arrayNesting.swift

/*
 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].

 https://leetcode.com/problems/array-nesting/
 */
class Solution {
    func arrayNesting(_ nums: [Int]) -> Int {
        var visited = [Bool](repeating: false, count: nums.count)
        var result = Int.min
        for i in nums.indices {
            guard !visited[i] else { continue }
            var start = nums[i], count = 0
            repeat {
                start = nums[start]
                count += 1
                visited[start] = true
            } while start != nums[i]
            result = max(result, count)
        }
        return result
    }
}