LeetCode 1920. Build Array from Permutation Solution in Java, C++, Python & More | Explanation + Code

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

Solution 1: Simulation

We can directly simulate the process described in the problem by constructing a new array ans. For each i, let ans[i] = nums[nums[i]].

The time complexity is O(n), where n is the length of the array nums. Ignoring the space consumption of the answer array, the space complexity is O(1).

PythonJavaC++GoTypeScriptRustJavaScriptC

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

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; } }(code-box)

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

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

function buildArray(nums: number[]): number[] { return nums.map(x => nums[x]); }(code-box)

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

/** * @param {number[]} nums * @return {number[]} */ var buildArray = function (nums) { return nums.map(x => nums[x]); };(code-box)

/** * Note: The returned array must be malloced, assume caller calls free(). */ int* buildArray(int* nums, int numsSize, int* returnSize) { int* ans = malloc(sizeof(int) * numsSize); for (int i = 0; i < numsSize; i++) { ans[i] = nums[nums[i]]; } *returnSize = numsSize; return ans; }(code-box)