;
        System.out.println("The minimum element is "
                           + findMin(arr3, 0, n3 - 1));

        int arr4[] = { 1, 2 };
        int n4 = arr4.length;
        System.out.println("The minimum element is "
                           + findMin(arr4, 0, n4 - 1));

        int arr5[] = { 2, 1 };
        int n5 = arr5.length;
        System.out.println("The minimum element is "
                           + findMin(arr5, 0, n5 - 1));

        int arr6[] = { 5, 6, 7, 1, 2, 3, 4 };
        int n6 = arr6.length;
        System.out.println("The minimum element is "
                           + findMin(arr6, 0, n6 - 1));

        int arr7[] = { 1, 2, 3, 4, 5, 6, 7 };
        int n7 = arr7.length;
        System.out.println("The minimum element is "
                           + findMin(arr7, 0, n7 - 1));

        int arr8[] = { 2, 3, 4, 5, 6, 7, 8, 1 };
        int n8 = arr8.length;
        System.out.println("The minimum element is "
                           + findMin(arr8, 0, n8 - 1));

        int arr9[] = { 3, 4, 5, 1, 2 };
        int n9 = arr9.length;
        System.out.println("The minimum element is "
                           + findMin(arr9, 0, n9 - 1));
    }
}

Python3

# Python program to find minimum element
# in a sorted and rotated array


def findMin(arr, low, 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
    mid = int((low + high)/2)

    # Check if element (mid+1) is minimum element. Consider
    # the cases like [3, 4, 5, 1, 2]
    if mid < high and arr[mid+1] < arr[mid]:
        return arr[mid+1]

    # Check if mid itself is minimum element
    if mid > low and 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
arr1 = [5, 6, 1, 2, 3, 4]
n1 = len(arr1)
print("The minimum element is " + str(findMin(arr1, 0, n1-1)))

arr2 = [1, 2, 3, 4]
n2 = len(arr2)
print("The minimum element is " + str(findMin(arr2, 0, n2-1)))

arr3 = [1]
n3 = len(arr3)
print("The minimum element is " + str(findMin(arr3, 0, n3-1)))

arr4 = [1, 2]
n4 = len(arr4)
print("The minimum element is " + str(findMin(arr4, 0, n4-1)))

arr5 = [2, 1]
n5 = len(arr5)
print("The minimum element is " + str(findMin(arr5, 0, n5-1)))

arr6 = [5, 6, 7, 1, 2, 3, 4]
n6 = len(arr6)
print("The minimum element is " + str(findMin(arr6, 0, n6-1)))

arr7 = [1, 2, 3, 4, 5, 6, 7]
n7 = len(arr7)
print("The minimum element is " + str(findMin(arr7, 0, n7-1)))

arr8 = [2, 3, 4, 5, 6, 7, 8, 1]
n8 = len(arr8)
print("The minimum element is " + str(findMin(arr8, 0, n8-1)))

arr9 = [3, 4, 5, 1, 2]
n9 = len(arr9)
print("The minimum element is " + str(findMin(arr9, 0, n9-1)))

# This code is contributed by Pratik Chhajer

C#

// C# program to find minimum element
// in a sorted and rotated array
using System;

class Minimum {

    static 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
        // (low + high)/2
        int mid = low + (high - low) / 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
    public static void Main()
    {
        int[] arr1 = { 5, 6, 1, 2, 3, 4 };
        int n1 = arr1.Length;
        Console.WriteLine("The minimum element is "
                          + findMin(arr1, 0, n1 - 1));

        int[] arr2 = { 1, 2, 3, 4 };
        int n2 = arr2.Length;
        Console.WriteLine("The minimum element is "
                          + findMin(arr2, 0, n2 - 1));

        int[] arr3 = { 1 };
        int n3 = arr3.Length;
        Console.WriteLine("The minimum element is "
                          + findMin(arr3, 0, n3 - 1));

        int[] arr4 = { 1, 2 };
        int n4 = arr4.Length;
        Console.WriteLine("The minimum element is "
                          + findMin(arr4, 0, n4 - 1));

        int[] arr5 = { 2, 1 };
        int n5 = arr5.Length;
        Console.WriteLine("The minimum element is "
                          + findMin(arr5, 0, n5 - 1));

        int[] arr6 = { 5, 6, 7, 1, 2, 3, 4 };
        int n6 = arr6.Length;
        Console.WriteLine("The minimum element is "
                          + findMin(arr6, 0, n1 - 1));

        int[] arr7 = { 1, 2, 3, 4, 5, 6, 7 };
        int n7 = arr7.Length;
        Console.WriteLine("The minimum element is "
                          + findMin(arr7, 0, n7 - 1));

        int[] arr8 = { 2, 3, 4, 5, 6, 7, 8, 1 };
        int n8 = arr8.Length;
        Console.WriteLine("The minimum element is "
                          + findMin(arr8, 0, n8 - 1));

        int[] arr9 = { 3, 4, 5, 1, 2 };
        int n9 = arr9.Length;
        Console.WriteLine("The minimum element is "
                          + findMin(arr9, 0, n9 - 1));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to find minimum
// element in a sorted and
// rotated array

function findMin($arr, $low,
                 $high)
{
    // This condition is needed
    // to handle the case when
    // array is not rotated at all
    if ($high < $low) return $arr[0];

    // If there is only
    // one element left
    if ($high == $low) return $arr[$low];

    // Find mid
    $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 Code
$arr1 = array(5, 6, 1, 2, 3, 4);
$n1 = sizeof($arr1);
echo "The minimum element is " .
    findMin($arr1, 0, $n1 - 1) . "\n";

$arr2 = array(1, 2, 3, 4);
$n2 = sizeof($arr2);
echo "The minimum element is " .
    findMin($arr2, 0, $n2 - 1) . "\n";

$arr3 = array(1);
$n3 = sizeof($arr3);
echo "The minimum element is " .
    findMin($arr3, 0, $n3 - 1) . "\n";

$arr4 = array(1, 2);
$n4 = sizeof($arr4);
echo "The minimum element is " .
    findMin($arr4, 0, $n4 - 1) . "\n";

$arr5 = array(2, 1);
$n5 = sizeof($arr5);
echo "The minimum element is " .
    findMin($arr5, 0, $n5 - 1) . "\n";

$arr6 = array(5, 6, 7, 1, 2, 3, 4);
$n6 = sizeof($arr6);
echo "The minimum element is " .
    findMin($arr6, 0, $n6 - 1) . "\n";

$arr7 = array(1, 2, 3, 4, 5, 6, 7);
$n7 = sizeof($arr7);
echo "The minimum element is " .
    findMin($arr7, 0, $n7 - 1) . "\n";

$arr8 = array(2, 3, 4, 5, 6, 7, 8, 1);
$n8 = sizeof($arr8);
echo "The minimum element is " .
    findMin($arr8, 0, $n8 - 1) . "\n";

$arr9 = array(3, 4, 5, 1, 2);
$n9 = sizeof($arr9);
echo "The minimum element is " .
    findMin($arr9, 0, $n9 - 1) . "\n";

// This code is contributed by ChitraNayal
?>

Javascript

<script>

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


    function findMin(arr,low,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
        let mid =low + Math.floor((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
    let arr1=[5, 6, 1, 2, 3, 4];
    let n1 = arr1.length;
    document.write("The minimum element is "+ findMin(arr1, 0, n1-1)+"<br>");

    let arr2=[1, 2, 3, 4];
    let n2 = arr2.length;
    document.write("The minimum element is "+ findMin(arr2, 0, n2-1)+"<br>");

    let arr3=[1];
    let n3 = arr3.length;
    document.write("The minimum element is "+ findMin(arr3, 0, n3-1)+"<br>");

    let arr4=[1,2];
    let n4 = arr4.length;
    document.write("The minimum element is "+ findMin(arr4, 0, n4-1)+"<br>");

    let arr5=[2,1];
    let n5 = arr5.length;
    document.write("The minimum element is "+ findMin(arr5, 0, n5-1)+"<br>");

    let arr6=[5, 6, 7, 1, 2, 3, 4];
    let n6 = arr6.length;
    document.write("The minimum element is "+ findMin(arr6, 0, n6-1)+"<br>");

    let arr7=[1, 2, 3, 4, 5, 6, 7];
    let n7 = arr7.length;
    document.write("The minimum element is "+ findMin(arr7, 0, n7-1)+"<br>");

    let arr8=[2, 3, 4, 5, 6, 7, 8, 1];
    let n8 = arr8.length;
    document.write("The minimum element is "+ findMin(arr8, 0, n8-1)+"<br>");

    let arr9=[3, 4, 5, 1, 2];
    let n9 = arr9.length;
    document.write("The minimum element is "+ findMin(arr9, 0, n9-1)+"<br>");

    // This code is contributed by avanitrachhadiya2155

</script>

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

Time Complexity: O(N) 
Auxiliary Space: O(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 containing 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;
}

// This code is contributed by Sania Kumari Gupta
// (kriSania804)

C

// C program to find minimum element in a sorted
// and rotated array containing duplicate elements.
#include <stdio.h>

// 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]);
    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;
}

// This code is contributed by Sania Kumari Gupta
// (kriSania804)

Java

// Java program to find minimum element
// in a sorted and rotated array containing
// duplicate elements.
import java.lang.*;
import java.util.*;

class GFG {

    // Function to find minimum element
    public static 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
    public static void main(String args[])
    {
        int arr1[] = { 5, 6, 1, 2, 3, 4 };
        int n1 = arr1.length;
        System.out.println("The minimum element is "
                           + findMin(arr1, 0, n1 - 1));

        int arr2[] = { 1, 2, 3, 4 };
        int n2 = arr2.length;
        System.out.println("The minimum element is "
                           + findMin(arr2, 0, n2 - 1));

        int arr3[] = { 1 };
        int n3 = arr3.length;
        System.out.println("The minimum element is "
                           + findMin(arr3, 0, n3 - 1));

        int arr4[] = { 1, 2 };
        int n4 = arr4.length;
        System.out.println("The minimum element is "
                           + findMin(arr4, 0, n4 - 1));

        int arr5[] = { 2, 1 };
        int n5 = arr5.length;
        System.out.println("The minimum element is "
                           + findMin(arr5, 0, n5 - 1));

        int arr6[] = { 5, 6, 7, 1, 2, 3, 4 };
        int n6 = arr6.length;
        System.out.println("The minimum element is "
                           + findMin(arr6, 0, n6 - 1));

        int arr7[] = { 1, 2, 3, 4, 5, 6, 7 };
        int n7 = arr7.length;
        System.out.println("The minimum element is "
                           + findMin(arr7, 0, n7 - 1));

        int arr8[] = { 2, 3, 4, 5, 6, 7, 8, 1 };
        int n8 = arr8.length;
        System.out.println("The minimum element is "
                           + findMin(arr8, 0, n8 - 1));

        int arr9[] = { 3, 4, 5, 1, 2 };
        int n9 = arr9.length;
        System.out.println("The minimum element is "
                           + findMin(arr9, 0, n9 - 1));
    }
}

// This is code is contributed by SoumikMondal

Python3

# Python3 program to find
# minimum element in a sorted
# and rotated array containing
# duplicate elements.

# Function to find minimum element


def findMin(arr, low, high):

    while (low < high):
        mid = low + (high - low) // 2

        if (arr[mid] == arr[high]):
            high -= 1
        elif (arr[mid] > arr[high]):
            low = mid + 1
        else:
            high = mid
    return arr[high]


# Driver code
if __name__ == '__main__':

    arr1 = [5, 6, 1, 2, 3, 4]
    n1 = len(arr1)
    print("The minimum element is ",
          findMin(arr1, 0, n1 - 1))

    arr2 = [1, 2, 3, 4]
    n2 = len(arr2)
    print("The minimum element is ",
          findMin(arr2, 0, n2 - 1))

    arr3 = [1]
    n3 = len(arr3)
    print("The minimum element is ",
          findMin(arr3, 0, n3 - 1))

    arr4 = [1, 2]
    n4 = len(arr4)
    print("The minimum element is ",
          findMin(arr4, 0, n4 - 1))

    arr5 = [2, 1]
    n5 = len(arr5)
    print("The minimum element is ",
          findMin(arr5, 0, n5 - 1))

    arr6 = [5, 6, 7, 1, 2, 3, 4]
    n6 = len(arr6)
    print("The minimum element is ",
          findMin(arr6, 0, n6 - 1))

    arr7 = [1, 2, 3, 4, 5, 6, 7]
    n7 = len(arr7)
    print("The minimum element is ",
          findMin(arr7, 0, n7 - 1))

    arr8 = [2, 3, 4, 5, 6, 7, 8, 1]
    n8 = len(arr8)
    print("The minimum element is ",
          findMin(arr8, 0, n8 - 1))

    arr9 = [3, 4, 5, 1, 2]
    n9 = len(arr9)
    print("The minimum element is ",
          findMin(arr9, 0, n9 - 1))

# This code is contributed by Princi Singh

C#

// C# program to find minimum element
// in a sorted and rotated array
// containing duplicate elements.
using System;

class GFG {

    // Function to find minimum element
    public static 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
    public static void Main(String[] args)
    {
        int[] arr1 = { 5, 6, 1, 2, 3, 4 };
        int n1 = arr1.Length;
        Console.WriteLine("The minimum element is "
                          + findMin(arr1, 0, n1 - 1));

        int[] arr2 = { 1, 2, 3, 4 };
        int n2 = arr2.Length;
        Console.WriteLine("The minimum element is "
                          + findMin(arr2, 0, n2 - 1));

        int[] arr3 = { 1 };
        int n3 = arr3.Length;
        Console.WriteLine("The minimum element is "
                          + findMin(arr3, 0, n3 - 1));

        int[] arr4 = { 1, 2 };
        int n4 = arr4.Length;
        Console.WriteLine("The minimum element is "
                          + findMin(arr4, 0, n4 - 1));

        int[] arr5 = { 2, 1 };
        int n5 = arr5.Length;
        Console.WriteLine("The minimum element is "
                          + findMin(arr5, 0, n5 - 1));

        int[] arr6 = { 5, 6, 7, 1, 2, 3, 4 };
        int n6 = arr6.Length;
        Console.WriteLine("The minimum element is "
                          + findMin(arr6, 0, n6 - 1));

        int[] arr7 = { 1, 2, 3, 4, 5, 6, 7 };
        int n7 = arr7.Length;
        Console.WriteLine("The minimum element is "
                          + findMin(arr7, 0, n7 - 1));

        int[] arr8 = { 2, 3, 4, 5, 6, 7, 8, 1 };
        int n8 = arr8.Length;
        Console.WriteLine("The minimum element is "
                          + findMin(arr8, 0, n8 - 1));

        int[] arr9 = { 3, 4, 5, 1, 2 };
        int n9 = arr9.Length;
        Console.WriteLine("The minimum element is "
                          + findMin(arr9, 0, n9 - 1));
    }
}

// This code is contributed by Amit Katiyar

Javascript

<script>
// JavaScript program to find minimum element in a sorted
// and rotated array containing duplicate elements.

// Function to find minimum element
function findMin(arr, low, high)
{
    while(low < high)
    {
        let mid = Math.floor(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];
}

var arr1 = [5, 6, 1, 2, 3, 4];
var n1 = arr1.length;
document.write("The minimum element is " + findMin(arr1, 0, n1-1) + "<br>");

var arr2 = [1, 2, 3, 4];
var n2 = arr2.length;
document.write("The minimum element is " + findMin(arr2, 0, n2-1) + "<br>");

var arr3 = [1];
var n3 = arr3.length;
document.write("The minimum element is " + findMin(arr3, 0, n3-1) + "<br>");

var arr4 = [1, 2];
var n4 = arr4.length;
document.write("The minimum element is " + findMin(arr4, 0, n4-1) + "<br>");

var arr5 = [2, 1];
var n5 = arr5.length;
document.write("The minimum element is " + findMin(arr5, 0, n5-1) + "<br>");

var arr6 = [5, 6, 7, 1, 2, 3, 4];
var n6 = arr6.length;
document.write("The minimum element is " + findMin(arr6, 0, n6-1) + "<br>");

var arr7 = [1, 2, 3, 4, 5, 6, 7];
var n7 = arr7.length;
document.write("The minimum element is " + findMin(arr7, 0, n7-1) + "<br>");

var arr8 = [2, 3, 4, 5, 6, 7, 8, 1];
var n8 = arr8.length;
document.write("The minimum element is " + findMin(arr8, 0, n8-1) + "<br>");

var arr9 = [3, 4, 5, 1, 2];
var n9 = arr9.length;
document.write("The minimum element is " + findMin(arr9, 0, n9-1) + "<br>");

// This code is contributed by probinsah
</script>

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

Time Complexity: O(log N) 
Auxiliary Space: O(1)

This article is contributed by Abhay Rathi. Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.