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// 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. |
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!
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
