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 <= 1051 <= 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 × log n), where m and n are the lengths of nums1 and nums2 respectively. The space complexity is O(1).
PythonJavaC++GoTypeScriptRustJavaScript
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(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
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
}
}(code-box)
/**
* @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;
};(code-box)
Solution 2
PythonJavaC++GoTypeScriptRustJavaScript
class Solution:
def maxDistance(self, nums1: List[int], nums2: List[int]) -> int:
m, n = len(nums1), len(nums2)
ans = i = j = 0
while i < m:
while j < n and nums1[i] <= nums2[j]:
j += 1
ans = max(ans, j - i - 1)
i += 1
return ans(code-box)
class Solution {
public int maxDistance(int[] nums1, int[] nums2) {
int m = nums1.length, n = nums2.length;
int ans = 0;
for (int i = 0, j = 0; i < m; ++i) {
while (j < n && nums1[i] <= nums2[j]) {
++j;
}
ans = Math.max(ans, j - i - 1);
}
return ans;
}
}(code-box)
class Solution {
public:
int maxDistance(vector<int>& nums1, vector<int>& nums2) {
int m = nums1.size(), n = nums2.size();
int ans = 0;
for (int i = 0, j = 0; i < m; ++i) {
while (j < n && nums1[i] <= nums2[j]) {
++j;
}
ans = max(ans, j - i - 1);
}
return ans;
}
};(code-box)
func maxDistance(nums1 []int, nums2 []int) int {
m, n := len(nums1), len(nums2)
ans := 0
for i, j := 0, 0; i < m; i++ {
for j < n && nums1[i] <= nums2[j] {
j++
}
if ans < j-i-1 {
ans = j - i - 1
}
}
return ans
}(code-box)
function maxDistance(nums1: number[], nums2: number[]): number {
let ans = 0;
const m = nums1.length;
const n = nums2.length;
for (let i = 0, j = 0; i < m; ++i) {
while (j < n && nums1[i] <= nums2[j]) {
j++;
}
ans = Math.max(ans, j - i - 1);
}
return ans;
}(code-box)
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;
let mut j = 0;
for i in 0..m {
while j < n && nums1[i] <= nums2[j] {
j += 1;
}
res = res.max((j - i - 1) as i32);
}
res
}
}(code-box)
/**
* @param {number[]} nums1
* @param {number[]} nums2
* @return {number}
*/
var maxDistance = function (nums1, nums2) {
let ans = 0;
const m = nums1.length;
const n = nums2.length;
for (let i = 0, j = 0; i < m; ++i) {
while (j < n && nums1[i] <= nums2[j]) {
j++;
}
ans = Math.max(ans, j - i - 1);
}
return ans;
};(code-box)
