Minimum Operations to Make a Subsequence
<p>You are given an array <code>target</code> that consists of <strong>distinct</strong> integers and another integer array <code>arr</code> that <strong>can</strong> have duplicates.</p>

<p>In one operation, you can insert any integer at any position in <code>arr</code>. For example, if <code>arr = [1,4,1,2]</code>, you can add <code>3</code> in the middle and make it <code>[1,4,<u>3</u>,1,2]</code>. Note that you can insert the integer at the very beginning or end of the array.</p>

<p>Return <em>the <strong>minimum</strong> number of operations needed to make </em><code>target</code><em> a <strong>subsequence</strong> of </em><code>arr</code><em>.</em></p>

<p>A <strong>subsequence</strong> of an array is a new array generated from the original array by deleting some elements (possibly none) without changing the remaining elements&#39; relative order. For example, <code>[2,7,4]</code> is a subsequence of <code>[4,<u>2</u>,3,<u>7</u>,2,1,<u>4</u>]</code> (the underlined elements), while <code>[2,4,2]</code> is not.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> target = [5,1,3], <code>arr</code> = [9,4,2,3,4]
<strong>Output:</strong> 2
<strong>Explanation:</strong> You can add 5 and 1 in such a way that makes <code>arr</code> = [<u>5</u>,9,4,<u>1</u>,2,3,4], then target will be a subsequence of <code>arr</code>.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> target = [6,4,8,1,3,2], <code>arr</code> = [4,7,6,2,3,8,6,1]
<strong>Output:</strong> 3
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= target.length, arr.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= target[i], arr[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>target</code> contains no duplicates.</li>
</ul>
Greedy; Array; Hash Table; Binary Search
class BinaryIndexedTree {
    private n: number;
    private c: number[];

    constructor(n: number) {
        this.n = n;
        this.c = Array(n + 1).fill(0);
    }

    update(x: number, v: number): void {
        for (; x <= this.n; x += x & -x) {
            this.c[x] = Math.max(this.c[x], v);
        }
    }

    query(x: number): number {
        let ans = 0;
        for (; x > 0; x -= x & -x) {
            ans = Math.max(ans, this.c[x]);
        }
        return ans;
    }
}

function minOperations(target: number[], arr: number[]): number {
    const m = target.length;
    const d: Map<number, number> = new Map();
    target.forEach((x, i) => d.set(x, i + 1));
    const nums: number[] = [];
    arr.forEach(x => {
        if (d.has(x)) {
            nums.push(d.get(x)!);
        }
    });
    const tree = new BinaryIndexedTree(m);
    let ans = 0;
    nums.forEach(x => {
        const v = tree.query(x - 1) + 1;
        ans = Math.max(ans, v);
        tree.update(x, v);
    });
    return m - ans;
}
