Check if an array is sorted and rotated

16 May, 2022
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++
Java
Python3
C#
PHP
Javascript
// 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 clockwisevoid 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 codeint main(){    int arr[] = { 3, 4, 5, 1, 2 };    int n = sizeof(arr) / sizeof(arr[0]);     // Function Call    checkIfSortRotated(arr, n);    return 0;}
// Java program to check if an// array is sorted and rotated// clockwiseimport 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 program to check if an# array is sorted and rotated clockwiseimport 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 codearr = [3, 4, 5, 1, 2]n = len(arr) # Function CallcheckIfSortRotated(arr, n) # This code is contributed# by Shrikant13
// C# program to check if an// array is sorted and rotated// clockwiseusing 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 program to check if an// array is sorted and rotated// clockwise // Function to check if an array// is sorted and rotated clockwisefunction 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.?>
<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>
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++
Java
Python3
C#
Javascript
#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;}
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");    }}
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 Callif (checkRotatedAndSorted(arr, n)):    print("YES")else:    print("NO") # This code is contributed by shivanisinghss2110
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
<script>// JavaScript Function to check if an array is// Sorted and rotated clockwisefunction 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>
YES
Time Complexity: O(N)Auxiliary Space: O(1)
inderDuMCA
shrikanth13
shubhadanayak93
gaurav joshi 8
SROTASWINI SAHOO
mukesh07
btech19eskcs099
umadevi9616
shivanisinghss2110
clintra
simmytarika5
rotation
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