[LeetCode Sync] Runtime - 163 ms (91.04%), Memory - 29.6 MB (88.73%)

Author: NamVr

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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+class Solution:
+    def maximumCandies(self, candies: List[int], k: int) -> int:
+        # get O(n) total number of candies
+        total = sum(candies)
+
+        if total < k:
+            # if total candies are less than the number of children, theres no equal division since some kid will be left hungry.
+            return 0
+        
+        # run a binary search between 1 and total/k, the ideal max size of candy pile for each kid
+        l, r = 1, total // k
+
+        # we will store our final result as 0 initially
+        res = 0
+
+        while l <= r:
+            m = (l + r) // 2
+
+            # count will store number of piles for this iteration
+            count = 0
+
+            # run a linear check to see if this result is valid or not
+            # check for each candy in the array, if you can make piles of value [mid]
+            for c in candies:
+                count += c // m
+
+                # minor optimization, if anytime the count is above k, we can break the inner loop
+                if count >= k:
+                    break
+            
+            # check if count is valid or not (arithmetic check with k)
+            if count >= k:
+                # the result is valid
+                res = m
+                l = m+1
+            else:
+                # the result is not valid and search in lower half
+                r = m-1
+                
+
+        return res