565. Array Nesting

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.

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}

Note:

1. N is an integer within the range [1, 20,000].

2. The elements of A are all distinct.

3. Each element of A is an integer within the range [0, N-1].

int arrayNesting(vector<int>& nums) { // time: O(n); space: O(n)
    if (nums.empty()) return 0;
    int n = nums.size(), res = 0;
    vector<bool> visited(n, false);
    for (int i = 0; i < n; ++i) {
        if (visited[i]) continue;
        int cnt = 0, j = i;
        while (cnt == 0 || j != i) {
            visited[j] = true;
            j = nums[j];
            ++cnt;
        }
        res = max(res, cnt);
    }
    return res;
}

If the input array is allowed to modify, then we can swap the number to its corresponding index to optimize space usage to O(1).

int arrayNesting(vector<int>& nums) { // time: O(n); space: O(1)
    int n = nums.size(), res = 0;
    for (int i = 0; i < n; ++i) {
        int cnt = 1;
        while (nums[i] != i && nums[i] != nums[nums[i]]) {
            swap(nums[i], nums[nums[i]]);
            ++cnt;
        }
        res = max(res, cnt);
    }
    return res;
}