<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 {
public:
    int totalFruit(vector<int>& fruits) {
        int l=0, r=0, ans=0;
        pair<int, int> p1 = {-1, 0}, p2 = {-1, 0};
        while(r<fruits.size()){
            
            if(p1.first == -1){
                p1.first = fruits[r];
                p1.second++;
                r++;
                ans = max(ans, p1.second + p2.second);
            }
            else if(fruits[r] != p1.first && p2.first == -1){
                p2.first = fruits[r];
                p2.second++;
                r++;
                ans = max(ans, p1.second + p2.second);
            }
            else if(fruits[r] == p1.first){
                p1.second++;
                r++;
                ans = max(ans, p1.second + p2.second);
            }
            else if(fruits[r] == p2.first){
                p2.second++;
                r++;
                
                ans = max(ans, p1.second + p2.second);
            }
            else{
                while(p1.second != 0 && p2.second != 0){
                    if(fruits[l] == p1.first)
                        p1.second--;
                    if(fruits[l] == p2.first)
                        p2.second--;
                    l++;
                }
                if(p1.second == 0){
                    p1.first = fruits[r];
                    p1.second++;
                    r++;
                    ans = max(ans, p1.second + p2.second);
                }
                else{
                    p2.first = fruits[r];
                    p2.second++;
                    r++;
                    ans = max(ans, p1.second + p2.second);
                }
            }
        }
        return ans;
    }
};
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