Given an array arr[] of N distinct elements, the task is to check if it is possible to make the array increasing or decreasing by rotating the array in any direction.
Examples:
Input: arr[] = {4, 5, 6, 2, 3}
Output: Yes
Array can be rotated as {2, 3, 4, 5, 6}
Input: arr[] = {1, 2, 4, 3, 5}
Output: No
Approach: There are four possibilities:
- If the array is already increasing then the answer is Yes.
- If the array is already decreasing then the answer is Yes.
- If the array can be made increasing, this can be possible if the given array is first increasing up to the maximum element and then decreasing.
- If the array can be made decreasing, this can be possible if the given array is first decreasing up to the minimum element and then increasing.
If it is not possible to make the array increasing or decreasing then print No.
Below is the implementation of the above approach:
Java
// Java implementation of the approachclass GFG{ // Function that returns true if the array// can be made increasing or decreasing// after rotating it in any directionstatic boolean isPossible(int a[], int n){ // If size of the array is less than 3 if (n <= 2) return true; int flag = 0; // Check if the array is already decreasing for (int i = 0; i < n - 2; i++) { if (!(a[i] > a[i + 1] && a[i + 1] > a[i + 2])) { flag = 1; break; } } // If the array is already decreasing if (flag == 0) return true; flag = 0; // Check if the array is already increasing for (int i = 0; i < n - 2; i++) { if (!(a[i] < a[i + 1] && a[i + 1] < a[i + 2])) { flag = 1; break; } } // If the array is already increasing if (flag == 0) return true; // Find the indices of the minimum // && the maximum value int val1 = Integer.MAX_VALUE, mini = -1, val2 = Integer.MIN_VALUE, maxi = 0; for (int i = 0; i < n; i++) { if (a[i] < val1) { mini = i; val1 = a[i]; } if (a[i] > val2) { maxi = i; val2 = a[i]; } } flag = 1; // Check if we can make array increasing for (int i = 0; i < maxi; i++) { if (a[i] > a[i + 1]) { flag = 0; break; } } // If the array is increasing upto max index // && minimum element is right to maximum if (flag == 1 && maxi + 1 == mini) { flag = 1; // Check if array increasing again or not for (int i = mini; i < n - 1; i++) { if (a[i] > a[i + 1]) { flag = 0; break; } } if (flag == 1) return true; } flag = 1; // Check if we can make array decreasing for (int i = 0; i < mini; i++) { if (a[i] < a[i + 1]) { flag = 0; break; } } // If the array is decreasing upto min index // && minimum element is left to maximum if (flag == 1 && maxi - 1 == mini) { flag = 1; // Check if array decreasing again or not for (int i = maxi; i < n - 1; i++) { if (a[i] < a[i + 1]) { flag = 0; break; } } if (flag == 1) return true; } // If it is not possible to make the // array increasing or decreasing return false;}// Driver codepublic static void main(String args[]){ int a[] = { 4, 5, 6, 2, 3 }; int n = a.length; if (isPossible(a, n)) System.out.println( "Yes"); else System.out.println( "No");}}// This code is contributed by Arnab Kundu |
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 it is possible to make array increasing or decreasing by rotating the array for more details!
