100% | Easy-to-understand Python Solution
IntuitionThe problem asks us to choose one element from either of the two arrays at every index such that the overall sequence is non-decreasing. Using dynamic
# Intuition
The problem asks us to choose one element from either of the two arrays at every index such that the overall sequence is non-decreasing. Using dynamic programming to maintain two states at each index: one representing the longest non-decreasing sequence ending at that index if we choose an element from nums1, and the other if we choose from nums2. We can then extend these sequences by comparing the current elements with the previous elements chosen from either array.

# Approach
We maintain two variables:

`prev0`: The length of the longest non-decreasing sequence ending at the previous index when choosing from `nums1`.
`prev1`: The length of the longest non-decreasing sequence ending at the previous index when choosing from `nums2`.

For each index `i` from `1` to `n-1`, we:
a. Initialize the current sequence lengths (`cur0` and `cur1`) to 1 (starting a new sequence).
b. Update `cur0` if the current element from `nums1` can extend the sequence from the previous index.
c. Update `cur1` similarly using the current element from `nums2`.
d. Update a global maximum sequence length (`max_so_far`).

Finally, we notice that since each state only depends on the immediate previous state, we can optimize the space by only storing values for the previous index.

# Complexity
- Time complexity: O(n)

- Space complexity: O(1)
- 
# Code
```python3 []
from typing import List

class Solution:
    def maxNonDecreasingLength(self, nums1: List[int], nums2: List[int]) -> int:
        n = len(nums1)
        if n == 0:
            return 0

        # Initialize the DP values for index 0
        prev0 = 1  # last value chosen from nums1
        prev1 = 1  # last value chosen from nums2
        max_so_far = 1

        # Iterate once over the arrays
        for i in range(1, n):
            # Cache current and previous elements for speed
            a, b = nums1[i], nums2[i]
            a_prev, b_prev = nums1[i-1], nums2[i-1]
            
            # Start new sequences by default
            cur0 = 1  
            cur1 = 1
            
            # Transition if we choose nums1[i]
            if a >= a_prev:
                cur0 = prev0 + 1
            if a >= b_prev:
                tmp = prev1 + 1
                if tmp > cur0:
                    cur0 = tmp
            
            # Transition if we choose nums2[i]
            if b >= a_prev:
                cur1 = prev0 + 1
            if b >= b_prev:
                tmp = prev1 + 1
                if tmp > cur1:
                    cur1 = tmp

            # Update the maximum sequence length so far
            if cur0 > max_so_far:
                max_so_far = cur0
            if cur1 > max_so_far:
                max_so_far = cur1

            # Update previous state for the next iteration
            prev0, prev1 = cur0, cur1

        return max_so_far

```
['Python3']