Fruits Into Baskets II
<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;= 100</code></li>
	<li><code>1 &lt;= fruits[i], baskets[i] &lt;= 1000</code></li>
</ul>
Segment Tree; Array; Binary Search; Ordered Set; Simulation
func numOfUnplacedFruits(fruits []int, baskets []int) int {
	n := len(fruits)
	ans := n
	vis := make([]bool, n)
	for _, x := range fruits {
		for i, y := range baskets {
			if y >= x && !vis[i] {
				vis[i] = true
				ans--
				break
			}
		}
	}
	return ans
}
