package com.leetcode.array;

/**

* Created by charles on 6/10/17.

* A zero-indexed array A consisting of N different integers is given. The array contains all integers in the range [0, N - 1].

Sets S[K] for 0 <= K < N are defined as follows:

S[K] = { A[K], A[A[K]], A[A[A[K]]], ... }.

Sets S[K] are finite for each K and should NOT contain duplicates.

Write a function that given an array A consisting of N integers, return the size of the largest set S[K] for this array.

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 array A is an integer within the range [0, N-1].

*/

public class ArrayNesting_565 {

/**

* naive solution is to iterate over all indices of given number array

* for each index i chosen, increment count and go for next index nums[i] and terminating when i == nums[i]

* since all elem in nums lie in the range (0..N-1) the new indices will never lie outside array size limits

*

* [1,2,3,4,5,0] worst case

* in worst case, all elem are added to set corresponding to all starting indices.

* but all these sets corresponds to same set of elements only, leading to redundant calculation

*

* for each set, same elems will be added to current set irrespective of

* first elem chose to be added to set out of these marked elements

*

* we can use visited auxiliary array to mark elems in same set, so as to avoid redundant calculation

*

* ## ultimate version, only O(1) space

* use Larger Integer outside of range (0, N-1) like Integer.MAX_VALUE to mark position

* bad thing is to modify given input array.

*/

public int arrayNesting(int[] nums) {

int res = 0;

int next = 0, prev = 0;

int count = 0;

for (int i = 0; i < nums.length; i++) {

if (nums[i] != Integer.MAX_VALUE) { // not visited

next = nums[i];

count = 0;

while (nums[next] != Integer.MAX_VALUE) {

prev = next;

next = nums[next]; // move to next

nums[prev] = Integer.MAX_VALUE; // update visited to prev

count++;

}

res = Math.max(res, count);

}

}

return res;

}

}