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
class Solution:
    def makeArrayIncreasing(self, arr1: List[int], arr2: List[int]) -> int:
        arr2.sort()
        m = 0
        for x in arr2:
            if m == 0 or x != arr2[m - 1]:
                arr2[m] = x
                m += 1
        arr2 = arr2[:m]
        arr = [-inf] + arr1 + [inf]
        n = len(arr)
        f = [inf] * n
        f[0] = 0
        for i in range(1, n):
            if arr[i - 1] < arr[i]:
                f[i] = f[i - 1]
            j = bisect_left(arr2, arr[i])
            for k in range(1, min(i - 1, j) + 1):
                if arr[i - k - 1] < arr2[j - k]:
                    f[i] = min(f[i], f[i - k - 1] + k)
        return -1 if f[n - 1] >= inf else f[n - 1]
