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].

# @lc code=start
using LeetCode

function array_nesting(nums::Vector{Int})
    visited = fill(false, length(nums))
    res = 0
    for (idx, num) in enumerate(nums)
        num += 1
        visited[idx] && continue
        tmp = 1
        visited[idx] = true
        while !visited[num]
            visited[num] = true
            num = nums[num] + 1
            tmp += 1
        end
        res = max(res, tmp)
    end
    return res
end
# @lc code=end

array_nesting (generic function with 1 method)

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