# 565. Array Nesting

## Description

You are given an integer array nums of length n where nums is a permutation of the numbers in the range [0, n - 1].

You should build a set s[k] = {nums[k], nums[nums[k]], nums[nums[nums[k]]], ... } subjected to the following rule:

• The first element in s[k] starts with the selection of the element nums[k] of index = k.
• The next element in s[k] should be nums[nums[k]], and then nums[nums[nums[k]]], and so on.
• We stop adding right before a duplicate element occurs in s[k].

Return the longest length of a set s[k].

Example 1:

Input: nums = [5,4,0,3,1,6,2]
Output: 4
Explanation:
nums[0] = 5, nums[1] = 4, nums[2] = 0, nums[3] = 3, nums[4] = 1, nums[5] = 6, nums[6] = 2.
One of the longest sets s[k]:
s[0] = {nums[0], nums[5], nums[6], nums[2]} = {5, 6, 2, 0}


Example 2:

Input: nums = [0,1,2]
Output: 1


Constraints:

• 1 <= nums.length <= 105
• 0 <= nums[i] < nums.length
• All the values of nums are unique.

## Solutions

• class Solution {
public int arrayNesting(int[] nums) {
int ans = 0, n = nums.length;
for (int i = 0; i < n; ++i) {
int cnt = 0;
int j = i;
while (nums[j] < n) {
int k = nums[j];
nums[j] = n;
j = k;
++cnt;
}
ans = Math.max(ans, cnt);
}
return ans;
}
}


• class Solution {
public:
int arrayNesting(vector<int>& nums) {
int ans = 0, n = nums.size();
for (int i = 0; i < n; ++i) {
int cnt = 0;
int j = i;
while (nums[j] < n) {
int k = nums[j];
nums[j] = n;
j = k;
++cnt;
}
ans = max(ans, cnt);
}
return ans;
}
};

• class Solution:
def arrayNesting(self, nums: List[int]) -> int:
ans, n = 0, len(nums)
for i in range(n):
cnt = 0
while nums[i] != n:
j = nums[i]
nums[i] = n
i = j
cnt += 1
ans = max(ans, cnt)
return ans


• func arrayNesting(nums []int) int {
ans, n := 0, len(nums)
for i := range nums {
cnt, j := 0, i
for nums[j] != n {
k := nums[j]
nums[j] = n
j = k
cnt++
}
if ans < cnt {
ans = cnt
}
}
return ans
}