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# C++ 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:

## 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;``}`

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.

### Approach Name: Linear Search

Steps:

1. Traverse the array from the second element to the last element and check if the current element is smaller than the previous element. If so, then the array is rotated.
2. If the array is rotated, then find the index of the minimum element in the array. This can be done by traversing the array again and keeping track of the minimum element and its index.
3. Check if the array is sorted by comparing the elements starting from the minimum element to the previous element in a circular manner. If all elements are in non-descending order, then the array is sorted and rotated.

## C++

 `#include ``using` `namespace` `std;` `bool` `isSortedAndRotated(``int` `arr[], ``int` `n) {``    ``bool` `rotated = ``false``;``    ``int` `min_index = 0;``    ``int` `min_element = arr;``    ``for` `(``int` `i = 1; i < n; i++) {``        ``if` `(arr[i] < arr[i - 1]) {``            ``rotated = ``true``;``        ``}``        ``if` `(arr[i] < min_element) {``            ``min_index = i;``            ``min_element = arr[i];``        ``}``    ``}``    ``if` `(!rotated) {``        ``return` `false``;``    ``}``    ``for` `(``int` `i = 1; i < n; i++) {``        ``int` `index = (min_index + i) % n;``        ``int` `prev_index = (min_index + i - 1) % n;``        ``if` `(arr[index] < arr[prev_index]) {``            ``return` `false``;``        ``}``    ``}``    ``return` `true``;``}` `int` `main() {``    ``int` `arr1[] = { 3, 4, 5, 1, 2 };``    ``int` `arr2[] = { 7, 9, 11, 12, 5 };``    ``int` `arr3[] = { 1, 2, 3 };``    ``int` `arr4[] = { 3, 4, 6, 1, 2, 5 };``    ``int` `n1 = ``sizeof``(arr1) / ``sizeof``(arr1);``    ``int` `n2 = ``sizeof``(arr2) / ``sizeof``(arr2);``    ``int` `n3 = ``sizeof``(arr3) / ``sizeof``(arr3);``    ``int` `n4 = ``sizeof``(arr4) / ``sizeof``(arr4);``    ``if` `(isSortedAndRotated(arr1, n1)) {``        ``cout << ``"YES\n"``;``    ``} ``else` `{``        ``cout << ``"NO\n"``;``    ``}``    ``if` `(isSortedAndRotated(arr2, n2)) {``        ``cout << ``"YES\n"``;``    ``} ``else` `{``        ``cout << ``"NO\n"``;``    ``}``    ``if` `(isSortedAndRotated(arr3, n3)) {``        ``cout << ``"YES\n"``;``    ``} ``else` `{``        ``cout << ``"NO\n"``;``    ``}``    ``if` `(isSortedAndRotated(arr4, n4)) {``        ``cout << ``"YES\n"``;``    ``} ``else` `{``        ``cout << ``"NO\n"``;``    ``}``    ``return` `0;``}`

Output

```YES
YES
NO
NO
```

Time Complexity: O(n), where n is the size of the input array.

Auxiliary Space: O(1)

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

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