Big O of this findRange method

Question

0.00/5 (No votes)

See more:

I'm trying to figure out the runtime complexity of findRange shown below, which finds the bottom and top index given a sorted array a and key key.
If key is not found then return -1 -1.

int[] a = new int[] { 1, 2, 2, 3, 3, 5, 5, 5, 5 };
findRange(a, 5) => 5 8

I'm coming up with the solution O(2 * log(#times_key_appears)* log n) => O((log n)^2)
Is this correct?
Is there a better algorithmic solution out there?

Java

public void findRange(int[] a, int key) {
    int bsLeft = bsLeft(a, 0, a.length , key, -1);
    System.out.print(bsLeft);
    System.out.print(" ");
    System.out.println(bsRight(a, bsLeft, a.length , key, -1));
}

public int bsLeft(int[] a, int startIndex, int lastIndex, int key, int found){
    int value = Arrays.binarySearch(a, startIndex, lastIndex, key);
    if (value < 0) {
        return found;
    }
    return bsLeft(a, startIndex, value -1, key, value);
}

public int bsRight(int[] a, int startIndex, int lastIndex, int key, int found){
    int value = Arrays.binarySearch(a, startIndex , lastIndex, key);
    if (value < 0) {
        return found;
    }
    return bsRight(a, value + 1, lastIndex, key, value);
}

Posted 3-May-15 5:21am

slothbury

Add a Solution

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Bill_Hallahan · Accepted Answer · 2015-05-03T08:19:00

I'd have to understand your problem to know the best solution.

Are you free to change the forms of your stored data? It might best to preprocess that data one time into the form I mention below.

For example, you could only store unique values and have a parallel container that stores the count at each index. Of course, rather than have a parallel container, you could have a structure that contains the item and the count.

Item: 1, 2, 3, 5
Count: 1, 2, 2, 5

You can find the item using a regular binary search in O(log2 N) time, and then merely take that index for the item you found (for 5, the index would be 3), and add the count to find the top of the range, which for your example would be 3 + 5 = 8.

The issue with the structure you have is that it makes using a binary search difficult - I'd rearrange the data to be better.

By the way, if the data changes, that is, you are inserting and/or removing items from the list often, then a red-black tree would be a better solution as constantly inserting and sorting the array would be inefficient. Even in that case, I think storing a structure with the item and the count of the item would be best.