Searching for an element in a circular sorted array

guirgis picture guirgis · May 14, 2010 · Viewed 27.8k times · Source

We want to search for a given element in a circular sorted array in complexity not greater than O(log n).
Example: Search for 13 in {5,9,13,1,3}.

My idea was to convert the circular array into a regular sorted array then do a binary search on the resulting array, but my problem was the algorithm I came up was stupid that it takes O(n) in the worst case:

for(i = 1; i < a.length; i++){
    if (a[i] < a[i-1]){
        minIndex = i; break;
    }
}

then the corresponding index of ith element will be determined from the following relation:

(i + minInex - 1) % a.length

it is clear that my conversion (from circular to regular) algorithm may take O(n), so we need a better one.

According to ire_and_curses idea, here is the solution in Java:

public int circularArraySearch(int[] a, int low, int high, int x){
    //instead of using the division op. (which surprisingly fails on big numbers)
    //we will use the unsigned right shift to get the average
    int mid = (low + high) >>> 1;
    if(a[mid] == x){
        return mid;
    }
    //a variable to indicate which half is sorted
    //1 for left, 2 for right
    int sortedHalf = 0;
    if(a[low] <= a[mid]){
        //the left half is sorted
        sortedHalf = 1;
        if(x <= a[mid] && x >= a[low]){
            //the element is in this half
            return binarySearch(a, low, mid, x);
        }
    }
    if(a[mid] <= a[high]){
        //the right half is sorted
        sortedHalf = 2;
        if(x >= a[mid] && x<= a[high] ){
            return binarySearch(a, mid, high, x);
        }
    }
    // repeat the process on the unsorted half
    if(sortedHalf == 1){
        //left is sorted, repeat the process on the right one
        return circularArraySearch(a, mid, high, x);
    }else{
        //right is sorted, repeat the process on the left
        return circularArraySearch(a, low, mid, x);
    }
}

Hopefully this will work.

Answer

ire_and_curses picture ire_and_curses · May 14, 2010

You can do this by taking advantage of the fact that the array is sorted, except for the special case of the pivot value and one of its neighbours.

  • Find the middle value of the array a.
  • If a[0] < a[mid], then all values in the first half of the array are sorted.
  • If a[mid] < a[last], then all values in the second half of the array are sorted.
  • Take the sorted half, and check whether your value lies within it (compare to the maximum idx in that half).
  • If so, just binary search that half.
  • If it doesn't, it must be in the unsorted half. Take that half and repeat this process, determining which half of that half is sorted, etc.