Build Array from Permutation

The provided code is a solution to a common problem on platforms like LeetCode, where you want to transform an input array based on a certain mapping. Let's break down the solution:

Problem Description:

The problem typically asks you to take an array of integers, nums, where each integer is an index, and build a new array, arr, such that:

• Each element in arr at index i is equal to the value located at the index given by nums[i].

Code Explanation:

The class Solution contains two methods:

1. buildArray Method:

This is the main method that performs the transformation:

1int[] buildArray(int[] nums) {
2    int[] arr = new int[nums.length];  // Create a new array to hold the result
3
4    for (int i = nums.length; i > 0; --i) {  // Iterate backwards through the original array
5        arr[i - 1] = nums[nums[i - 1]];  // Set each element in the new array using the mapping
6    }
7
8    return arr;  // Return the transformed array
9}

• It first initializes a new array arr with the same length as nums.
• It then iterates through the array nums in reverse using for (int i = nums.length; i > 0; --i).
• For each index i-1 of arr, it assigns the value at nums[nums[i-1]], which means it first looks up nums[i-1] to get an index, and then uses that index to get a value from nums to populate arr[i-1].
• After processing all elements, it returns the newly created array arr.

The choice of iterating backwards (from nums.length down to 1) is a stylistic choice here; the loop could have also been written to iterate forwards without affecting correctness.

2. test Method:

This method is for testing the functionality of buildArray:

1void test() {
2    int[][] testInputs = {
3        {0, 2, 1, 5, 3, 4},
4        {0, 1, 2, 4, 5, 3}
5    };
6    int[][] testOutputs = {
7        {0, 1, 2, 4, 5, 3},
8        {4, 5, 0, 1, 2, 3}
9    };
10    for (int i = 0; i < testInputs.length; i++) {
11        int[] testInput = testInputs[i];
12        int[] testOutput = testOutputs[i];
13        int[] testcase = buildArray(testInput);  // Call buildArray with the test input
14        for (int j = 0; j < testInput.length; j++) {
15            assert testcase[j] == testOutput[j];  // Verify the output matches the expected output
16        }
17    }
18}

• Here, testInputs contains various input arrays that you want to test.
• testOutputs contains the expected results for the corresponding inputs.
• The for loop iterates through each test case, passing the test input to buildArray and comparing the returned result with the expected output.
• The assert statement checks if the computed array matches the expected one; if it doesn't, the program will throw an assertion error.

3. main Method:

This is the entry point of the program:

public static void main(String[] args) {
    new Solution().test();  // Create an instance of Solution and run the tests
}

• It creates an instance of Solution and calls the test method to run all the defined test cases.

Summary:

In summary, this Java class provides a solution to transform an input array based on index mapping and includes a testing mechanism to validate its correctness. The method efficiently constructs the new array using the original indices provided in the nums array.