565. Array Nesting

Problem

You are given an integer array nums of length n where nums is a permutation of the numbers in the range [0, n - 1].

You should build a set s[k] = {nums[k], nums[nums[k]], nums[nums[nums[k]]], ... } subjected to the following rule:

• The first element in s[k] starts with the selection of the element nums[k] of index = k.

• The next element in s[k] should be nums[nums[k]], and then nums[nums[nums[k]]], and so on.

• We stop adding right before a duplicate element occurs in s[k].

Return the longest length of a set s[k].

  Example 1:

Input: nums = [5,4,0,3,1,6,2]
Output: 4
Explanation: 
nums[0] = 5, nums[1] = 4, nums[2] = 0, nums[3] = 3, nums[4] = 1, nums[5] = 6, nums[6] = 2.
One of the longest sets s[k]:
s[0] = {nums[0], nums[5], nums[6], nums[2]} = {5, 6, 2, 0}

Example 2:

Input: nums = [0,1,2]
Output: 1

  Constraints:

• 1 <= nums.length <= 10^5

• 0 <= nums[i] < nums.length

• All the values of nums are unique.

Solution (Java)

class Solution {
    public int arrayNesting(int[] nums) {
        int index;
        int value;
        int maxLen = 0;
        int len;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != -1) {
                index = i;
                len = 0;
                while (nums[index] != -1) {
                    value = nums[index];
                    nums[index] = -1;
                    index = value;
                    len++;
                }
                maxLen = Math.max(len, maxLen);
            }
        }
        return maxLen;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).