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 &lt;= i &lt; nums.length and return it.

A zero-based permutation nums is an array of distinct integers from 0 to nums.length - 1 (inclusive).

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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]

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Constraints:


	1 &lt;= nums.length &lt;= 1000
	0 &lt;= nums[i] &lt; nums.length
	The elements in nums are distinct.


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Follow-up: Can you solve it without using an extra space (i.e., O(1) memory)?

class Solution:
	#As maximum value of the array element is 1000, this solution would work
    def buildArray(self, nums: List[int]) -> List[int]:
        for i in range(len(nums)):            
            if nums[nums[i]] <= len(nums):
                nums[i] = nums[nums[i]] * 1000 + nums[i]
            else:
                nums[i] = mod(nums[nums[i]],1000) * 1000 + nums[i]
            
        for i in range(len(nums)):
            nums[i]  = nums[i] // 1000
            
        return nums
class Solution {
    public int[] buildArray(int[] nums) {
        int n=nums.length;
        for(int i=0;i<n;i++)
            nums[i] = nums[i] + n*(nums[nums[i]]%n);
        for(int i=0;i<n;i++)
			nums[i] = nums[i]/n;
		
		return nums;
    }
}
class Solution {
public:
    vector<int> buildArray(vector<int>& nums) {
        int n=nums.size();
        vector<int> ans(n);
        for(int i=0;i<n;i++){
            ans[i]=nums[nums[i]];
        }
        return ans;
    }
};
var buildArray = function(nums) {
    return nums.map(a=>nums[a]);
};
Build Array from Permutation