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# 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 ``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

 ``

## Javascript

 ``

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 ``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

 ``

Output

`YES`

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