# 1920. Build Array from Permutation

## Description

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)?

## Solutions

• class Solution {
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;
}
}

• class Solution {
public:
vector<int> buildArray(vector<int>& nums) {
vector<int> ans;
for (int& num : nums) {
ans.push_back(nums[num]);
}
return ans;
}
};

• class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
return [nums[num] for num in nums]


• func buildArray(nums []int) []int {
ans := make([]int, len(nums))
for i, num := range nums {
ans[i] = nums[num]
}
return ans
}

• function buildArray(nums: number[]): number[] {
return nums.map(v => nums[v]);
}


• /**
* @param {number[]} nums
* @return {number[]}
*/
var buildArray = function (nums) {
let ans = [];
for (let i = 0; i < nums.length; ++i) {
ans[i] = nums[nums[i]];
}
return ans;
};


• impl Solution {
pub fn build_array(nums: Vec<i32>) -> Vec<i32> {
nums.iter()
.map(|&v| nums[v as usize])
.collect()
}
}