IndexOutOfBoundsException and quicksort algorithm

Question

I am having trouble figuring out why my code has an IndexOutOfBoundsException . I was wondering if someone could be my extra pair of eyes to help me catch the error. I am trying this out by hand but I think I am missing something as I go through the code.

code:

package Algorithms;

import java.util.ArrayList;

public class quickSort {

    public static ArrayList<Integer> mergeArrayList(ArrayList<Integer> min, int pivot, ArrayList<Integer> max) {
        ArrayList<Integer> mergedList = new ArrayList<Integer>();
        //mergedList.addAll(min);
        for (int i : min) {
            mergedList.add(i);
        }
        mergedList.add(pivot);
        //mergedList.addAll(max);
        for (int j : max) {
            mergedList.add(j);
        }
        return mergedList;
    }


    public static ArrayList<Integer> quicksort(ArrayList<Integer> array, int min, int max) {
        if (array.size() <= 1) {
            return array;
        }
        int pivot = (min + max)/2;
        ArrayList<Integer> less = new ArrayList<Integer>();
        ArrayList<Integer> greater = new ArrayList<Integer>();
        for (int i : array) {
            if (i <= array.get(pivot)) {
                less.add(i);
            }
            else {
                greater.add(i);
            }
        }
        return mergeArrayList(quicksort(less,min,pivot - 1), array.get(pivot), quicksort(greater, pivot + 1, max));
    }

    public static void main(String[] args) {
        ArrayList<Integer> arr1 = new ArrayList<Integer>();
        int[] arr = {1,2,3,4,5};
        for (int i : arr) {
            arr1.add(i);
        }
        System.out.println(quicksort(arr1,0,arr1.size() - 1));
    }
}

error:

Exception in thread "main" java.lang.IndexOutOfBoundsException: Index: 3, Size: 2
    at java.util.ArrayList.rangeCheck(ArrayList.java:635)
    at java.util.ArrayList.get(ArrayList.java:411)
    at Algorithms.quickSort.quicksort(quickSort.java:30)
    at Algorithms.quickSort.quicksort(quickSort.java:37)
    at Algorithms.quickSort.main(quickSort.java:46)

Answer 1

Add this line for debugging purposes at the start of your quicksort method.

public static ArrayList<Integer> quicksort(ArrayList<Integer> array, int min, int max) 
{
    System.out.println(Arrays.toString(array.toArray()) + " -> " + min + " " + max);
    ...
}

You will see exactly what is happening and when/where it goes wrong.

Answer 2

quicksort(greater,pivot + 1, max) in the return will result in:

int pivot = (max + pivot + 1) /2; //pseudocode

pivot will be greater than the length of the array,(which is half the original one) resulting in an out of bounds exception when you ask for array.get(pivot)

Answer 3

the problem is the min & max

I understand that you use them as indexes in the array that you send, But - you send the less & greater which are smaller than the original array.

so, for your example , after the first call to quicksort , pivot = 2 this part quicksort(greater, pivot + 1, max) will send the greater with value of 3,4 so this create you exception