Check if an array is sorted and rotated

Given an array of N distinct integers. The task is to write a program to check if this array is sorted and rotated clockwise. 
Note: A sorted array is not considered as sorted and rotated, i.e., there should be at least one rotation

Examples:

Input: arr[] = { 3, 4, 5, 1, 2 }
Output: YES
Explanation: The above array is sorted and rotated
Sorted array: {1, 2, 3, 4, 5}
Rotating this sorted array clockwise 
by 3 positions, we get: { 3, 4, 5, 1, 2}

Input: arr[] = {3, 4, 6, 1, 2, 5}
Output: NO

Check if an array is sorted and rotated by finding the pivot element(minimum element):

To solve the problem follow the below idea:

Find the minimum element in the array and if the array has been sorted and rotated then the elements before and after the minimum element will be in sorted order only

Follow the given steps to solve the problem:

• Find the minimum element in the array.
• Now, if the array is sorted and then rotated, then all the elements before the minimum element will be in increasing order and all elements after the minimum element will also be in increasing order.
• Check if all elements before the minimum element are in increasing order.
• Check if all elements after the minimum element are in increasing order.
• Check if the last element of the array is smaller than the starting element, as this would make sure that the array has been rotated at least once.
• If all of the above three conditions satisfies then print YES otherwise print NO

Below is the implementation of the above approach:

C++

// CPP program to check if an array is
// sorted and rotated clockwise
#include <bits/stdc++.h>
using namespace std;

// Function to check if an array is
// sorted and rotated clockwise
void checkIfSortRotated(int arr[], int n)
{
    int minEle = INT_MAX;
    int maxEle = INT_MIN;

    int minIndex = -1;

    // Find the minimum element
    // and it's index
    for (int i = 0; i < n; i++) {
        if (arr[i] < minEle) {
            minEle = arr[i];
            minIndex = i;
        }
    }

    int flag1 = 1;

    // Check if all elements before minIndex
    // are in increasing order
    for (int i = 1; i < minIndex; i++) {
        if (arr[i] < arr[i - 1]) {
            flag1 = 0;
            break;
        }
    }

    int flag2 = 1;

    // Check if all elements after minIndex
    // are in increasing order
    for (int i = minIndex + 1; i < n; i++) {
        if (arr[i] < arr[i - 1]) {
            flag2 = 0;
            break;
        }
    }

    // Check if last element of the array
    // is smaller than the element just
    // starting element of the array
    // for arrays like [3,4,6,1,2,5] - not circular array
    if (flag1 && flag2 && (arr[n - 1] < arr[0]))
        cout << "YES";
    else
        cout << "NO";
}

// Driver code
int main()
{
    int arr[] = { 3, 4, 5, 1, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    checkIfSortRotated(arr, n);
    return 0;
}

Java

// Java program to check if an
// array is sorted and rotated
// clockwise
import java.io.*;

class GFG {

    // Function to check if an array is
    // sorted and rotated clockwise
    static void checkIfSortRotated(int arr[], int n)
    {
        int minEle = Integer.MAX_VALUE;
        int maxEle = Integer.MIN_VALUE;

        int minIndex = -1;

        // Find the minimum element
        // and it's index
        for (int i = 0; i < n; i++) {
            if (arr[i] < minEle) {
                minEle = arr[i];
                minIndex = i;
            }
        }

        boolean flag1 = true;

        // Check if all elements before
        // minIndex are in increasing order
        for (int i = 1; i < minIndex; i++) {
            if (arr[i] < arr[i - 1]) {
                flag1 = false;
                break;
            }
        }

        boolean flag2 = true;

        // Check if all elements after
        // minIndex are in increasing order
        for (int i = minIndex + 1; i < n; i++) {
            if (arr[i] < arr[i - 1]) {
                flag2 = false;
                break;
            }
        }

        // check if the minIndex is 0, i.e the first element
        // is the smallest , which is the case when array is
        // sorted but not rotated.
        if (minIndex == 0) {
            System.out.print("NO");
            return;
        }
        // Check if last element of the array
        // is smaller than the element just
        // before the element at minIndex
        // starting element of the array
        // for arrays like [3,4,6,1,2,5] - not sorted
        // circular array
        if (flag1 && flag2 && (arr[n - 1] < arr[0]))
            System.out.println("YES");
        else
            System.out.print("NO");
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 3, 4, 5, 1, 2 };

        int n = arr.length;

        // Function Call
        checkIfSortRotated(arr, n);
    }
}

// This code is contributed
// by inder_verma.

Python3

# Python3 program to check if an
# array is sorted and rotated clockwise
import sys

# Function to check if an array is
# sorted and rotated clockwise


def checkIfSortRotated(arr, n):
    minEle = sys.maxsize
    maxEle = -sys.maxsize - 1
    minIndex = -1

    # Find the minimum element
    # and it's index
    for i in range(n):
        if arr[i] < minEle:
            minEle = arr[i]
            minIndex = i
    flag1 = 1

    # Check if all elements before
    # minIndex are in increasing order
    for i in range(1, minIndex):
        if arr[i] < arr[i - 1]:
            flag1 = 0
            break
    flag2 = 2

    # Check if all elements after
    # minIndex are in increasing order
    for i in range(minIndex + 1, n):
        if arr[i] < arr[i - 1]:
            flag2 = 0
            break

    # Check if last element of the array
    # is smaller than the element just
    # before the element at minIndex
    # starting element of the array
    # for arrays like [3,4,6,1,2,5] - not sorted circular array
    if (flag1 and flag2 and
            arr[n - 1] < arr[0]):
        print("YES")
    else:
        print("NO")


# Driver code
if __name__ == "__main__":
    arr = [3, 4, 5, 1, 2]
    n = len(arr)

    # Function Call
    checkIfSortRotated(arr, n)

# This code is contributed
# by Shrikant13

C#

// C# program to check if an
// array is sorted and rotated
// clockwise
using System;

class GFG {

    // Function to check if an array is
    // sorted and rotated clockwise
    static void checkIfSortRotated(int[] arr, int n)
    {
        int minEle = int.MaxValue;
        // int maxEle = int.MinValue;

        int minIndex = -1;

        // Find the minimum element
        // and it's index
        for (int i = 0; i < n; i++) {
            if (arr[i] < minEle) {
                minEle = arr[i];
                minIndex = i;
            }
        }

        bool flag1 = true;

        // Check if all elements before
        // minIndex are in increasing order
        for (int i = 1; i < minIndex; i++) {
            if (arr[i] < arr[i - 1]) {
                flag1 = false;
                break;
            }
        }

        bool flag2 = true;

        // Check if all elements after
        // minIndex are in increasing order
        for (int i = minIndex + 1; i < n; i++) {
            if (arr[i] < arr[i - 1]) {
                flag2 = false;
                break;
            }
        }

        // Check if last element of the array
        // is smaller than the element just
        // before the element at minIndex
        // starting element of the array
        // for arrays like [3,4,6,1,2,5] - not circular
        // array
        if (flag1 && flag2 && (arr[n - 1] < arr[0]))
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
    }

    // Driver code
    public static void Main()
    {
        int[] arr = { 3, 4, 5, 1, 2 };

        int n = arr.Length;

        // Function Call
        checkIfSortRotated(arr, n);
    }
}

// This code is contributed
// by inder_verma.

PHP

<?php
// PHP program to check if an
// array is sorted and rotated
// clockwise

// Function to check if an array
// is sorted and rotated clockwise
function checkIfSortRotated($arr, $n)
{
    $minEle = PHP_INT_MAX;
    $maxEle = PHP_INT_MIN;

    $minIndex = -1;

    // Find the minimum element
    // and it's index
    for ($i = 0; $i <$n; $i++)
    {
        if ($arr[$i] < $minEle)
        {
            $minEle = $arr[$i];
            $minIndex = $i;
        }
    }

    $flag1 = 1;

    // Check if all elements before
    // minIndex are in increasing order
    for ( $i = 1; $i <$minIndex; $i++)
    {
        if ($arr[$i] < $arr[$i - 1])
        {
            $flag1 = 0;
            break;
        }
    }

    $flag2 = 1;

    // Check if all elements after
    // minIndex are in increasing order
    for ($i = $minIndex + 1; $i <$n; $i++)
    {
        if ($arr[$i] < $arr[$i - 1])
        {
            $flag2 = 0;
            break;
        }
    }

    // Check if last element of the array
    // is smaller than the element just
    // starting element of the array
    // for arrays like [3,4,6,1,2,5] - not sorted circular array
    if ($flag1 && $flag2 &&
       ($arr[$n - 1] < $arr[0]))
        echo( "YES");
    else
        echo( "NO");
}

// Driver code
$arr = array(3, 4, 5, 1, 2);

//Function Call
$n = count($arr);

checkIfSortRotated($arr, $n);

// This code is contributed
// by inder_verma.
?>

Javascript

<script>
    // Javascript program to check if an
    // array is sorted and rotated
    // clockwise

    // Function to check if an array is
    // sorted and rotated clockwise
    function checkIfSortRotated(arr, n)
    {
        let minEle = Number.MAX_VALUE;
        // int maxEle = int.MinValue;

        let minIndex = -1;

        // Find the minimum element
        // and it's index
        for (let i = 0; i < n; i++) {
            if (arr[i] < minEle) {
                minEle = arr[i];
                minIndex = i;
            }
        }

        let flag1 = true;

        // Check if all elements before
        // minIndex are in increasing order
        for (let i = 1; i < minIndex; i++) {
            if (arr[i] < arr[i - 1]) {
                flag1 = false;
                break;
            }
        }

        let flag2 = true;

        // Check if all elements after
        // minIndex are in increasing order
        for (let i = minIndex + 1; i < n; i++) {
            if (arr[i] < arr[i - 1]) {
                flag2 = false;
                break;
            }
        }

        // Check if last element of the array
        // is smaller than the element just
        // before the element at minIndex
        // starting element of the array
        // for arrays like [3,4,6,1,2,5] - not circular
        // array
        if (flag1 && flag2 && (arr[n - 1] < arr[0]))
            document.write("YES");
        else
            document.write("NO");
    }

    let arr = [ 3, 4, 5, 1, 2 ];

    let n = arr.length;

    // Function Call
    checkIfSortRotated(arr, n);

// This code is contributed by mukesh07.
</script>

Output

YES

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

Check if an array is sorted and rotated by counting adjacent inversions:

To solve the problem follow the below idea:

As the array is sorted and rotated, so the case when the previous element is greater than the current element will occur only once. If this occurs zero times or more than one times then the array is not properly sorted and rotated

Follow the given steps to solve the problem:

• Take two variables to say x and y, initialized as 0
• Now traverse the array
• If the previous element is smaller than the current, increment x by one
• Else increment y by one.
• After traversal, if y is not equal to 1, return false.
• Then compare the last element with the first element (0th element as current, and last element as previous.) i.e. if the last element is greater increase y by one else increase x by one
• Again check if y equals one return true. i.e. Array is sorted and rotated. Else return false

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;

bool checkRotatedAndSorted(int arr[], int n)
{

    // Your code here
    // Initializing two variables x,y as zero.
    int x = 0, y = 0;

    // Traversing array 0 to last element.
    // n-1 is taken as we used i+1.
    for (int i = 0; i < n - 1; i++) {
        if (arr[i] < arr[i + 1])
            x++;
        else
            y++;
    }

    // If till now both x,y are greater than 1 means
    // array is not sorted. If both any of x,y is zero
    // means array is not rotated.
    if (y == 1) {
        // Checking for last element with first.
        if (arr[n - 1] < arr[0])
            x++;
        else
            y++;

        // Checking for final result.
        if (y == 1)
            return true;
    }

    // If still not true then definitely false.
    return false;
}

//  Driver Code Starts.

int main()
{
    int arr[] = { 4, 5, 1, 2, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    if (checkRotatedAndSorted(arr, n))
        cout << "YES" << endl;
    else
        cout << "NO" << endl;

    return 0;
}

Java

import java.io.*;

class GFG {

    // Function to check if an array is
    // Sorted and rotated clockwise
    static boolean checkIfSortRotated(int arr[], int n)
    {
        // Initializing two variables x,y as zero.
        int x = 0, y = 0;

        // Traversing array 0 to last element.
        // n-1 is taken as we used i+1.
        for (int i = 0; i < n - 1; i++) {
            if (arr[i] < arr[i + 1])
                x++;
            else
                y++;
        }

        // If till now both x,y are greater
        // than 1 means array is not sorted.
        // If both any of x,y is zero means
        // array is not rotated.
        if (y == 1) {
            // Checking for last element with first.
            if (arr[n - 1] < arr[0])
                x++;
            else
                y++;

            // Checking for final result.
            if (y == 1)
                return true;
        }
        // If still not true then definitely false.
        return false;
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 5, 1, 2, 3, 4 };

        int n = arr.length;

        // Function Call
        boolean x = checkIfSortRotated(arr, n);
        if (x == true)
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}

Python3

def checkRotatedAndSorted(arr, n):

    # Your code here
    # Initializing two variables x,y as zero.
    x = 0
    y = 0

    # Traversing array 0 to last element.
    # n-1 is taken as we used i+1.
    for i in range(n-1):
        if (arr[i] < arr[i + 1]):
            x += 1
        else:
            y += 1

    # If till now both x,y are greater than 1 means
    # array is not sorted. If both any of x,y is zero
    # means array is not rotated.
    if (y == 1):
        # Checking for last element with first.
        if (arr[n - 1] < arr[0]):
            x += 1
        else:
            y += 1

        # Checking for final result.
        if (y == 1):
            return True

    # If still not true then definitely false.
    return False


# Driver Code Starts.
arr = [4, 5, 1, 2, 3]
n = len(arr)

# Function Call
if (checkRotatedAndSorted(arr, n)):
    print("YES")
else:
    print("NO")

# This code is contributed by shivanisinghss2110

C#

using System;
public class GFG {

    // Function to check if an array is
    // Sorted and rotated clockwise
    static bool checkIfSortRotated(int[] arr, int n)
    {
        // Initializing two variables x,y as zero.
        int x = 0, y = 0;

        // Traversing array 0 to last element.
        // n-1 is taken as we used i+1.
        for (int i = 0; i < n - 1; i++) {
            if (arr[i] < arr[i + 1])
                x++;
            else
                y++;
        }

        // If till now both x,y are greater
        // than 1 means array is not sorted.
        // If both any of x,y is zero means
        // array is not rotated.
        if (y == 1) {
            // Checking for last element with first.
            if (arr[n - 1] < arr[0])
                x++;
            else
                y++;

            // Checking for readonly result.
            if (y == 1)
                return true;
        }
        // If still not true then definitely false.
        return false;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = { 5, 1, 2, 3, 4 };

        int n = arr.Length;

        // Function Call
        bool x = checkIfSortRotated(arr, n);
        if (x == true)
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
    }
}

// This code is contributed by umadevi9616

Javascript

<script>
// JavaScript Function to check if an array is
// Sorted and rotated clockwise
function checkIfSortRotated(arr, n)
    {

        // Initializing two variables x,y as zero.
        var x = 0, y = 0;

        // Traversing array 0 to last element.
        // n-1 is taken as we used i+1.
        for (var i = 0; i < n - 1; i++) {
            if (arr[i] < arr[i + 1])
                x++;
            else
                y++;
        }

        // If till now both x,y are greater
        // than 1 means array is not sorted.
        // If both any of x,y is zero means
        // array is not rotated.
        if (y == 1) {
            // Checking for last element with first.
            if (arr[n - 1] < arr[0])
                x++;
            else
                y++;

            // Checking for final result.
            if (y == 1)
                return true;
        }
        // If still not true then definitely false.
        return false;
    }

    // Driver code
        var arr = [ 5, 1, 2, 3, 4 ];

        var n = arr.length;

        // Function Call
        var x = checkIfSortRotated(arr, n);
        if (x == true)
            document.write("YES");
        else
            document.write("NO");

// This code is contributed by shivanisinghss2110
</script>

Output

YES

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