Build Array from Permutation

Author: anantkaushik

1920-build-array-from-permutation.py

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+"""
+Problem Link: https://leetcode.com/problems/build-array-from-permutation/
+
+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)?
+"""
+class Solution:
+ def buildArray(self, nums: List[int]) -> List[int]:
+ l = len(nums)
+ 
+ for index in range(l):
+ nums[index] += l * (nums[nums[index]] % l)
+ 
+ for index in range(l):
+ nums[index] //= l
+ 
+ return nums
+ 
+ 
+class Solution1:
+ def buildArray(self, nums: List[int]) -> List[int]:
+ return [nums[index] for index in nums]
+