<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>

from bisect import bisect_left
class Solution:
    def makeArrayIncreasing(self, arr1: List[int], arr2: List[int]) -> int:

        # arr2 = list(set(sorted(arr2)))
        arr2.sort()
        n,m = len(arr1), len(arr2)
        dp = {}
        def dfs(i, prev):
            if i == n:
                return 0
            if (i,prev) in dp:
                return dp[(i,prev)]
            
            opt1, opt2 = float("inf"), float("inf")
            k = bisect_left(arr2, prev+1)
            if k < m:
                opt1 = 1 + dfs(i+1, arr2[k])
            if arr1[i] > prev:
                opt2 = dfs(i+1, arr1[i])
            res = min(opt1, opt2)
            dp[(i,prev)] = res
            return res
            
        res = dfs(0,float("-inf"))
        return res if res != float("inf") else -1
        # n, m = len(arr1), len(arr2)
        # arr2.sort()
        
        # @cache
        # def dfs(i: int, p: int) -> int:
        #     if i == n: return 0
        #     j = bisect_left(arr2, p + 1)
        #     return min(
        #         dfs(i + 1, arr2[j]) + 1 if j < m else inf,
        #         dfs(i + 1, arr1[i]) if i == 0 or arr1[i] > p else inf
        #     )

        # res = dfs(0, min(arr2) - 1)
        # return res if res < inf else -1

            

            


        


        
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