Find the Rotation Count In Rotated Sorted Array - Java Solution

1. Introduction

This blog post addresses the problem of determining the rotation count in a rotated sorted array. When a sorted array is rotated around a pivot, our task is to find out how many times the array has been rotated.

Problem

Given an array of distinct numbers sorted in ascending order, which has been rotated k times around a pivot, find the value of k (the number of rotations).

Example 1:

Input: a = [5, 7, 9, 1, 3]

Output: 3

Explanation: The original array must be [1, 3, 5, 7, 9], rotated 3 times.

Example 2:

Input: a = [4, 6, 8, 10, 0, 1, 2]

Output: 4

Explanation: The original array would be [0, 1, 2, 4, 6, 8, 10], rotated 4 times.

Example 3:

Input: a = [5, 10, 15, 20]

Output: 0

Explanation: The array has not been rotated.

2. Solution Steps

1. Apply binary search to find the rotation point (pivot) where the array has been rotated.

2. The index of the minimum element in the array gives the rotation count.

3. Handle the edge case where the array is not rotated at all.

3. Code Program

public class Solution {

    // Main method for testing
    public static void main(String[] args) {
        int[] a1 = {5, 7, 9, 1, 3};
        System.out.println("Rotation count: " + findRotationCount(a1));

        int[] a2 = {4, 6, 8, 10, 0, 1, 2};
        System.out.println("Rotation count: " + findRotationCount(a2));

        int[] a3 = {5, 10, 15, 20};
        System.out.println("Rotation count: " + findRotationCount(a3));
    }

    // Method to find the rotation count in the array
    public static int findRotationCount(int[] a) {
        int start = 0, end = a.length - 1;

        while (start < end) {
            int mid = start + (end - start) / 2;

            if (a[mid] > a[end]) {
                start = mid + 1;
            } else {
                end = mid;
            }
        }
        return start; // The index of the minimum element is the rotation count
    }
}

Output:

Rotation count: 3
Rotation count: 4
Rotation count: 0

Explanation:

The findRotationCount method uses a modified binary search to find the index of the smallest element, which is the pivot point of rotation. 

In the first example, [5, 7, 9, 1, 3], the smallest element is 1 at index 3, indicating the array was rotated 3 times. 

Similarly, for [4, 6, 8, 10, 0, 1, 2], the rotation count is 4. When the array is not rotated like in [5, 10, 15, 20], the rotation count is 0.