[06 Aug 2020] Sorted and Rotated Problems

 

Check if an array is sorted and rotated using Binary Search

Given an array arr[] of N distinct integers, the task is to check if this array is sorted when rotated counter-clockwise. A sorted array is not considered sorted and rotated, i.e., there should at least one rotation.

Examples:

Input: arr[] = { 3, 4, 5, 1, 2 }
Output: true
Explanation:
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: true

In this article, an approach using Binary Search concept is mentioned.

1. To apply a binary search, the array needs to follow some order by which at every iteration, one half of the array can be eliminated.
2. Therefore, the order followed by an array which is sorted and rotated array is that all the elements to the left of the pivot(the point at which the array is rotated) are in descending order and all the elements to the right of the pivot would be in ascending order.

   This can be visualized from the illustration below:

   

3. Therefore, the pivot can be found using Binary Search and recursion in the following way:
   • Base Cases: The base case will be either when the pivot has been found or if the pivot cannot be found in the given array. The pivot cannot be found when the right index is less than the left index. -1 is returned in these cases. And when high and low are pointing to the same element, then the element at low is the pivot and that element is returned.

     if (high < low)
          return -1;
     if (high == low)
          return low;

     Apart from this, another base case is when mid((low + high) / 2) is a pivot. The element at mid is considered when that element is less than the next element or greater than the previous element.

     if (mid < high && arr[mid + 1] < arr[mid])
         reutrn mid;
     if (mid > low && arr[mid] < arr[mid - 1])
         return mid - 1;
     

   • Recursive Case: When none of the base cases satisfies, then a decision has to be made whether to ignore the first half or second half. This decision is taken by checking if the element at the first index (low) is greater than the element at the middle index or not. If it is, then the pivot for sure lies in the first half. Else, the pivot lies in the second half.

     if (arr[low] > arr[mid]) 
         return findPivot(arr, low, mid - 1);    
     else 
         return findPivot(arr, mid + 1, high);
     
     

4. Once the pivot is found, then either traverse to the left of the array from the pivot and check if all the elements are in descending order, or traverse to the right of the array and check if all the elements are in ascending order or not.

Below is the implementation of the above approach:

#include <bits/stdc++.h>    using namespace std;    // Function to return the // index of the pivot int findPivot(int arr[], int low, int high) {     // Base cases     if (high < low)         return -1;     if (high == low)         return low;        int mid = (low + high) / 2;     if (mid < high && arr[mid + 1] < arr[mid])      {         return mid;     }        // Check if element at (mid - 1) is pivot     // Consider the cases like {4, 5, 1, 2, 3}     if (mid > low && arr[mid] < arr[mid - 1])     {         return mid - 1;     }        // Decide whether we need to go to     // the left half or the right half     if (arr[low] > arr[mid])     {         return findPivot(arr, low, mid - 1);     }     else      {         return findPivot(arr, mid + 1, high);     } }    // Function to check if a given array // is sorted rotated or not bool isRotated(int arr[], int n) {     int l = 0;     int r = n - 1;     int pivot = -1;     if (arr[l] > arr[r])     {         pivot = findPivot(arr, l, r);            // To check if the elements to the left         // of the pivot are in descending or not         if (l < pivot)         {             while (pivot > l)             {                 if (arr[pivot] < arr[pivot - 1])                 {                     return false;                 }                 pivot--;             }         }            // To check if the elements to the right         // of the pivot are in ascending or not         else {             pivot++;             while (pivot < r) {                 if (arr[pivot] > arr[pivot + 1]) {                     return false;                 }                 pivot++;             }         }            // If any of the above if or else is true         // Then the array is sorted rotated         return true;     }        // Else the array is not sorted rotated     else {         return false;     } }    // Driver code int main() {     int arr[] = { 3, 4, 5, 1, 2 };     if (isRotated(arr, 5)) cout<<"true";     else     cout<<"false";     return 0; }    // This code is contributed by mohit kumar 29

Output:

true

Time Complexity: O(N) as:

• The pivot element is being found using Binary Search in O(log N)
• But in order to check if the left part or right part is in descending or ascending order, O(N) time is needed in worst case scenario.
• Therefore the overall time complexity is O(N)