Sync LeetCode submission Runtime - 367 ms (50.80%), Memory - 29.7 MB (88.94%)

Author: jimit105

1335-maximum-candies-allocated-to-k-children/README.md

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+<p>You are given a <strong>0-indexed</strong> integer array <code>candies</code>. Each element in the array denotes a pile of candies of size <code>candies[i]</code>. You can divide each pile into any number of <strong>sub piles</strong>, but you <strong>cannot</strong> merge two piles together.</p>
+
+<p>You are also given an integer <code>k</code>. You should allocate piles of candies to <code>k</code> children such that each child gets the <strong>same</strong> number of candies. Each child can be allocated candies from <strong>only one</strong> pile of candies and some piles of candies may go unused.</p>
+
+<p>Return <em>the <strong>maximum number of candies</strong> each child can get.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> candies = [5,8,6], k = 3
+<strong>Output:</strong> 5
+<strong>Explanation:</strong> We can divide candies[1] into 2 piles of size 5 and 3, and candies[2] into 2 piles of size 5 and 1. We now have five piles of candies of sizes 5, 5, 3, 5, and 1. We can allocate the 3 piles of size 5 to 3 children. It can be proven that each child cannot receive more than 5 candies.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> candies = [2,5], k = 11
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> There are 11 children but only 7 candies in total, so it is impossible to ensure each child receives at least one candy. Thus, each child gets no candy and the answer is 0.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= candies.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= candies[i] &lt;= 10<sup>7</sup></code></li>
+	<li><code>1 &lt;= k &lt;= 10<sup>12</sup></code></li>
+</ul>

1335-maximum-candies-allocated-to-k-children/solution.py

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+# Approach: Binary Search on The Answer
+
+# n = len(candies), m = max(candies)
+# Time: O(n log m)
+# Space: O(1)
+
+class Solution:
+    def maximumCandies(self, candies: List[int], k: int) -> int:
+        max_candies_in_pile = max(candies)
+
+        left = 0
+        right = max_candies_in_pile
+
+        while left < right:
+            mid = (left + right + 1) // 2
+
+            if self._can_allocate_candies(candies, k, mid):
+                # If possible, move to the upper half to search for a larger number
+                left = mid
+            else:
+                right = mid - 1
+
+        return left
+
+    def _can_allocate_candies(self, candies, k, num_of_candies):
+        max_num_of_children = 0
+
+        for pile in candies:
+            max_num_of_children += pile // num_of_candies
+
+        return max_num_of_children >= k