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            "title": "Maximum Fruits Harvested After at Most K Steps",
            "titleSlug": "maximum-fruits-harvested-after-at-most-k-steps",
            "content": "<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>\n\n<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>\n\n<p>Return <em>the <strong>maximum total number</strong> of fruits you can harvest</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/11/21/1.png\" style=\"width: 472px; height: 115px;\" />\n<pre>\n<strong>Input:</strong> fruits = [[2,8],[6,3],[8,6]], startPos = 5, k = 4\n<strong>Output:</strong> 9\n<strong>Explanation:</strong> \nThe optimal way is to:\n- Move right to position 6 and harvest 3 fruits\n- Move right to position 8 and harvest 6 fruits\nYou moved 3 steps and harvested 3 + 6 = 9 fruits in total.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/11/21/2.png\" style=\"width: 512px; height: 129px;\" />\n<pre>\n<strong>Input:</strong> fruits = [[0,9],[4,1],[5,7],[6,2],[7,4],[10,9]], startPos = 5, k = 4\n<strong>Output:</strong> 14\n<strong>Explanation:</strong> \nYou can move at most k = 4 steps, so you cannot reach position 0 nor 10.\nThe optimal way is to:\n- Harvest the 7 fruits at the starting position 5\n- Move left to position 4 and harvest 1 fruit\n- Move right to position 6 and harvest 2 fruits\n- Move right to position 7 and harvest 4 fruits\nYou moved 1 + 3 = 4 steps and harvested 7 + 1 + 2 + 4 = 14 fruits in total.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/11/21/3.png\" style=\"width: 476px; height: 100px;\" />\n<pre>\n<strong>Input:</strong> fruits = [[0,3],[6,4],[8,5]], startPos = 3, k = 2\n<strong>Output:</strong> 0\n<strong>Explanation:</strong>\nYou can move at most k = 2 steps and cannot reach any position with fruits.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= fruits.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>fruits[i].length == 2</code></li>\n\t<li><code>0 &lt;= startPos, position<sub>i</sub> &lt;= 2 * 10<sup>5</sup></code></li>\n\t<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>\n\t<li><code>1 &lt;= amount<sub>i</sub> &lt;= 10<sup>4</sup></code></li>\n\t<li><code>0 &lt;= k &lt;= 2 * 10<sup>5</sup></code></li>\n</ul>\n",
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            "difficulty": "Hard",
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                    "name": "Array",
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                    "name": "Binary Search",
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                    "code": "class Solution {\npublic:\n    int maxTotalFruits(vector<vector<int>>& fruits, int startPos, int k) {\n        \n    }\n};",
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                    "code": "class Solution(object):\n    def maxTotalFruits(self, fruits, startPos, k):\n        \"\"\"\n        :type fruits: List[List[int]]\n        :type startPos: int\n        :type k: int\n        :rtype: int\n        \"\"\"\n        ",
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                    "code": "public class Solution {\n    public int MaxTotalFruits(int[][] fruits, int startPos, int k) {\n        \n    }\n}",
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                    "code": "/**\n * @param {number[][]} fruits\n * @param {number} startPos\n * @param {number} k\n * @return {number}\n */\nvar maxTotalFruits = function(fruits, startPos, k) {\n    \n};",
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                    "code": "# @param {Integer[][]} fruits\n# @param {Integer} start_pos\n# @param {Integer} k\n# @return {Integer}\ndef max_total_fruits(fruits, start_pos, k)\n    \nend",
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                    "code": "class Solution {\n    func maxTotalFruits(_ fruits: [[Int]], _ startPos: Int, _ k: Int) -> Int {\n        \n    }\n}",
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                    "lang": "Go",
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                    "code": "func maxTotalFruits(fruits [][]int, startPos int, k int) int {\n    \n}",
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                    "code": "object Solution {\n    def maxTotalFruits(fruits: Array[Array[Int]], startPos: Int, k: Int): Int = {\n        \n    }\n}",
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                    "lang": "Kotlin",
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                    "code": "class Solution {\n    fun maxTotalFruits(fruits: Array<IntArray>, startPos: Int, k: Int): Int {\n        \n    }\n}",
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                    "lang": "Rust",
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                    "code": "impl Solution {\n    pub fn max_total_fruits(fruits: Vec<Vec<i32>>, start_pos: i32, k: i32) -> i32 {\n        \n    }\n}",
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                    "code": "class Solution {\n\n    /**\n     * @param Integer[][] $fruits\n     * @param Integer $startPos\n     * @param Integer $k\n     * @return Integer\n     */\n    function maxTotalFruits($fruits, $startPos, $k) {\n        \n    }\n}",
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                    "code": "function maxTotalFruits(fruits: number[][], startPos: number, k: number): number {\n\n};",
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                    "lang": "Erlang",
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                    "code": "-spec max_total_fruits(Fruits :: [[integer()]], StartPos :: integer(), K :: integer()) -> integer().\nmax_total_fruits(Fruits, StartPos, K) ->\n  .",
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                    "code": "defmodule Solution do\n  @spec max_total_fruits(fruits :: [[integer]], start_pos :: integer, k :: integer) :: integer\n  def max_total_fruits(fruits, start_pos, k) do\n\n  end\nend",
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            "hints": [
                "Does an optimal path have very few patterns? For example, could a path that goes left, turns and goes right, then turns again and goes left be any better than a path that simply goes left, turns, and goes right?",
                "The optimal path turns at most once. That is, the optimal path is one of these: to go left only; to go right only; to go left, turn and go right; or to go right, turn and go left.",
                "Moving x steps left then k-x steps right gives you a range of positions that you can reach.",
                "Use prefix sums to get the sum of all fruits for each possible range.",
                "Use a similar strategy for all the paths that go right, then turn and go left."
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