Dynamic Programming approach
To make your solution more efficient, we can optimize it further by eliminating redundant computations and reducing the recursion depth where possible. Here's a
To make your solution more efficient, we can optimize it further by eliminating redundant computations and reducing the recursion depth where possible. Here's an optimized version of your solution:

Backtracking

```python
from functools import lru_cache
from typing import List

class Solution:
    def maxNonDecreasingLength(self, nums1: List[int], nums2: List[int]) -> int:
        n = len(nums1)
        result = 1

        @lru_cache(None)
        def solve(index: int, prev: int) -> int:
            if index >= n:
                return 0

            # Calculate the maximum length for each option
            option1 = solve(index + 1, nums1[index]) + 1 if prev <= nums1[index] else 0
            option2 = solve(index + 1, nums2[index]) + 1 if prev <= nums2[index] else 0

            return max(option1, option2)

        # Iterate through all starting points
        for i in range(n):
            result = max(result, solve(i, 0))

        solve.cache_clear()  # Clear the cache after usage
        return result
```
DP
```python
from typing import List

class Solution:
    def maxNonDecreasingLength(self, nums1: List[int], nums2: List[int]) -> int:
        n = len(nums1)

        # Initialize dynamic programming arrays
        dp1 = [1] * n  # Tracks max length ending at i using nums1
        dp2 = [1] * n  # Tracks max length ending at i using nums2

        for i in range(1, n):
            # Extend sequences from previous index if conditions are met
            if nums1[i] >= nums1[i - 1]:
                dp1[i] = max(dp1[i], dp1[i - 1] + 1)
            if nums1[i] >= nums2[i - 1]:
                dp1[i] = max(dp1[i], dp2[i - 1] + 1)
            if nums2[i] >= nums1[i - 1]:
                dp2[i] = max(dp2[i], dp1[i - 1] + 1)
            if nums2[i] >= nums2[i - 1]:
                dp2[i] = max(dp2[i], dp2[i - 1] + 1)

        # Return the maximum length from both arrays
        return max(max(dp1), max(dp2))
```

### Explanation of Improvements:
1. **Dynamic Programming**:
   - Instead of using recursion and caching, we employ a dynamic programming (DP) approach to store intermediate results directly in arrays (`dp1` and `dp2`).
   - This reduces the overhead of recursive calls and cache lookups, making the solution more efficient.

2. **Iterative Approach**:
   - Replaced the recursive approach with a loop, which is generally faster and avoids issues with recursion depth for large inputs.

3. **Space Optimization**:
   - The DP arrays (`dp1` and `dp2`) track the maximum non-decreasing subsequence length at each index for `nums1` and `nums2`, respectively, reducing the need for additional memory.

4. **Avoided Redundant Computations**:
   - The iterative method directly updates the results based on valid transitions between indices, eliminating redundant work.

This implementation has a time complexity of **O(n)** and space complexity of **O(n)** due to the DP arrays. However, if further optimization is required, we can attempt to reduce the space complexity to **O(1)** by keeping only the last state instead of full arrays.
['Python3']