Sync LeetCode submission Runtime - 12599 ms (10.72%), Memory - 36.8 MB (83.48%)

Author: jimit105

3791-fruits-into-baskets-iii/README.md

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+<p>You are given two arrays of integers, <code>fruits</code> and <code>baskets</code>, each of length <code>n</code>, where <code>fruits[i]</code> represents the <strong>quantity</strong> of the <code>i<sup>th</sup></code> type of fruit, and <code>baskets[j]</code> represents the <strong>capacity</strong> of the <code>j<sup>th</sup></code> basket.</p>
+
+<p>From left to right, place the fruits according to these rules:</p>
+
+<ul>
+	<li>Each fruit type must be placed in the <strong>leftmost available basket</strong> with a capacity <strong>greater than or equal</strong> to the quantity of that fruit type.</li>
+	<li>Each basket can hold <b>only one</b> type of fruit.</li>
+	<li>If a fruit type <b>cannot be placed</b> in any basket, it remains <b>unplaced</b>.</li>
+</ul>
+
+<p>Return the number of fruit types that remain unplaced after all possible allocations are made.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">fruits = [4,2,5], baskets = [3,5,4]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li><code>fruits[0] = 4</code> is placed in <code>baskets[1] = 5</code>.</li>
+	<li><code>fruits[1] = 2</code> is placed in <code>baskets[0] = 3</code>.</li>
+	<li><code>fruits[2] = 5</code> cannot be placed in <code>baskets[2] = 4</code>.</li>
+</ul>
+
+<p>Since one fruit type remains unplaced, we return 1.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">fruits = [3,6,1], baskets = [6,4,7]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">0</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li><code>fruits[0] = 3</code> is placed in <code>baskets[0] = 6</code>.</li>
+	<li><code>fruits[1] = 6</code> cannot be placed in <code>baskets[1] = 4</code> (insufficient capacity) but can be placed in the next available basket, <code>baskets[2] = 7</code>.</li>
+	<li><code>fruits[2] = 1</code> is placed in <code>baskets[1] = 4</code>.</li>
+</ul>
+
+<p>Since all fruits are successfully placed, we return 0.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == fruits.length == baskets.length</code></li>
+	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= fruits[i], baskets[i] &lt;= 10<sup>9</sup></code></li>
+</ul>

3791-fruits-into-baskets-iii/solution.py

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+import math
+
+class Solution:
+    def numOfUnplacedFruits(self, fruits: List[int], baskets: List[int]) -> int:
+        n = len(baskets)
+        m = int(math.sqrt(n))
+        section = (n + m - 1) // m
+        count = 0
+        maxV = [0] * section
+
+        for i in range(n):
+            maxV[i // m] = max(maxV[i // m], baskets[i])
+
+        for fruit in fruits:
+            unset = 1
+            for sec in range(section):
+                if maxV[sec] < fruit:
+                    continue
+                choose = 0
+                maxV[sec] = 0
+                for i in range(m):
+                    pos = sec * m + i
+                    if pos < n and baskets[pos] >= fruit and not choose:
+                        baskets[pos] = 0
+                        choose = 1
+                    if pos < n:
+                        maxV[sec] = max(maxV[sec], baskets[pos])
+                unset = 0
+                break
+            count += unset
+        return count
+