Array Nesting

Problem

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:

  • 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].

Accepted

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class Solution {
public:
    int arrayNesting(vector<int>& nums) {
      int longest = 0;
      int visited[nums.size()] = {0};
      for (int i = 0; i < nums.size(); i++) {
        // if we have alreayd visited this then starting from this index
        // will always be shorter, so skip
        if (visited[i]) continue;
        int j = i, count = 0;
        while (j >=0 && j < nums.size() && !visited[j]) {
          count++;
          visited[j] = 1;
          if (j < 0 || j >= nums.size()) break;
          j = nums[j];
        }
        longest = max(longest, count);
      }
      return longest;
    }
};