Find the Minimum element in a Sorted and Rotated Array

Difficulty Level : Medium
Last Updated : 26 Mar, 2023

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Given a sorted array arr[] (may be distinct or may contain duplicates) of size N that is rotated at some unknown point, the task is to find the minimum element in it.

Examples:

Input: arr[] = {5, 6, 1, 2, 3, 4}
Output: 1
Explanation: 1 is the minimum element present in the array.

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

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

Find the minimum element in a sorted and rotated array using Linear Search:

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

Follow the steps mentioned below to implement the idea:

Declare a variable (say min_ele) to store the minimum value and initialize it with arr[0].
Traverse the array from the start.
- Update the minimum value (min_ele) if the current element is less than it.
Return the final value of min_ele as the required answer.

Below is the implementation of the above approach.

C++

// C++ code  to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum value
int findMin(int arr[], int n)
{
    int min_ele = arr[0];
 
    // Traversing over array to
    // find minimum element
    for (int i = 0; i < n; i++) {
        if (arr[i] < min_ele) {
            min_ele = arr[i];
        }
    }
 
    return min_ele;
}
 
// Driver code
int main()
{
    int arr[] = { 5, 6, 1, 2, 3, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    cout << findMin(arr, N) << endl;
    return 0;
}

Java

/*package whatever //do not write package name here */
import java.io.*;
 
class GFG {
 
  // Function to find the minimum value
  static int findMin(int arr[], int n)
  {
    int min_ele = arr[0];
 
    // Traversing over array to
    // find minimum element
    for (int i = 0; i < n; i++) {
      if (arr[i] < min_ele) {
        min_ele = arr[i];
      }
    }
 
    return min_ele;
  }
 
  public static void main (String[] args) {
    int arr[] = { 5, 6, 1, 2, 3, 4 };
    int N = arr.length;
    System.out.println(findMin(arr, N));
  }
}
 
// This code is contributed by aadityaburujwale.

Python3

# python3 code  to implement the approach
 
def findMin(arr, N):
     
    min_ele = arr[0];
 
    # Traversing over array to
    # find minimum element
    for i in range(N) :
        if arr[i] < min_ele :
            min_ele = arr[i]
 
    return min_ele;
 
# Driver program
arr = [5, 6, 1, 2, 3, 4]
N = len(arr)
 
print(findMin(arr,N))
 
# This code is contributed by aditya942003patil

C#

// C# code to implement above approach
using System;
  
class Minimum {
  
    static int findMin(int[] arr, int N)
    {
        int min_ele = arr[0];
         
        // Traversing over array to
        // find minimum element
        for (int i = 0; i < N; i++) {
            if (arr[i] < min_ele) {
                min_ele = arr[i];
            }
        }
         
        return min_ele;
    }
  
    // Driver Program
    public static void Main()
    {
        int[] arr = { 5, 6, 1, 2, 3, 4 };
        int N = arr.Length;
        
        Console.WriteLine(findMin(arr, N));
  
    }
}
  
// This code is contributed by aditya942003patil.

Javascript

// JS code to implement the approach
 
// Function to find the minimum value
function findMin(arr, n) {
    let min_ele = arr[0];
 
    // Traversing over array to
    // find minimum element
    for (let i = 0; i < n; i++) {
        if (arr[i] < min_ele) {
            min_ele = arr[i];
        }
    }
 
    return min_ele;
}
 
// Driver code
let arr = [5, 6, 1, 2, 3, 4];
let N = arr.length;
 
// Function call
console.log(findMin(arr, N));
 
// This code is contributed by adityamaharshi21.

Output

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

The Easy Way using STL:

The Approach:

The is verry simple here we have array/vector we use *min_element function in stl and print the minimum element in the vector/array.

C++

#include <iostream>
#include<bits/stdc++.h>
using namespace std;
 
int main() {
    vector<int>v{5, 6, 1, 2, 3, 4};
    cout<<"The Minimum Element in the vector is: ";
    cout<<*min_element(v.begin(),v.end())<<endl;
    return 0;
}

Java

/*package whatever //do not write package name here */
 
import java.util.*;
 
class GFG {
    public static void main(String[] args)
    {
        List<Integer> v = new ArrayList<>(
            Arrays.asList(5, 6, 1, 2, 3, 4));
        System.out.print(
            "The Minimum Element in the vector is: ");
        System.out.println(Collections.min(v));
    }
}
// contributed by akashish__

Python3

from typing import List
 
def find_min(arr: List[int]) -> int:
    return min(arr)
 
v = [5, 6, 1, 2, 3, 4]
print("The Minimum Element in the vector is: ", find_min(v))

C#

// C# program to find minimum elementin array
using System;
using System.Linq;
class GFG
{
  static int find_min(int []arr)
      {
          return arr.Min();
      }
 
  // Driver code
  public static void Main(String[] args)
      {
        int []arr = {5, 6, 1, 2, 3, 4};
        Console.WriteLine("The Minimum Element in the vector is: " + find_min(arr));
      }
}
 
// This code is contributed by Pratik Gupta (guptapratik)

Javascript

function findMin(arr) {
  return Math.min(...arr);
}
 
let v = [5, 6, 1, 2, 3, 4];
console.log("The Minimum Element in the vector is: ", findMin(v));

Output

The Minimum Element in the vector is: 1

Time Complexity: O(N),in the worst case.
Auxiliary Space: O(1)

Find the minimum element in a sorted and rotated array using Binary Search:

This approach is 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.

Follow the steps below to solve the given problem:

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 arr[] = { 5, 6, 1, 2, 3, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);
   
    // Function call
    cout << "The minimum element is "
         << findMin(arr, 0, N - 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. (low + high)/2
    int mid = low + (high - low) / 2;
 
    // Check if element (mid+1) is minimum element. 
    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 arr[] = { 5, 6, 1, 2, 3, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);
   
    // Function call
    printf("The minimum element is %d\n",
           findMin(arr, 0, N - 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 arr[] = { 5, 6, 1, 2, 3, 4 };
        int N = arr.length;
        System.out.println("The minimum element is "
                           + findMin(arr, 0, N - 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
if __name__ == '__main__':
    arr = [5, 6, 1, 2, 3, 4]
    N = len(arr)
    print("The minimum element is " + \
          str(findMin(arr, 0, N-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[] arr = { 5, 6, 1, 2, 3, 4 };
        int N = arr.Length;
       
        Console.WriteLine("The minimum element is "
                          + findMin(arr, 0, N - 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
$arr = array(5, 6, 1, 2, 3, 4);
$N = sizeof($arr);
echo "The minimum element is " . 
    findMin($arr, 0, $N - 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 arr=[5, 6, 1, 2, 3, 4];
    let N = arr.length;
    document.write("The minimum element is "+ findMin(arr, 0, N-1)+"<br>");
     
    // This code is contributed by avanitrachhadiya2155
     
</script>

Output

The minimum element is 1

Time Complexity: O(logN), using binary search
Auxiliary Space: O(1)

Find the minimum element in a sorted and rotated array using Modified Binary Search:

The findMin function takes three arguments: arr (the input array), low (the lowest index of the array), and high (the highest index of the array).
If the array is not rotated, i.e., the first element is less than or equal to the last element, then the first element is returned as the minimum element.
Otherwise, a binary search algorithm is implemented to find the minimum element.
The binary search loop continues until the low index is less than or equal to the high index.
Inside the loop, the middle index mid is calculated as the average of the low and high indices.
If the element at the mid index is less than the element at mid-1 index, then the element at the mid index is returned as the minimum element.
If the element at the mid index is greater than the element at the high index, then the minimum element must be in the left half of the array. So, the low index is updated to mid+1.
Otherwise, the minimum element must be in the right half of the array. So, the high index is updated to mid-1.
If no minimum element is found during the binary search, then the findMin function returns None.
In the driver program, an input array arr is defined with rotated and sorted elements, and the findMin function is called with low and high indices set to 0 and N-1, respectively, where N is the length of the input array.
The minimum element is printed using the print function with the help of the str function to convert the returned value to a string

Illustration:

Let the array be arr = [7, 8, 9, 1, 2, 3, 4, 5, 6]. Here, the minimum element in the array is 1.

We start with low = 0 and high = 8.

First iteration:

mid = (low + high) // 2 = 4

arr[mid] = 2, arr[mid-1] = 1

arr[mid] < arr[mid-1], so we return arr[mid] which is 2.

So the minimum element is found at index 3 and arr[3] = 1.

Below is the code implementation of the above approach:

Python3

def findMin(arr, low, high):
    # If the array is not rotated
    if arr[low] <= arr[high]:
        return arr[low]
 
    # Binary search
    while low <= high:
        mid = (low + high) // 2
 
        # Check if mid is the minimum element
        if arr[mid] < arr[mid-1]:
            return arr[mid]
 
        # If the right half is sorted, the minimum element must be in the left half
        if arr[mid] > arr[high]:
            low = mid + 1
 
        # If the left half is sorted, the minimum element must be in the right half
        else:
            high = mid - 1
 
    # If no minimum element is found, return None
    return None
# Driver program to test above functions
if __name__ == '__main__':
    arr = [5, 6, 1, 2, 3, 4]
    N = len(arr)
    print("The minimum element is " + \
          str(findMin(arr, 0, N-1)))

Output

The minimum element is 1

Time complexity: O(log n) – where n is the number of elements in the array. This is because the algorithm uses binary search, which has a logarithmic time complexity.

Auxiliary Space: O(1) – the algorithm uses a constant amount of extra space to store variables such as low, high, and mid, regardless of the size of the input array.

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