How to Sort an Array in Java Based on Another Array Using the indexOf Method

How can I sort one array in Java according to the order specified by another array using the indexOf method?

// Example code snippet provided below
String[] order = {"banana", "apple", "orange"};
String[] fruits = {"apple", "orange", "banana", "grape"};

Arrays.sort(fruits, (a, b) -> {
    return Integer.compare(Arrays.asList(order).indexOf(a), Arrays.asList(order).indexOf(b));
});

In Java, to sort one array based on the order defined by another array, you can leverage the `indexOf` method. This method finds the index of an element, allowing you to compare elements’ positions based on a predefined order. Below, you will find a detailed explanation and example code to achieve this.

import java.util.Arrays;

public class ArraySorter {
    public static void main(String[] args) {
        String[] order = {"banana", "apple", "orange"};
        String[] fruits = {"apple", "orange", "banana", "grape"};

        Arrays.sort(fruits, (a, b) -> {
            return Integer.compare(Arrays.asList(order).indexOf(a), Arrays.asList(order).indexOf(b));
        });

        System.out.println(Arrays.toString(fruits)); // Output: [banana, apple, orange, grape]
    }
}

Causes

• Lack of understanding of how to compare elements in arrays.
• Difficulty in processing two arrays simultaneously for sorting.

Solutions

• Use the `Arrays.sort()` method with a custom comparator to compare the indices of elements in the order array.
• Ensure that all elements of the sorting array are present in the order array to avoid `-1` results from indexOf.

Mistake: Forgetting to check if all elements to be sorted exist in the order array.

Solution: Always validate that the elements in the array being sorted are present in the ordering array to prevent unexpected behaviors.

Mistake: Using `indexOf` within the sorting comparator which can lead to inefficient sorting.

Solution: Consider storing indices outside the comparator if performance is a priority.