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# Java Program for 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:

## 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.`

Output

`YES`

Time Complexity: O(N), where N represents the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Please refer complete article on Check if an array is sorted and rotated for more details!

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