# Leetcode: Array Nesting

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:

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].

Github: code.dennyzhang.com

Credits To: leetcode.com

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

• Solution:
```// Blog link: 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
}
```

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