<p>You are visiting a farm that has a single row of fruit trees arranged from left to right. The trees are represented by an integer array <code>fruits</code> where <code>fruits[i]</code> is the <strong>type</strong> of fruit the <code>i<sup>th</sup></code> tree produces.</p>

<p>You want to collect as much fruit as possible. However, the owner has some strict rules that you must follow:</p>

<ul>
	<li>You only have <strong>two</strong> baskets, and each basket can only hold a <strong>single type</strong> of fruit. There is no limit on the amount of fruit each basket can hold.</li>
	<li>Starting from any tree of your choice, you must pick <strong>exactly one fruit</strong> from <strong>every</strong> tree (including the start tree) while moving to the right. The picked fruits must fit in one of your baskets.</li>
	<li>Once you reach a tree with fruit that cannot fit in your baskets, you must stop.</li>
</ul>

<p>Given the integer array <code>fruits</code>, return <em>the <strong>maximum</strong> number of fruits you can pick</em>.</p>

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

<pre>
<strong>Input:</strong> fruits = [<u>1,2,1</u>]
<strong>Output:</strong> 3
<strong>Explanation:</strong> We can pick from all 3 trees.
</pre>

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

<pre>
<strong>Input:</strong> fruits = [0,<u>1,2,2</u>]
<strong>Output:</strong> 3
<strong>Explanation:</strong> We can pick from trees [1,2,2].
If we had started at the first tree, we would only pick from trees [0,1].
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> fruits = [1,<u>2,3,2,2</u>]
<strong>Output:</strong> 4
<strong>Explanation:</strong> We can pick from trees [2,3,2,2].
If we had started at the first tree, we would only pick from trees [1,2].
</pre>

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

<ul>
	<li><code>1 &lt;= fruits.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= fruits[i] &lt; fruits.length</code></li>
</ul>

class Solution:
    def totalFruit(self, fruits: List[int]) -> int:
        #so the num at an index in fruits is just the type of fruit
        #So we want to essentially see whats the longest subarray whose elements are of only 2 possible values.

        #Trying a more sliding window approach - it works and is quite fast
        #the idea is that we increase right by 1 and if we encounter something that's not one of our 2 types, we decrement left
        #from right - 1 as long as there is a continuous sequence of the elem at right - 1. Every iteration we increase right tho.

        left = 0
        right = 0
        types = set()
        max_fruits = 0

        while right < len(fruits):
            curr = fruits[right]
            if len(types) < 2:
                types.add(curr)
            else:
                if curr not in types:
                    max_fruits = max(max_fruits, right - left)
                    left = right - 1
                    print(right)
                    print(left)
                    print("\n")
                    while left > 0 and fruits[left-1] == fruits[right - 1]:
                        left -= 1
                    types.add(fruits[right])
                    types.remove(fruits[left-1])                    
            right += 1

        max_fruits = max(max_fruits, right - left)
        return max_fruits



        #not really working overcomplicated approach

        max_fruits = 0
        most_recently_picked = -1
        fruit_dict = {}
        num_fruits = 0
    
        for index in range(len(fruits)):
            curr = fruits[index]
            if len(fruit_dict) < 2:
                #then we can add this fruit
                most_recently_picked = curr
                fruit_dict[curr] = fruit_dict.get(curr, 0) + 1
                num_fruits += 1
                print("added to a new basket")
                print(fruit_dict)
                print("\n")
            else:
                #we can't add a new fruit type
                if curr in fruit_dict:
                    #then we can pick this fruit up and keep going
                    fruit_dict[curr] += 1
                    most_recently_picked = curr
                    num_fruits += 1
                    print("added to existing basket")
                    print(fruit_dict)
                    print("\n")
                else:
                    #then we should update maxes and remove all fruits the first fruit type we picked up of the 2 fruit types we have currently
                    max_fruits = max(max_fruits, num_fruits)
                    for key in fruit_dict.keys():
                        if key != most_recently_picked:
                            break
                    #so now we have the least recently picked fruit
                    num_fruits -= fruit_dict[key]
                    fruit_dict.pop(key)
                    fruit_dict[curr] = 1
                    num_fruits += 1
                    most_recently_picked = curr
                    # print("removed from basket")
                    # print(fruit_dict)
                    # print(max_fruits)
                    # print("\n")

        return max(num_fruits, max_fruits)
23200