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
func makeArrayIncreasing(arr1 []int, arr2 []int) int {
	sort.Ints(arr2)
	m := 0
	for _, x := range arr2 {
		if m == 0 || x != arr2[m-1] {
			arr2[m] = x
			m++
		}
	}
	arr2 = arr2[:m]
	const inf = 1 << 30
	arr1 = append([]int{-inf}, arr1...)
	arr1 = append(arr1, inf)
	n := len(arr1)
	f := make([]int, n)
	for i := range f {
		f[i] = inf
	}
	f[0] = 0
	for i := 1; i < n; i++ {
		if arr1[i-1] < arr1[i] {
			f[i] = f[i-1]
		}
		j := sort.SearchInts(arr2, arr1[i])
		for k := 1; k <= min(i-1, j); k++ {
			if arr1[i-k-1] < arr2[j-k] {
				f[i] = min(f[i], f[i-k-1]+k)
			}
		}
	}
	if f[n-1] >= inf {
		return -1
	}
	return f[n-1]
}