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javac++algorithmdata-structuresbinary-search

Binary search to find the rotation point in a rotated sorted list

I have a sorted list which is rotated and would like to do a binary search on that list to find the minimum element.

Lets suppose initial list is {1,2,3,4,5,6,7,8} rotated list can be like {5,6,7,8,1,2,3,4}

Normal binary search doesn't work in this case. Any idea how to do this.

-- Edit

I have one another condition. What if the list is not sorted??

Solution

  • A slight modification on the binary search algorithm is all you need; here's the solution in complete runnable Java (see Serg's answer for Delphi implementation, and tkr's answer for visual explanation of the algorithm).

    import java.util.*;
public class BinarySearch {
    static int findMinimum(Integer[] arr) {
        int low = 0;
        int high = arr.length - 1;
        while (arr[low] > arr[high]) {
            int mid = (low + high) >>> 1;
            if (arr[mid] > arr[high]) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        return low;
    }
    public static void main(String[] args) {
        Integer[] arr = { 1, 2, 3, 4, 5, 6, 7 };
        // must be in sorted order, allowing rotation, and contain no duplicates

        for (int i = 0; i < arr.length; i++) {
            System.out.print(Arrays.toString(arr));
            int minIndex = findMinimum(arr);
            System.out.println(" Min is " + arr[minIndex] + " at " + minIndex);
            Collections.rotate(Arrays.asList(arr), 1);
        }
    }
}

    This prints:

    [1, 2, 3, 4, 5, 6, 7] Min is 1 at 0
[7, 1, 2, 3, 4, 5, 6] Min is 1 at 1
[6, 7, 1, 2, 3, 4, 5] Min is 1 at 2
[5, 6, 7, 1, 2, 3, 4] Min is 1 at 3
[4, 5, 6, 7, 1, 2, 3] Min is 1 at 4
[3, 4, 5, 6, 7, 1, 2] Min is 1 at 5
[2, 3, 4, 5, 6, 7, 1] Min is 1 at 6

    See also

    On duplicates

    Note that duplicates makes it impossible to do this in O(log N). Consider the following bit array consisting of many 1, and one 0:

      (sorted)
  01111111111111111111111111111111111111111111111111111111111111111
  ^

  (rotated)
  11111111111111111111111111111111111111111111101111111111111111111
                                               ^

  (rotated)
  11111111111111101111111111111111111111111111111111111111111111111
                 ^

    This array can be rotated in N ways, and locating the 0 in O(log N) is impossible, since there's no way to tell if it's in the left or right side of the "middle".

    I have one another condition. What if the list is not sorted??

    Then, unless you want to sort it first and proceed from there, you'll have to do a linear search to find the minimum.

    See also