# Check if an array is sorted and rotated

• Difficulty Level : Easy
• Last Updated : 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++

 `// CPP program to check if an array is``// sorted and rotated clockwise``#include ``#include ` `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))``        ``cout << ``"YES"``;``    ``else``        ``cout << ``"NO"``;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 3, 4, 5, 1, 2 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``// 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))``            ``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

 ``

## Javascript

 ``
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 ``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)``            ``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);` `    ``// 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)``                ``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

 ``
Output
`YES`

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

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