Sunday, March 31, 2013

No. 47 - Search in a Rotation of an Array


Question: When some elements at the beginning of an array are moved to the end, it gets a rotation of the original array. Please implement a function to search a number in a rotation of an increasingly sorted array. Assume there are no duplicated numbers in the array.
For example, array {3, 4, 5, 1, 2} is a rotation of array {1, 2, 3, 4, 5}. If the target number to be searched is 4, the index of the number 4 in the rotation 1 should be returned. If the target number to be searched is 6, -1 should be returned because the number does not exist in the rotated array.
Analysis: Binary search is suitable for sorted arrays. Let us try to utilize it on a rotation of a sorted array. Notice that a rotation of a sorted array can be partitioned into two sorted sub-arrays, and numbers in the first sub-array are greater than numbers in the second one.
Two pointers P1 and P2 are utilized. P1 references to the first element in the array, and P2 references to the last element. According to the rotation rule, the first element should be greater than the last one.
The algorithm always compares the number in middle with numbers pointed by P1 and P2 during binary search. If the middle number is in the first increasingly sorted sub-array, it is greater than the number pointed by P1.
If the value of target number to be search is between the number pointed by P1 and the middle number, we then search the target number in the first half sub-array. In such a case the first half sub-array is in the first increasing sub-array, we could utilize the binary search algorithm. For example, if we search the number 4 in a rotation {3, 4, 5, 1, 2}, we could search the target number 4 in the sub-array {3, 4, 5} because 4 is between the first number 3 and the middle number 5.
If the value of target number is not between the number pointed by P1 and the middle number, we search the target in the second half sub-array. Notice that the second half sub-array also contains two increasing sub-array and itself is also a rotation, so we could search recursively with the same strategy. For example, if we search the number 1 in a rotation {3, 4, 5, 1, 2}, we could search the target number 1 in the sub-array {5, 1, 2} recursively.
The analysis above is for two cases when the middle number is in the first increasing sub-array. Please analyze the other two cases when the middle number is in the second increasing sub-array yourself, when the middle number is less than the number pointed by P2.
The code implementing this algorithm is listed below, in C/C++:
int searchInRotation(int numbers[], int length, int k)
{
    if(numbers == NULL || length <= 0)
        return -1;
   
    return searchInRotation(numbers, k, 0, length - 1);
}
int searchInRotation(int numbers[], int k, int start, int end)
{
    if(start > end)
        return -1;
       
    int middle = start + (end - start) / 2;
    if(numbers[middle] == k)
        return middle;
   
    // the middle number is in the first increasing sub-array
    if(numbers[middle] >= numbers[start])
    {
        if(k >= numbers[start] && k < numbers[middle])
            return binarySearch(numbers, k, start, middle - 1);
        return searchInRotation(numbers, k, middle + 1, end);
    }
    // the middle number is in the second increasing sub-array
    else if(numbers[middle] <= numbers[end])
    {
        if(k > numbers[middle] && k <= numbers[end])
            return binarySearch(numbers, k, middle + 1, end);
        return searchInRotation(numbers, k, start, middle - 1);
    }
   
    // It should never reach here if the input is valid
    assert(false);
}
Since the function binarySearch is for the classic binary search algorithm, it is not listed here. You might implement your own binary search code if you are interested.
In each round of search, half of the array is excluded for the next round, so the time complexity is O(logn).
You may wonder why we assume there are no duplications in the input array. We determine whether the middle number is in the first or second sub-array by comparing the middle number and the numbers pointed by P1 or P2. When the middle number, the number pointed by P1 and P2 are identical, we don’t know whether the middle number is in the first or second increasing sub-array.
Let’s look at some examples. Two arrays {1, 0, 1, 1, 1} and {1, 1, 1, 0, 1} are both rotations of an increasingly sorted array {0, 1, 1, 1, 1}, which are visualized in Figure 1.



Figure 1: Two rotations of an increasingly sorted array {0, 1, 1, 1, 1}: {1, 0, 1, 1, 1} and {1, 1, 1, 0, 1}. Elements with gray background are in the second increasing sub-array.
In Figure 1, the elements pointed by P1 and P2, as well as the middle element are all 1. The middle element with index 2 is in the second sub-array in Figure 1 (a), while the middle element is in the first sub-array in Figure 1 (b).
Since we can’t determine whether the middle number in the first or second increasing sub-array, we have to search sequentially for such cases, and our code listed above should be revised.

More coding interview questions are discussed in my book <Coding Interviews: Questions, Analysis & Solutions>. You may find the details of this book on Amazon.com, or Apress.

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8 comments:

  1. This doesn't work when k = 14 for this input array {15, 16, 19, 20, 25, 1, 3, 4, 5, 7, 10, 14};

    ReplyDelete
    Replies
    1. it works, I have tested, the solution is correct

      Delete
  2. BTW, i used my version of binary search

    static int binarySearch(int[] a, int object, int left, int right) {
    int l = left;
    int r = right;

    while (l <= r) {
    int mid = l + (r - l)/2;
    if (a[mid] == object) return mid;
    else if (object < a[mid]) {
    r = mid - 1;
    } else {
    l = mid + 1;
    }
    }

    return -1;
    }

    ReplyDelete
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  4. Have a look at my solution. Please give feedback. htpp://knavite.blogspot.com/2013/11/searching-element-in-rotated-sorted.html

    ReplyDelete
  5. function find(list, n)
    {
    var start = 0;
    for(var i = 0; i + 1 < list.length; i++)
    {
    if (list[i] > list[i + 1])
    {
    start = i + 1;
    break;
    }
    }

    for(var a = 0, b = list.length; a < b;)
    {
    var r = Math.floor((b - a) / 2);
    var j = (a + r + start) % list.length;
    console.log("j=" + j + " list[j]=" + list[j]);
    if (list[j] == n) return true;
    if (list[j] > n) b = a + r;
    else a += r + 1;
    if (r == 0) break;
    }

    return false;
    }

    ReplyDelete