Build Array from Permutation (LeetCode)

1920. Build Array from Permutation (Easy)

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.

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)?

Hints:

Just apply what’s said in the statement.

Notice that you can’t apply it on the same array directly since some elements will change after application

Build Array from Permutation (Java)

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;
    }
}

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