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 {
    int[] nums;
    int[] tr;

    public SegmentTree(int[] nums) {
        this.nums = nums;
        int n = nums.length;
        this.tr = new int[n << 2];
        build(1, 1, n);
    }

    public void build(int u, int l, int r) {
        if (l == r) {
            tr[u] = nums[l - 1];
            return;
        }
        int mid = (l + r) >> 1;
        build(u << 1, l, mid);
        build(u << 1 | 1, mid + 1, r);
        pushup(u);
    }

    public void modify(int u, int l, int r, int i, int v) {
        if (l == r) {
            tr[u] = v;
            return;
        }
        int mid = (l + r) >> 1;
        if (i <= mid) {
            modify(u << 1, l, mid, i, v);
        } else {
            modify(u << 1 | 1, mid + 1, r, i, v);
        }
        pushup(u);
    }

    public int query(int u, int l, int r, int v) {
        if (tr[u] < v) {
            return -1;
        }
        if (l == r) {
            return l;
        }
        int mid = (l + r) >> 1;
        if (tr[u << 1] >= v) {
            return query(u << 1, l, mid, v);
        }
        return query(u << 1 | 1, mid + 1, r, v);
    }

    public void pushup(int u) {
        tr[u] = Math.max(tr[u << 1], tr[u << 1 | 1]);
    }
}

class Solution {
    public int numOfUnplacedFruits(int[] fruits, int[] baskets) {
        SegmentTree tree = new SegmentTree(baskets);
        int n = baskets.length;
        int ans = 0;
        for (int x : fruits) {
            int i = tree.query(1, 1, n, x);
            if (i < 0) {
                ans++;
            } else {
                tree.modify(1, 1, n, i, 0);
            }
        }
        return ans;
    }
}
