Today I have done leetcode's May LeetCoding Challenge with cpp.

May LeetCoding Challenge 6

Description

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)?

Solution

class Solution {
public:
  vector<int> buildArray(vector<int>& nums) {
    vector<int> answer(nums.size());
    for(int i = 0; i < nums.size(); ++i) {
      answer[i] = nums[nums[i]];
    }
    return answer;
  }
};

// Accepted
// 140/140 cases passed (4 ms)
// Your runtime beats 5.44 % of cpp submissions
// Your memory usage beats 39.42 % of cpp submissions (20.5 MB)