565. Array Nesting

565. Array Nesting (Medium)

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].

Related Topics:
Array

Similar Questions:

• Nested List Weight Sum (Easy)
• Flatten Nested List Iterator (Medium)
• Nested List Weight Sum II (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/array-nesting/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int arrayNesting(vector<int>& A) {
        int ans = 0;
        for (int i = 0; i < A.size(); ++i) {
            if (A[i] == -1) continue;
            int cnt = 0, j = i;
            while (A[j] != -1) {
                int k = j;
                j = A[j];
                A[k] = -1;
                ++cnt;
            }
            ans = max(ans, cnt);
        }
        return ans;
    }
};