You are given two non-increasing 0-indexed integer arrays nums1​​​​​​ and nums2​​​​​​.

A pair of indices (i, j), where 0 <= i < nums1.length and 0 <= j < nums2.length, is valid if both i <= j and nums1[i] <= nums2[j]. The distance of the pair is j - i​​​​.

Return the maximum distance of any valid pair (i, j). If there are no valid pairs, return 0.

An array arr is non-increasing if arr[i-1] >= arr[i] for every 1 <= i < arr.length.

Example 1:

Input: nums1 = [55,30,5,4,2], nums2 = [100,20,10,10,5]
Output: 2
Explanation: The valid pairs are (0,0), (2,2), (2,3), (2,4), (3,3), (3,4), and (4,4).
The maximum distance is 2 with pair (2,4).


Example 2:

Input: nums1 = [2,2,2], nums2 = [10,10,1]
Output: 1
Explanation: The valid pairs are (0,0), (0,1), and (1,1).
The maximum distance is 1 with pair (0,1).


Example 3:

Input: nums1 = [30,29,19,5], nums2 = [25,25,25,25,25]
Output: 2
Explanation: The valid pairs are (2,2), (2,3), (2,4), (3,3), and (3,4).
The maximum distance is 2 with pair (2,4).


Example 4:

Input: nums1 = [5,4], nums2 = [3,2]
Output: 0
Explanation: There are no valid pairs, so return 0.


Constraints:

• 1 <= nums1.length <= 105
• 1 <= nums2.length <= 105
• 1 <= nums1[i], nums2[j] <= 105
• Both nums1 and nums2 are non-increasing.

## Solution: Two Pointers

For each i, find the largest j such that nums[j] >= nums[i].

Time complexity: O(n + m)
Space complexity: O(1)

## C++

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