Fruits Into Baskets III
<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;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= fruits[i], baskets[i] &lt;= 10<sup>9</sup></code></li>
</ul>
Segment Tree; Array; Binary Search; Ordered Set
class SegmentTree:
    __slots__ = ["nums", "tr"]

    def __init__(self, nums):
        self.nums = nums
        n = len(nums)
        self.tr = [0] * (n << 2)
        self.build(1, 1, n)

    def build(self, u, l, r):
        if l == r:
            self.tr[u] = self.nums[l - 1]
            return
        mid = (l + r) >> 1
        self.build(u << 1, l, mid)
        self.build(u << 1 | 1, mid + 1, r)
        self.pushup(u)

    def modify(self, u, l, r, i, v):
        if l == r:
            self.tr[u] = v
            return
        mid = (l + r) >> 1
        if i <= mid:
            self.modify(u << 1, l, mid, i, v)
        else:
            self.modify(u << 1 | 1, mid + 1, r, i, v)
        self.pushup(u)

    def query(self, u, l, r, v):
        if self.tr[u] < v:
            return -1
        if l == r:
            return l
        mid = (l + r) >> 1
        if self.tr[u << 1] >= v:
            return self.query(u << 1, l, mid, v)
        return self.query(u << 1 | 1, mid + 1, r, v)

    def pushup(self, u):
        self.tr[u] = max(self.tr[u << 1], self.tr[u << 1 | 1])


class Solution:
    def numOfUnplacedFruits(self, fruits: List[int], baskets: List[int]) -> int:
        tree = SegmentTree(baskets)
        n = len(baskets)
        ans = 0
        for x in fruits:
            i = tree.query(1, 1, n, x)
            if i < 0:
                ans += 1
            else:
                tree.modify(1, 1, n, i, 0)
        return ans
