<p>Fruits are available at some positions on an infinite x-axis. You are given a 2D integer array <code>fruits</code> where <code>fruits[i] = [position<sub>i</sub>, amount<sub>i</sub>]</code> depicts <code>amount<sub>i</sub></code> fruits at the position <code>position<sub>i</sub></code>. <code>fruits</code> is already <strong>sorted</strong> by <code>position<sub>i</sub></code> in <strong>ascending order</strong>, and each <code>position<sub>i</sub></code> is <strong>unique</strong>.</p>

<p>You are also given an integer <code>startPos</code> and an integer <code>k</code>. Initially, you are at the position <code>startPos</code>. From any position, you can either walk to the <strong>left or right</strong>. It takes <strong>one step</strong> to move <strong>one unit</strong> on the x-axis, and you can walk <strong>at most</strong> <code>k</code> steps in total. For every position you reach, you harvest all the fruits at that position, and the fruits will disappear from that position.</p>

<p>Return <em>the <strong>maximum total number</strong> of fruits you can harvest</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/11/21/1.png" style="width: 472px; height: 115px;" />
<pre>
<strong>Input:</strong> fruits = [[2,8],[6,3],[8,6]], startPos = 5, k = 4
<strong>Output:</strong> 9
<strong>Explanation:</strong> 
The optimal way is to:
- Move right to position 6 and harvest 3 fruits
- Move right to position 8 and harvest 6 fruits
You moved 3 steps and harvested 3 + 6 = 9 fruits in total.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/11/21/2.png" style="width: 512px; height: 129px;" />
<pre>
<strong>Input:</strong> fruits = [[0,9],[4,1],[5,7],[6,2],[7,4],[10,9]], startPos = 5, k = 4
<strong>Output:</strong> 14
<strong>Explanation:</strong> 
You can move at most k = 4 steps, so you cannot reach position 0 nor 10.
The optimal way is to:
- Harvest the 7 fruits at the starting position 5
- Move left to position 4 and harvest 1 fruit
- Move right to position 6 and harvest 2 fruits
- Move right to position 7 and harvest 4 fruits
You moved 1 + 3 = 4 steps and harvested 7 + 1 + 2 + 4 = 14 fruits in total.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/11/21/3.png" style="width: 476px; height: 100px;" />
<pre>
<strong>Input:</strong> fruits = [[0,3],[6,4],[8,5]], startPos = 3, k = 2
<strong>Output:</strong> 0
<strong>Explanation:</strong>
You can move at most k = 2 steps and cannot reach any position with fruits.
</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>fruits[i].length == 2</code></li>
	<li><code>0 &lt;= startPos, position<sub>i</sub> &lt;= 2 * 10<sup>5</sup></code></li>
	<li><code>position<sub>i-1</sub> &lt; position<sub>i</sub></code> for any <code>i &gt; 0</code>&nbsp;(<strong>0-indexed</strong>)</li>
	<li><code>1 &lt;= amount<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= k &lt;= 2 * 10<sup>5</sup></code></li>
</ul>

class Solution:
    def maxTotalFruits(self, fruits: List[List[int]], startPos: int, k: int) -> int:
        
        n = max(fruits[-1][0], startPos)
        towards_rt_sum = [0] * (n + 2) # prefix
        towards_lf_sum = [0] * (n + 2) # suffix

        for idx, n_fruits in fruits:
            towards_rt_sum[idx+1] = n_fruits
            towards_lf_sum[idx] = n_fruits
        
        #print("OG:\n", towards_lf_sum)
        if k == 0:
            return towards_lf_sum[startPos]

        for i in range(1, n+2):
            towards_rt_sum[i] += towards_rt_sum[i-1]
            towards_lf_sum[n+1-i] += towards_lf_sum[n+2-i]
        
        
        #print(towards_lf_sum)
        #print(towards_rt_sum)

        # going left then turning right
        lf_reach = startPos - k
        rt_reach = startPos + k
        max_fruits = 0

        # as long as left remaining > right possibility
        #print(startPos, towards_lf_sum[startPos], towards_rt_sum[startPos])

        for mov in range(1, k+1):
            lf_move = max(startPos-mov, 0)
            rt_move = min(startPos+mov+1, n+1)
            lf_poss = max(min(lf_reach + (2 * mov), startPos), 0)
            rt_poss = min(max(rt_reach - (2 * mov)+1, startPos+1), n+1)

            #print(f"Left Gain: {towards_lf_sum[lf_move] - towards_lf_sum[startPos+1]}; Right Poss: {towards_rt_sum[rt_poss] - towards_rt_sum[startPos]}")
            #print(f"Right Gain: {towards_rt_sum[rt_move] - towards_rt_sum[startPos-1]}; Left Possi: {towards_lf_sum[lf_poss] - towards_lf_sum[startPos]}")


            max_fruits = max(
                max_fruits,
                towards_lf_sum[lf_move] - towards_lf_sum[startPos+1] + towards_rt_sum[rt_poss] - towards_rt_sum[startPos+1],
                towards_rt_sum[rt_move] - towards_rt_sum[startPos] + towards_lf_sum[lf_poss] - towards_lf_sum[startPos]
            )

        return max_fruits
        
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