Make Array Strictly Increasing
<p>Given two integer arrays&nbsp;<code>arr1</code> and <code>arr2</code>, return the minimum number of operations (possibly zero) needed&nbsp;to make <code>arr1</code> strictly increasing.</p>

<p>In one operation, you can choose two indices&nbsp;<code>0 &lt;=&nbsp;i &lt; arr1.length</code>&nbsp;and&nbsp;<code>0 &lt;= j &lt; arr2.length</code>&nbsp;and do the assignment&nbsp;<code>arr1[i] = arr2[j]</code>.</p>

<p>If there is no way to make&nbsp;<code>arr1</code>&nbsp;strictly increasing,&nbsp;return&nbsp;<code>-1</code>.</p>

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

<pre>
<strong>Input:</strong> arr1 = [1,5,3,6,7], arr2 = [1,3,2,4]
<strong>Output:</strong> 1
<strong>Explanation:</strong> Replace <code>5</code> with <code>2</code>, then <code>arr1 = [1, 2, 3, 6, 7]</code>.
</pre>

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

<pre>
<strong>Input:</strong> arr1 = [1,5,3,6,7], arr2 = [4,3,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Replace <code>5</code> with <code>3</code> and then replace <code>3</code> with <code>4</code>. <code>arr1 = [1, 3, 4, 6, 7]</code>.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr1 = [1,5,3,6,7], arr2 = [1,6,3,3]
<strong>Output:</strong> -1
<strong>Explanation:</strong> You can&#39;t make <code>arr1</code> strictly increasing.</pre>

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

<ul>
	<li><code>1 &lt;= arr1.length, arr2.length &lt;= 2000</code></li>
	<li><code>0 &lt;= arr1[i], arr2[i] &lt;= 10^9</code></li>
</ul>

<p>&nbsp;</p>
Array; Binary Search; Dynamic Programming; Sorting
function makeArrayIncreasing(arr1: number[], arr2: number[]): number {
    arr2.sort((a, b) => a - b);
    let m = 0;
    for (const x of arr2) {
        if (m === 0 || x !== arr2[m - 1]) {
            arr2[m++] = x;
        }
    }
    arr2 = arr2.slice(0, m);
    const inf = 1 << 30;
    arr1 = [-inf, ...arr1, inf];
    const n = arr1.length;
    const f: number[] = new Array(n).fill(inf);
    f[0] = 0;
    const search = (arr: number[], x: number): number => {
        let l = 0;
        let r = arr.length;
        while (l < r) {
            const mid = (l + r) >> 1;
            if (arr[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    for (let i = 1; i < n; ++i) {
        if (arr1[i - 1] < arr1[i]) {
            f[i] = f[i - 1];
        }
        const j = search(arr2, arr1[i]);
        for (let k = 1; k <= Math.min(i - 1, j); ++k) {
            if (arr1[i - k - 1] < arr2[j - k]) {
                f[i] = Math.min(f[i], f[i - k - 1] + k);
            }
        }
    }
    return f[n - 1] >= inf ? -1 : f[n - 1];
}
