<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:
        
        sz =fruits[-1][0]+1
        # sz =10**5
        arr = [0]*(sz)
        
        if startPos-k > fruits[-1][0]:return 0
        
        if k==0:
            return sum([x[1] for x in fruits if x[0]==startPos])
        for a,b in fruits:
            arr[a] = b
        
        prefix = list(accumulate(arr))
        def rangeq(a,b):
            # if a<0 or b<0 or a>=sz or b>=sz:
            #     return 0
            a = max(a,0)
            a = min(a,sz-1)
            
            b = max(b,0)
            b = min(b,sz-1)
            
            if a==0:
                return prefix[b]
            return prefix[b] - prefix[a-1]
        res = 0
        
        for i in range(startPos,sz):
            balance = k
            val = 0
            range1 = i-startPos
            if range1<=balance:
                # print("at",i,range1,rangeq(startPos,i))
                val+=rangeq(startPos,i)
                balance-=range1
            idx = i-balance
            idx = max(idx,0)
            if idx<startPos:
                val+=rangeq(idx,startPos-1)
            res = max(res,val)
        for i in range(startPos,-1,-1):
            balance = k
            val = 0
            range1 = startPos - i
            if range1>0 and range1<=balance:
                # print("at",i,range1,rangeq(startPos,i))
                val+=rangeq(i,startPos)
                balance-=range1
            idx = i+balance
            idx = min(idx,len(arr)-1)
            if startPos<idx:
                val+=rangeq(startPos+1,idx)
            res = max(res,val)
        return res
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