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