Array Nesting

Similar Problems:

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].

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

- Solution:

// https://code.dennyzhang.com/array-nesting // Basic Ideas: BFS // It could be a circle or a list // Two pass // Complexity: Time O(n), Space O(n) func arrayNesting(nums []int) int { l := make([]int, len(nums)) res, count := 0, 0 for i, _ := range nums { if l[i] == 0 { j := i count = 0 for j < len(nums) && l[j] == 0 { count++ l[j] = count j = nums[j] } if count != 1 { j = i for l[j] != count { l[j] = count j = nums[j] } } if count > res { res = count } } } return res }