<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>. You have to partition the array into one or more <strong>contiguous</strong> subarrays.</p>

<p>We call a partition of the array <strong>valid</strong> if each of the obtained subarrays satisfies <strong>one</strong> of the following conditions:</p>

<ol>
	<li>The subarray consists of <strong>exactly</strong> <code>2,</code> equal elements. For example, the subarray <code>[2,2]</code> is good.</li>
	<li>The subarray consists of <strong>exactly</strong> <code>3,</code> equal elements. For example, the subarray <code>[4,4,4]</code> is good.</li>
	<li>The subarray consists of <strong>exactly</strong> <code>3</code> consecutive increasing elements, that is, the difference between adjacent elements is <code>1</code>. For example, the subarray <code>[3,4,5]</code> is good, but the subarray <code>[1,3,5]</code> is not.</li>
</ol>

<p>Return <code>true</code><em> if the array has <strong>at least</strong> one valid partition</em>. Otherwise, return <code>false</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,4,4,5,6]
<strong>Output:</strong> true
<strong>Explanation:</strong> The array can be partitioned into the subarrays [4,4] and [4,5,6].
This partition is valid, so we return true.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,1,2]
<strong>Output:</strong> false
<strong>Explanation:</strong> There is no valid partition for this array.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution:
    def validPartition(self, nums: List[int]) -> bool:
        """
        neetcode solution.
        https://www.youtube.com/watch?v=OxXPiwWFdTI
        
        Good scenarios:
        (1) '22'  (size 2)
        (2) '444' (size 3)
        (3) '456' (size 3 and increasing by 1)
            check mid-1, mid, mid+1

        E.g.,
                   i 
                  [4,4,4,5,6] i=0
                    /   \\
        i=2     [4,4]   [4,4,4] i=3 
                /           \\
        i=5  [4,5,6]           [5,6] 
        
            Where @i=2
                i
            [4,4][4,5,6]

            (Q1) Are 4,5 equal? no
            (Q2) Are 4,5,6 equal? no
            (Q3) Are 4,5,6 increasing by 1? yes!
            move i=2 to i=5
            We found a valid solution.

            Where @i=3
                    i
            [4,4,4][5,6]
            (Q1) are 5,6 equal? no
            Cannot even ask Q2 and Q3 because
            there are not 3 numbers left.

        How did we get to @i? does not really matter
        It could be with 10 differnent partitions or whatever.
        Given the fact that we were able to partition 
        up to @i, we only care about the remaining numbers
        (remaining = subproblem!)
        This subprob is the same regarless of how we got there.
        --> memoization! with the index i that defines the 
        subproblem we are dealing with currently.

        i will map to True/False
        The # of subprob = O(n)
        Each subprob is O(1) operation because checking the
        fixed size array 2 or 3 is a constant operation.

        Time: O(n)
        Space: O(n)
        """

        def dfs(i):
            # base case: out of bounds
            # return True because 
            # we are able to split all the way
            if i >= len(nums):
                return True

            # base case: memo 
            if i in memo:
                return memo[i]

            # recursive case 

            check = False
            # do we have two equal values?
            if (
                # prevent out of bounds error
                i < len(nums) - 1 and 
                nums[i] == nums[i+1]
            ):
                # good, we now need to check
                # the subprob
                check = dfs(i+2)

            # prevent out of bounds error
            if i < len(nums) - 2:
                if (
                    # do we have three equal values?
                    (nums[i] == nums[i+1] == nums[i+2]) or 
                    # do we have three equal values?
                    (nums[i] + 1 == nums[i+1] == nums[i+2] - 1)
                ):
                    # good, we now need to check
                    # the subprob
                    check = check or dfs(i+3)
            
            memo[i] = check
            return check

        memo = {}
        return dfs(0)
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