Array Nesting

Description

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].

Solution(javascript)

// 这题暴力求解的话是 O(n^2), 需要用 DP 缓存重复计算的值
// const arrayNesting = function (nums = []) {
//   const map = {}
//   const aux = (index, visited = {}) => {
//     if (map[index] !== undefined) {
//       return map[index]
//     }
//     if (visited[index]) {
//       return 0
//     }
//     visited[index] = true
//     map[index] = aux(nums[index], visited) + 1
//     return map[index]
//   }
//   let max = 0
//   nums.forEach((n, index) => {
//     map[index] = aux(index)
//     max = Math.max(max, map[index])
//   })
//   return max
// }


const arrayNesting = function (nums = []) {
  const visited = {}
  const aux = (index) => {
    if (visited[index]) {
      return 0
    }
    visited[index] = true
    return aux(nums[index], visited) + 1
  }
  let max = 0
  nums.forEach((n, index) => {
    if (!visited[index]) {
      max = Math.max(max, aux(index))
    }
  })
  return max
}