The problem statement
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).
Understanding the problem
The question is asking us to return an array which takes the iterated item and sets the value to the index of the value i.e. nums[loopedItem].
There is nothing mentioned about ordering so we just do it inline sequentially as we trace down the array.
It asks for a bonus answer with O(1) space. But for now O(n) is good enough.
Examples
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]
Code
function buildArray(nums: number[]): number[] {
return nums.map((val) => nums[val]);
}