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 counter-clockwise. A sorted array is not considered as sorted and rotated, i.e., there should at least one rotation.
Examples: 

Input : arr[] = { 3, 4, 5, 1, 2 }
Output : YES
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[] = {7, 9, 11, 12, 5}
Output: YES

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

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

Approach: 

• Find the minimum element in the array.
• Now, if the array is sorted and then rotate 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 minimum element are in increasing order.
• Check if all elements after minimum element are in increasing order.
• Check if the last element of the array is smaller than the starting element.
• If all of the above three conditions satisfies then print YES otherwise print NO.

Below is the implementation of the above idea: 

C++

// CPP program to check if an array is
// sorted and rotated clockwise
#include <climits>
#include <iostream>

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
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

Method 2: 

Approach:

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

Explanation:

• The idea is simple. If array is sorted and rotated , element are either increasing or decreasing, but with one exception. So we count how many times the element is greater then its previous element, and how many times the element is smaller then its previous element. 
• The special case is when array is sorted but not rotated. for this check last element with first element specially.

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 (x == 1 || y == 1) {
        // Checking for last element with first.
        if (arr[n - 1] < arr[0])
            x++;
        else
            y++;

        // Checking for final result.
        if (x == 1 || 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 (x == 1 || y == 1) {
            // Checking for last element with first.
            if (arr[n - 1] < arr[0])
                x++;
            else
                y++;

            // Checking for final result.
            if (x == 1 || 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 (x == 1 or y == 1):
        # Checking for last element with first.
        if (arr[n - 1] < arr[0]):
            x += 1
        else:
            y += 1

        # Checking for final result.
        if (x == 1 or 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 (x == 1 || y == 1) {
            // Checking for last element with first.
            if (arr[n - 1] < arr[0])
                x++;
            else
                y++;

            // Checking for readonly result.
            if (x == 1 || 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 (x == 1 || y == 1) {
            // Checking for last element with first.
            if (arr[n - 1] < arr[0])
                x++;
            else
                y++;

            // Checking for final result.
            if (x == 1 || 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)