Formatted question description: https://leetcode.ca/all/1855.html
1855. Maximum Distance Between a Pair of Values
Level
Medium
Description
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 <= 10^5
1 <= nums2.length <= 10^5
1 <= nums1[i], nums2[j] <= 10^5
- Both
nums1
andnums2
are non-increasing.
Solution
Use two pointers. Loop over index2
from nums2.length - 1
to 0. For each index2
, find the smallest index1
such that nums1[index1] <= nums2[index2]
and calculate index2 - index1
as the distance. Since the distance is at least 0, if index1 > index2
, the distance is negative and will never exceed 0. Only the non-negative instances are stored to maintain the maximum distance.
class Solution {
public int maxDistance(int[] nums1, int[] nums2) {
int maxDistance = 0;
int length1 = nums1.length, length2 = nums2.length;
int index1 = length1 - 1, index2 = length2 - 1;
while (index1 >= 0 && index2 >= 0) {
while (index1 >= 0 && nums1[index1] <= nums2[index2]) {
maxDistance = Math.max(maxDistance, index2 - index1);
index1--;
}
index2--;
}
return maxDistance;
}
}