Left Right (with.Java)

This is a post about left-right (with.Java).

We’ll be solving coding test questions, reflecting on the problems we’ve solved, and learning about other ways to solve them.

Let’s start with the problem.

Problem

A list of strings, str_list, contains a number of the four strings “u”, “d”, “l”, and “r”.

Complete the solution function to return an ordered list of strings to the left of “l” and “r” in str_list if “l” comes first, and an ordered list of strings to the right of “r” if “r” comes first.

If there is no “l” or “r”, return an empty list.

Example input and output

My solution to the problem

class Solution {
    public String[] solution(String[] str_list) {
        int lIndex = -1;
        int rIndex = -1;

        for (int i = 0; i < str_list.length; i++) {
            if (str_list[i].equals("l")) {
                lIndex = i;
                break;
            } else if (str_list[i].equals("r")) {
                rIndex = i;
                break;
            }
        }
        if (lIndex == -1 && rIndex == -1) {
            return new String[0];
        }

        if (lIndex != -1) {
            String[] answer = new String[lIndex];
            for (int j = 0; j < lIndex; j++) {
                answer[j] = str_list[j];
            }
            } return answer;
        }
        else {
            String[] answer = new String[str_list.length - rIndex - 1];
            for (int k = rIndex + 1; k < str_list.length; k++) {
                answer[k - rIndex - 1] = str_list[k];
            }
            } return answer;
        }
    }
}

solution description

int lIndex = -1;, int rIndex = -1;: Initialize the index of the “l” character and the index of the “r” character. The initial values are set to -1.

First iteration (for (int i = 0; i < str_list.length; i++) {): Loops through the array str_list, looking for the character “l” or “r”. If an “l” is found, store its index in lIndex and exit the loop; if an “r” is found, store its index in rIndex and exit the loop.

if (lIndex == -1 && rIndex == -1) { return new String[0]; }: If neither “l” nor “r” is found, return an empty string array.

if (lIndex != -1) { … } else { … }: Performs different logic depending on if either “l” or “r” was found.

If “l” is found:

String[] answer = new String[lIndex];: Create an array answer to store the strings before the “l” character.

Use a loop to copy the strings before the “l” character into the answer array.

Return the answer array.

If an “r” is found:

String[] answer = new String[str_list.length - rIndex - 1];: Create an array answer to store the strings after the “r” character.

Use a loop to copy the strings after the “r” character into the answer array.

Return the answer array.