Optimization insights to avoid TLE

This is my 923rd problem solved on LC: Array Nesting. Problem is down below. Basically the author is looking for the longest sequence a[index], a[a[index]], a[a[a[index]]], and so on, without dupes. I tried an N^2, hitting TLE. The insight that did the trick was that once we have tried a sequence, for example, {a,b,c,d,e}, it is pointless to try a sequence starting with b, c, d or e since the length will be shorter than that tried for a. Barely made the pass after that. Code is down below, cheers, ACC.


565. Array Nesting
Medium

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.


public int ArrayNesting(int[] nums)
{
 int retVal = 0;
 Hashtable indexAlreadyTried = new Hashtable();
 for (int i = 0; i < nums.Length; i++) { Hashtable seen = new Hashtable(); int last = nums[i]; if (!indexAlreadyTried.ContainsKey(last)) { seen.Add(last, true); indexAlreadyTried.Add(last, true); for (; ; ) { last = nums[last]; if (!indexAlreadyTried.ContainsKey(last)) indexAlreadyTried.Add(last, true); if (seen.ContainsKey(last)) break; seen.Add(last, true); } retVal = Math.Max(retVal, seen.Count); if (retVal == nums.Length) break; } } return retVal; }

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