Sync LeetCode submission Runtime - 670 ms (34.28%), Memory - 52.5 MB (50.71%)

Author: jimit105

2229-maximum-fruits-harvested-after-at-most-k-steps/README.md

@@ -0,0 +1,54 @@
+<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>

2229-maximum-fruits-harvested-after-at-most-k-steps/solution.py

@@ -0,0 +1,34 @@
+# Approach 1: Binary Search
+
+# Time: O(n + k log n)
+# Space: O(n)
+
+class Solution:
+    def maxTotalFruits(self, fruits: List[List[int]], startPos: int, k: int) -> int:
+        n = len(fruits)
+        sum_ = [0] * (n + 1)
+        indices = [0] * n
+
+        for i in range(n):
+            sum_[i + 1] = sum_[i] + fruits[i][1]
+            indices[i] = fruits[i][0]
+
+        ans = 0
+        for x in range(k // 2 + 1):
+            # Move left x steps, then right (k - 2x) steps
+            y = k - 2 * x
+            left = startPos - x
+            right = startPos + y
+            start = bisect_left(indices, left)
+            end = bisect_right(indices, right)
+            ans = max(ans, sum_[end] - sum_[start])
+
+            # Move right x steps, then left (k - 2x) steps
+            y = k - 2 * x
+            left = startPos - y
+            right = startPos + x
+            start = bisect_left(indices, left)
+            end = bisect_right(indices, right)
+            ans = max(ans, sum_[end] - sum_[start])
+
+        return ans