[LeetCode Sync] Runtime - 1 ms (68.00%), Memory - 13.4 MB (16.38%)

Author: NamVr

2692-take-gifts-from-the-richest-pile/README.md

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+<p>You are given an integer array <code>gifts</code> denoting the number of gifts in various piles. Every second, you do the following:</p>
+
+<ul>
+	<li>Choose the pile with the maximum number of gifts.</li>
+	<li>If there is more than one pile with the maximum number of gifts, choose any.</li>
+	<li>Leave behind the floor of the square root of the number of gifts in the pile. Take the rest of the gifts.</li>
+</ul>
+
+<p>Return <em>the number of gifts remaining after </em><code>k</code><em> seconds.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> gifts = [25,64,9,4,100], k = 4
+<strong>Output:</strong> 29
+<strong>Explanation:</strong> 
+The gifts are taken in the following way:
+- In the first second, the last pile is chosen and 10 gifts are left behind.
+- Then the second pile is chosen and 8 gifts are left behind.
+- After that the first pile is chosen and 5 gifts are left behind.
+- Finally, the last pile is chosen again and 3 gifts are left behind.
+The final remaining gifts are [5,8,9,4,3], so the total number of gifts remaining is 29.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> gifts = [1,1,1,1], k = 4
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> 
+In this case, regardless which pile you choose, you have to leave behind 1 gift in each pile. 
+That is, you can&#39;t take any pile with you. 
+So, the total gifts remaining are 4.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= gifts.length &lt;= 10<sup>3</sup></code></li>
+	<li><code>1 &lt;= gifts[i] &lt;= 10<sup>9</sup></code></li>
+	<li><code>1 &lt;= k &lt;= 10<sup>3</sup></code></li>
+</ul>

2692-take-gifts-from-the-richest-pile/solution.cpp

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+class Solution {
+public:
+    long long pickGifts(vector<int>& gifts, int k) {
+        priority_queue<int> h(gifts.begin(), gifts.end());
+
+        for (int i = 0; i < k; i++) {
+            int maxElement = h.top();
+            h.pop();
+
+            h.push(sqrt(maxElement)); // heapify
+        }
+
+        long long res = 0;
+        while (!h.empty()) {
+            res += h.top();
+            h.pop();
+        }
+
+        return res;
+    }
+};