# 1855. Maximum Distance Between a Pair of Values

## 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).


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 $O(1)$.

• class Solution {
public int maxDistance(int[] nums1, int[] nums2) {
int ans = 0;
int m = nums1.length, n = nums2.length;
for (int i = 0; i < m; ++i) {
int left = i, right = n - 1;
while (left < right) {
int mid = (left + right + 1) >> 1;
if (nums2[mid] >= nums1[i]) {
left = mid;
} else {
right = mid - 1;
}
}
ans = Math.max(ans, left - i);
}
return ans;
}
}

• class Solution {
public:
int maxDistance(vector<int>& nums1, vector<int>& nums2) {
int ans = 0;
reverse(nums2.begin(), nums2.end());
for (int i = 0; i < nums1.size(); ++i) {
int j = nums2.size() - (lower_bound(nums2.begin(), nums2.end(), nums1[i]) - nums2.begin()) - 1;
ans = max(ans, j - i);
}
return ans;
}
};

• class Solution:
def maxDistance(self, nums1: List[int], nums2: List[int]) -> int:
ans = 0
nums2 = nums2[::-1]
for i, v in enumerate(nums1):
j = len(nums2) - bisect_left(nums2, v) - 1
ans = max(ans, j - i)
return ans


• func maxDistance(nums1 []int, nums2 []int) int {
ans, n := 0, len(nums2)
for i, num := range nums1 {
left, right := i, n-1
for left < right {
mid := (left + right + 1) >> 1
if nums2[mid] >= num {
left = mid
} else {
right = mid - 1
}
}
if ans < left-i {
ans = left - i
}
}
return ans
}

• function maxDistance(nums1: number[], nums2: number[]): number {
let ans = 0;
let m = nums1.length;
let n = nums2.length;
for (let i = 0; i < m; ++i) {
let left = i;
let right = n - 1;
while (left < right) {
const mid = (left + right + 1) >> 1;
if (nums2[mid] >= nums1[i]) {
left = mid;
} else {
right = mid - 1;
}
}
ans = Math.max(ans, left - i);
}
return ans;
}


• /**
* @param {number[]} nums1
* @param {number[]} nums2
* @return {number}
*/
var maxDistance = function (nums1, nums2) {
let ans = 0;
let m = nums1.length;
let n = nums2.length;
for (let i = 0; i < m; ++i) {
let left = i;
let right = n - 1;
while (left < right) {
const mid = (left + right + 1) >> 1;
if (nums2[mid] >= nums1[i]) {
left = mid;
} else {
right = mid - 1;
}
}
ans = Math.max(ans, left - i);
}
return ans;
};


• impl Solution {
pub fn max_distance(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {
let m = nums1.len();
let n = nums2.len();
let mut res = 0;
for i in 0..m {
let mut left = i;
let mut right = n;
while left < right {
let mid = left + (right - left) / 2;
if nums2[mid] >= nums1[i] {
left = mid + 1;
} else {
right = mid;
}
}
res = res.max((left - i - 1) as i32);
}
res
}
}