565. Array Nesting

1. Description

A zero-indexed array A of length N contains all integers from 0 to N-1. Find and return the longest length of set S, where S[i] = {A[i], A[A[i]], A[A[A[i]]], … } subjected to the rule below.
Suppose the first element in S starts with the selection of element A[i] of index = i, the next element in S should be A[A[i]], and then A[A[A[i]]]… By that analogy, we stop adding right before a duplicate element occurs in S.

2. Example

Example 1:
Input: A = [5,4,0,3,1,6,2]
Output: 4
Explanation:
A[0] = 5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2.
One of the longest S[K]:
S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}

3. Note

• N is an integer within the range [1, 20,000].
• The elements of A are all distinct.
• Each element of A is an integer within the range [0, N-1].

4. Solutions

My Accepted Solution

n = m_nums.size()
Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    // int arrayNesting(vector<int>& nums)
    int arrayNesting(vector<int>& m_nums) {
        int maxNestLength = 0;
        for (int i = 0; i < m_nums.size(); ++i) {
            if (m_nums[i] >= 0) {
                int index = i, length = 0;
                while (m_nums[index] >= 0) {
                    int oldIndex = index;
                    index = m_nums[index];
                    m_nums[oldIndex] = INT_MIN;

                    length++;
                }

                maxNestLength = max(length, maxNestLength);
                if (maxNestLength >= (m_nums.size() + 1) / 2) {
                    break;
                }
            }
        }

        return maxNestLength;
    }
};