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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 <= 1051 <= nums2.length <= 1051 <= nums1[i], nums2[j] <= 105nums1 and nums2 are non-increasing.class Solution:
def maxDistance(self, nums1: List[int], nums2: List[int]) -> int:
res = 0
for i in range(len(nums1)):
l, r = i, len(nums2) - 1
while l <= r:
mid = (l + r) >> 1
if nums2[mid] >= nums1[i]:
res = max(res, mid - i)
l = mid + 1
else:
r = mid - 1
return res
class Solution {
public int maxDistance(int[] nums1, int[] nums2) {
int res = 0;
for (int i = 0; i < nums1.length; ++i) {
int l = i, r = nums2.length - 1;
while (l <= r) {
int mid = (l + r) >>> 1;
if (nums2[mid] >= nums1[i]) {
res = Math.max(res, mid - i);
l = mid + 1;
} else {
r = mid - 1;
}
}
}
return res;
}
}
class Solution {
public:
int maxDistance(vector<int>& nums1, vector<int>& nums2) {
int res = 0;
for (int i = 0; i < nums1.size(); ++i) {
int l = i, r = nums2.size() - 1;
while (l <= r) {
int mid = (l + r) >> 1;
if (nums2[mid] >= nums1[i]) {
res = max(res, mid - i);
l = mid + 1;
} else {
r = mid - 1;
}
}
}
return res;
}
};
/**
* @param {number[]} nums1
* @param {number[]} nums2
* @return {number}
*/
var maxDistance = function(nums1, nums2) {
let res = 0;
for (let i = 0; i < nums1.length; i++) {
let left = 0, right = nums2.length - 1;
while (left <= right) {
mid = (left + right) >> 1;
if (nums2[mid] >= nums1[i]) {
res = Math.max(res, mid - i);
left = mid + 1;
} else {
right = mid - 1;
}
}
}
return res;
};
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