Maximum Distance Between a Pair of Values

#leetcode #dsa #theabbie

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.

SOLUTION:

# class Solution:
#     def maxDistance(self, nums1: List[int], nums2: List[int]) -> int:
#         mdiff = 0
#         m = len(nums1)
#         n = len(nums2)
#         nums1min = [0] * m
#         nums2max = [0] * n
#         curr = 0
#         for i in range(m):
#             if nums1[i] < nums1[curr]:
#                 curr = i
#             nums1min[i] = curr
#         curr = n - 1
#         for i in range(n - 1, -1, -1):
#             if nums2[i] > nums2[curr]:
#                 curr = i
#             nums2max[i] = curr
#         print()
#         k = min(m, n)
#         for i in range(k):
#             mdiff = max(mdiff, nums2max[i] - nums1min[i])
#         return mdiff

import bisect

class Solution:
    def maxDistance(self, nums1: List[int], nums2: List[int]) -> int:
        mdiff = 0
        m = len(nums1)
        n = len(nums2)
        nums2.reverse()
        for i in range(m):
            j = n - bisect.bisect_left(nums2, nums1[i]) - 1
            mdiff = max(mdiff, j - i)
        return mdiff