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leetcode/problem/count-collisions-of-monkeys-on-a-polygon.html

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+<p>There is a regular convex polygon with <code>n</code> vertices. The vertices are labeled from <code>0</code> to <code>n - 1</code> in a clockwise direction, and each vertex has <strong>exactly one monkey</strong>. The following figure shows a convex polygon of <code>6</code> vertices.</p>
+<img alt="" src="https://assets.leetcode.com/uploads/2023/01/22/hexagon.jpg" style="width: 300px; height: 293px;" />
+<p>Each monkey moves simultaneously to a neighboring vertex. A neighboring vertex for a vertex <code>i</code> can be:</p>
+
+<ul>
+	<li>the vertex <code>(i + 1) % n</code> in the clockwise direction, or</li>
+	<li>the vertex <code>(i - 1 + n) % n</code> in the counter-clockwise direction.</li>
+</ul>
+
+<p>A <strong>collision</strong> happens if at least two monkeys reside on the same vertex after the movement.</p>
+
+<p>Return <em>the number of ways the monkeys can move so that at least <strong>one collision</strong></em> <em> happens</em>. Since the answer may be very large, return it modulo <code>10<sup>9 </sup>+ 7</code>.</p>
+
+<p><strong>Note</strong> that each monkey can only move once.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 3
+<strong>Output:</strong> 6
+<strong>Explanation:</strong> There are 8 total possible movements.
+Two ways such that they collide at some point are:
+- Monkey 1 moves in a clockwise direction; monkey 2 moves in an anticlockwise direction; monkey 3 moves in a clockwise direction. Monkeys 1 and 2 collide.
+- Monkey 1 moves in an anticlockwise direction; monkey 2 moves in an anticlockwise direction; monkey 3 moves in a clockwise direction. Monkeys 1 and 3 collide.
+It can be shown 6 total movements result in a collision.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 4
+<strong>Output:</strong> 14
+<strong>Explanation:</strong> It can be shown that there are 14 ways for the monkeys to collide.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>3 &lt;= n &lt;= 10<sup>9</sup></code></li>
+</ul>

leetcode/problem/count-distinct-numbers-on-board.html

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+<p>You are given a positive integer <code>n</code>, that is initially placed on a board. Every day, for <code>10<sup>9</sup></code> days, you perform the following procedure:</p>
+
+<ul>
+	<li>For each number <code>x</code> present on the board, find all numbers <code>1 &lt;= i &lt;= n</code> such that <code>x % i == 1</code>.</li>
+	<li>Then, place those numbers on the board.</li>
+</ul>
+
+<p>Return<em> the number of <strong>distinct</strong> integers present on the board after</em> <code>10<sup>9</sup></code> <em>days have elapsed</em>.</p>
+
+<p><strong>Note:</strong></p>
+
+<ul>
+	<li>Once a number is placed on the board, it will remain on it until the end.</li>
+	<li><code>%</code>&nbsp;stands&nbsp;for the modulo operation. For example,&nbsp;<code>14 % 3</code> is <code>2</code>.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 5
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> Initially, 5 is present on the board. 
+The next day, 2 and 4 will be added since 5 % 2 == 1 and 5 % 4 == 1. 
+After that day, 3 will be added to the board because 4 % 3 == 1. 
+At the end of a billion days, the distinct numbers on the board will be 2, 3, 4, and 5. 
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 3
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> 
+Since 3 % 2 == 1, 2 will be added to the board. 
+After a billion days, the only two distinct numbers on the board are 2 and 3. 
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 100</code></li>
+</ul>

leetcode/problem/count-increasing-quadruplets.html

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+<p>Given a <strong>0-indexed</strong> integer array <code>nums</code> of size <code>n</code> containing all numbers from <code>1</code> to <code>n</code>, return <em>the number of increasing quadruplets</em>.</p>
+
+<p>A quadruplet <code>(i, j, k, l)</code> is increasing if:</p>
+
+<ul>
+	<li><code>0 &lt;= i &lt; j &lt; k &lt; l &lt; n</code>, and</li>
+	<li><code>nums[i] &lt; nums[k] &lt; nums[j] &lt; nums[l]</code>.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,3,2,4,5]
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> 
+- When i = 0, j = 1, k = 2, and l = 3, nums[i] &lt; nums[k] &lt; nums[j] &lt; nums[l].
+- When i = 0, j = 1, k = 2, and l = 4, nums[i] &lt; nums[k] &lt; nums[j] &lt; nums[l]. 
+There are no other quadruplets, so we return 2.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,2,3,4]
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> There exists only one quadruplet with i = 0, j = 1, k = 2, l = 3, but since nums[j] &lt; nums[k], we return 0.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>4 &lt;= nums.length &lt;= 4000</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= nums.length</code></li>
+	<li>All the integers of <code>nums</code> are <strong>unique</strong>. <code>nums</code> is a permutation.</li>
+</ul>

leetcode/problem/put-marbles-in-bags.html

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+<p>You have <code>k</code> bags. You are given a <strong>0-indexed</strong> integer array <code>weights</code> where <code>weights[i]</code> is the weight of the <code>i<sup>th</sup></code> marble. You are also given the integer <code>k.</code></p>
+
+<p>Divide the marbles into the <code>k</code> bags according to the following rules:</p>
+
+<ul>
+	<li>No bag is empty.</li>
+	<li>If the <code>i<sup>th</sup></code> marble and <code>j<sup>th</sup></code> marble are in a bag, then all marbles with an index between the <code>i<sup>th</sup></code> and <code>j<sup>th</sup></code> indices should also be in that same bag.</li>
+	<li>If a bag consists of all the marbles with an index from <code>i</code> to <code>j</code> inclusively, then the cost of the bag is <code>weights[i] + weights[j]</code>.</li>
+</ul>
+
+<p>The <strong>score</strong> after distributing the marbles is the sum of the costs of all the <code>k</code> bags.</p>
+
+<p>Return <em>the <strong>difference</strong> between the <strong>maximum</strong> and <strong>minimum</strong> scores among marble distributions</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> weights = [1,3,5,1], k = 2
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> 
+The distribution [1],[3,5,1] results in the minimal score of (1+1) + (3+1) = 6. 
+The distribution [1,3],[5,1], results in the maximal score of (1+3) + (5+1) = 10. 
+Thus, we return their difference 10 - 6 = 4.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> weights = [1, 3], k = 2
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> The only distribution possible is [1],[3]. 
+Since both the maximal and minimal score are the same, we return 0.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= k &lt;= weights.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= weights[i] &lt;= 10<sup>9</sup></code></li>
+</ul>