565. Array Nesting

Problem statement:

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].

Solution:

This is a DFS solution. I solved it by employing the DFS template with a returned length of S[k].

For each S[k], it forms a circle. It means any element in this circle returns the same length.

For the purpose of pruning, we set a visited array to denote whether current element has been visited before.

Time complexity is O(n).

Space complexity is O(n).

class Solution {

public:

    int arrayNesting(vector<int>& nums) {

        int largest = ;

        vector<int> visited(nums.size(), );

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

            largest = max(largest, largest_nesting(nums, visited, nums[i], ));

        }

        return largest;

    }

    int largest_nesting(vector<int>& nums, vector<int>& visited, int idx, int size){

        if(visited[nums[idx]] == ){

            visited[nums[idx]] = ;

            return largest_nesting(nums, visited, nums[idx], size + );

        } else {

            return size;

        }

    }

};

巴特西

565. Array Nesting

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