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
type SegmentTree struct {
	nums, tr []int
}

func NewSegmentTree(nums []int) *SegmentTree {
	n := len(nums)
	tree := &SegmentTree{
		nums: nums,
		tr:   make([]int, n*4),
	}
	tree.build(1, 1, n)
	return tree
}

func (st *SegmentTree) build(u, l, r int) {
	if l == r {
		st.tr[u] = st.nums[l-1]
		return
	}
	mid := (l + r) >> 1
	st.build(u*2, l, mid)
	st.build(u*2+1, mid+1, r)
	st.pushup(u)
}

func (st *SegmentTree) modify(u, l, r, i, v int) {
	if l == r {
		st.tr[u] = v
		return
	}
	mid := (l + r) >> 1
	if i <= mid {
		st.modify(u*2, l, mid, i, v)
	} else {
		st.modify(u*2+1, mid+1, r, i, v)
	}
	st.pushup(u)
}

func (st *SegmentTree) query(u, l, r, v int) int {
	if st.tr[u] < v {
		return -1
	}
	if l == r {
		return l
	}
	mid := (l + r) >> 1
	if st.tr[u*2] >= v {
		return st.query(u*2, l, mid, v)
	}
	return st.query(u*2+1, mid+1, r, v)
}

func (st *SegmentTree) pushup(u int) {
	st.tr[u] = max(st.tr[u*2], st.tr[u*2+1])
}

func numOfUnplacedFruits(fruits []int, baskets []int) (ans int) {
	tree := NewSegmentTree(baskets)
	n := len(baskets)
	for _, x := range fruits {
		i := tree.query(1, 1, n, x)
		if i < 0 {
			ans++
		} else {
			tree.modify(1, 1, n, i, 0)
		}
	}
	return
}
