Mastering Daily DSA Problems: A Step-by-Step Guide

What I Learned Today :

How to detect if an array is sorted and rotated by counting the number of order breaks.

The importance of checking both ascending and descending patterns to handle all rotation cases.

Used edge comparisons (arr[n-1] vs arr[0]) to verify rotation wrap-around.

Improved logical thinking around order consistency and handling circular conditions in arrays.

1. Given an array arr[] of N distinct integers, check if this array is Sorted (non-increasing or non-decreasing) and Rotated counter-clockwise. Note that input array may be sorted in either increasing or decreasing order, then rotated. A sorted array is not considered as sorted and rotated, i.e., there should be at least one rotation. Input: N = 4 arr[] = {3,4,1,2} Output: Yes

   Explanation: The array is sorted (1, 2, 3, 4) and rotated twice (3, 4, 1, 2).

   ```java int ascBreak = 0;

   int descBreak = 0;

   for (int i = 0; i < arr.length-1; i++) { // iterating the array if (arr[i] < arr[i+1]) { // if the next number is higher than current number //ascending order break ex: {1,2,3,4,2} 4,1 ascending break if (arr[i] > arr[i+1]) { ascBreak++; }

   // if the next number is lesser that current number descending order break //ex: {4,3,2,1,5} 1, descending break if (arr[i] < arr[i+1]) { descBreak++; } }

   } // edge cases for ascending if (arr[num-1] < arr[0]) { ascBreak++; } // edge cases for descending if (arr[0] < arr[num-1]) { descBreak++; } // either one of order has to be there and both should not be zero if (ascBreak == 1 || descBreak == 1 && ascBreak != 0 && descBreak != 0 ) { return true; } return false;

```