BuildArray.java

//Given a zero-based permutation nums (0-indexed), build an array ans of the same length where ans[i] = nums[nums[i]] for each 0 <= i < nums.length and return it.
//
//A zero-based permutation nums is an array of distinct integers from 0 to nums.length - 1 (inclusive).
//
//
//
//Example 1:
//
//Input: nums = [0,2,1,5,3,4]
//Output: [0,1,2,4,5,3]
//Explanation: The array ans is built as follows:
//ans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]]
//    = [nums[0], nums[2], nums[1], nums[5], nums[3], nums[4]]
//    = [0,1,2,4,5,3]
//Example 2:
//
//Input: nums = [5,0,1,2,3,4]
//Output: [4,5,0,1,2,3]
//Explanation: The array ans is built as follows:
//ans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]]
//    = [nums[5], nums[0], nums[1], nums[2], nums[3], nums[4]]
//    = [4,5,0,1,2,3]
//
//
//Constraints:
//
//1 <= nums.length <= 1000
//0 <= nums[i] < nums.length
//The elements in nums are distinct.
//
//
//Follow-up: Can you solve it without using an extra space (i.e., O(1) memory)?

import java.util.Arrays;

public class BuildArray {
    public int[] buildArray(int[] nums) {
        int[] ans = new int[nums.length];
        for(int i = 0; i < nums.length; i++){
            ans[i] = nums[nums[i]];
        }
        return ans;
    }

    public static void main(String[] args) {
        BuildArray b = new BuildArray();
        int[] nums = new int[]{0, 2, 1, 5, 3, 4};
        System.out.println(Arrays.toString(b.buildArray(nums)));
    }
}