RUST GYM v0.2.11

565. Array Nesting

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].

Rust Solution

struct Solution;

impl Solution {
    fn array_nesting(nums: Vec<i32>) -> i32 {
        let mut res = 0;
        let n = nums.len();
        let mut visited = vec![false; n];
        for i in 0..n {
            if visited[i] {
                continue;
            }
            let mut j = i;
            let mut length = 0;
            while !visited[j] {
                visited[j] = true;
                j = nums[j] as usize;
                length += 1;
            }
            res = res.max(length);
        }
        res
    }
}

#[test]
fn test() {
    let nums = vec![5, 4, 0, 3, 1, 6, 2];
    let res = 4;
    assert_eq!(Solution::array_nesting(nums), res);
}

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