#
# Create by Hua on 3/21/22.
#

"""
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)?
core math?
https://leetcode.com/problems/build-array-from-permutation/discuss/1315926/Python-O(n)-Time-O(1)-Space-w-Full-Explanation
"""


class Solution(object):
    def buildArray(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]

        thought: do as statement
        03/21/2022 17:54	Accepted	117 ms	13.7 MB	python

        easy 1 min, math?

        """
        ret = list()
        for i in range(len(nums)):
            ret.append(nums[nums[i]])

        return ret
