<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>

class Solution:
    def makeArrayIncreasing(self, arr1: List[int], arr2: List[int]) -> int:
        
        # Sort arr2 to make it easy to look for an element to swap
        arr2.sort()

        # Initialize a cache which will be used for memoization. Initially, 
        # our recursive function will reach all the way to the end of the array 
        # before returning. Therefore, we can store the future dfs calls instead
        # of having to calculate them again. 
        cache = {}

        def dfs(idx: int, prev: int) -> int:
            
            # If we have reached the end of the array, then we hit a base case
            # and return 0 since we don't need to swap anything.
            if idx == len(arr1):
                return 0
            
            # If this function call is already stored in the cache, then we can
            # just return that instead.
            if (idx, prev) in cache:
                return cache[(idx, prev)]
            
            # We initialize the number of swaps to be infinite. This will help
            # us check if we can't make arr1 strictly increasing.
            swaps = float('inf')

            # If we don't need to swap the current element, we explore the rest
            # of the array
            if arr1[idx] > prev:
                swaps = dfs(idx+1, arr1[idx])
            
            # This is the case where we do perform a swap, either because we need
            # to or because we can. 

            # We get the index of the smallest element in arr2 which is greater
            # than arr1[idx-1].
            swap_idx = bisect.bisect_right(arr2, prev)

            # If this index is in range, then we have found an element and now
            # we can explore the rest of the array with 1 + dfs(idx+1, arr2[idx])
            # swaps. Finally, we take the minimum of the two choices.
            if swap_idx < len(arr2):
                swaps = min(swaps, 1+dfs(idx+1, arr2[swap_idx]))

            # We store the current result in the cache
            cache[(idx, prev)] = swaps
            
            # We return the cost for the higher levels of recursion
            return swaps

        # We call the function from the start of the array and return the number
        # of swaps if it if not infinity (since that indicates the problem can't
        # be solved)
        return dfs(0, -1) if dfs(0, -1) != float('inf') else -1     
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