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).
Constraints:
1 <= nums1.length, nums2.length <= 105
1 <= nums1[i], nums2[j] <= 105
Both nums1 and nums2 are non-increasing.
Solutions
Solution 1: Binary Search
Assume the lengths of $nums1$ and $nums2$ are $m$ and $n$ respectively.
Traverse array $nums1$, for each number $nums1[i]$, perform a binary search for numbers in $nums2$ in the range $[i,n)$, find the last position $j$ that is greater than or equal to $nums1[i]$, calculate the distance between this position and $i$, and update the maximum distance value $ans$.
The time complexity is $O(m \times \log n)$, where $m$ and $n$ are the lengths of $nums1$ and $nums2$ respectively. The space complexity is $O(1)$.