human
Solve in Python:
Given two integer arrays arr1 and arr2, return the minimum number of operations (possibly zero) needed to make arr1 strictly increasing.
In one operation, you can choose two indices 0 <= i < arr1.length and 0 <= j < arr2.length and do the assignment arr1[i] = arr2[j].
If there is no way to make arr1 strictly increasing, return -1.
 
Example 1:
Input: arr1 = [1,5,3,6,7], arr2 = [1,3,2,4]
Output: 1
Explanation: Replace 5 with 2, then arr1 = [1, 2, 3, 6, 7].

Example 2:
Input: arr1 = [1,5,3,6,7], arr2 = [4,3,1]
Output: 2
Explanation: Replace 5 with 3 and then replace 3 with 4. arr1 = [1, 3, 4, 6, 7].

Example 3:
Input: arr1 = [1,5,3,6,7], arr2 = [1,6,3,3]
Output: -1
Explanation: You can't make arr1 strictly increasing.
 
Constraints:

1 <= arr1.length, arr2.length <= 2000
0 <= arr1[i], arr2[i] <= 10^9
 Can you provide a the Python program you requested?
assistant
#Here is the Python program you requested.

import bisect
from functools import lru_cache

class Solution:
    def makeArrayIncreasing(self, A: List[int], B: List[int]) -> int:
        B = sorted(set(B))
        b2idx = {b: i for i, b in enumerate(B)}
        
        @lru_cache(None)
        def find_larger_value_in_B(val):
            if val >= B[-1]:
                return float('inf')
            if val in b2idx:
                return B[b2idx[val] + 1]
            return B[bisect.bisect_right(B, val)]

        @lru_cache(None)
        def min_last_value_given_operations(n, ops):
            if ops < 0:
                return float('inf')
            elif n == 0:
                return float('-inf')
            elif ops > n:
                return min_last_value_given_operations(n, n)

            skip_op = min_last_value_given_operations(n - 1, ops)
            if skip_op == float('inf'):
                return float('inf')
            return min(
                A[n - 1] if skip_op < A[n - 1] else float('inf'),
                find_larger_value_in_B(min_last_value_given_operations(n - 1, ops - 1)),
            )

        last_success = -1
        for ops in range(min(len(A), len(B)), -1, -1):
            if min_last_value_given_operations(len(A), ops) == float('inf'):
                break
            last_success = ops

        return last_success