给你两个整数数组 arr1
, arr2
和一个整数 d
,请你返回两个数组之间的 距离值 。
「距离值」 定义为符合此距离要求的元素数目:对于元素 arr1[i]
,不存在任何元素 arr2[j]
满足 |arr1[i]-arr2[j]| <= d
。
示例 1:
输入:arr1 = [4,5,8], arr2 = [10,9,1,8], d = 2 输出:2 解释: 对于 arr1[0]=4 我们有: |4-10|=6 > d=2 |4-9|=5 > d=2 |4-1|=3 > d=2 |4-8|=4 > d=2 所以 arr1[0]=4 符合距离要求 对于 arr1[1]=5 我们有: |5-10|=5 > d=2 |5-9|=4 > d=2 |5-1|=4 > d=2 |5-8|=3 > d=2 所以 arr1[1]=5 也符合距离要求 对于 arr1[2]=8 我们有: |8-10|=2 <= d=2 |8-9|=1 <= d=2 |8-1|=7 > d=2 |8-8|=0 <= d=2 存在距离小于等于 2 的情况,不符合距离要求 故而只有 arr1[0]=4 和 arr1[1]=5 两个符合距离要求,距离值为 2
示例 2:
输入:arr1 = [1,4,2,3], arr2 = [-4,-3,6,10,20,30], d = 3 输出:2
示例 3:
输入:arr1 = [2,1,100,3], arr2 = [-5,-2,10,-3,7], d = 6 输出:1
提示:
1 <= arr1.length, arr2.length <= 500
-10^3 <= arr1[i], arr2[j] <= 10^3
0 <= d <= 100
方法一:暴力枚举
由于 arr1
和 arr2
的长度不超过 500,因此可以直接暴力遍历。
时间复杂度 arr1
的长度,$n$ 为 arr2
的长度。
方法二:二分查找
对于 arr1
中的每个元素 a
,若在 arr2
中存在 b
,使得 b ∈ [a - d, a + d]
,那么就符合距离要求,不进行累加。
因此,可以先对 arr2
进行排序。然后对于每个元素 a
,二分枚举 arr2
判断是否存在符合距离要求的 b
。
时间复杂度
class Solution:
def findTheDistanceValue(self, arr1: List[int], arr2: List[int], d: int) -> int:
return sum(all(abs(a - b) > d for b in arr2) for a in arr1)
class Solution:
def findTheDistanceValue(self, arr1: List[int], arr2: List[int], d: int) -> int:
def check(a):
idx = bisect_left(arr2, a - d)
if idx != len(arr2) and arr2[idx] <= a + d:
return False
return True
arr2.sort()
return sum(check(a) for a in arr1)
class Solution {
public int findTheDistanceValue(int[] arr1, int[] arr2, int d) {
int ans = 0;
for (int a : arr1) {
if (check(arr2, a, d)) {
++ans;
}
}
return ans;
}
private boolean check(int[] arr, int a, int d) {
for (int b : arr) {
if (Math.abs(a - b) <= d) {
return false;
}
}
return true;
}
}
class Solution {
public int findTheDistanceValue(int[] arr1, int[] arr2, int d) {
Arrays.sort(arr2);
int ans = 0;
for (int a : arr1) {
if (check(arr2, a, d)) {
++ans;
}
}
return ans;
}
private boolean check(int[] arr, int a, int d) {
int left = 0, right = arr.length;
while (left < right) {
int mid = (left + right) >> 1;
if (arr[mid] >= a - d) {
right = mid;
} else {
left = mid + 1;
}
}
if (left != arr.length && arr[left] <= a + d) {
return false;
}
return true;
}
}
class Solution {
public:
int findTheDistanceValue(vector<int>& arr1, vector<int>& arr2, int d) {
int ans = 0;
for (int& a : arr1)
ans += check(arr2, a, d);
return ans;
}
bool check(vector<int>& arr, int a, int d) {
for (int& b : arr)
if (abs(a - b) <= d)
return false;
return true;
}
};
class Solution {
public:
int findTheDistanceValue(vector<int>& arr1, vector<int>& arr2, int d) {
sort(arr2.begin(), arr2.end());
int ans = 0;
for (int& a : arr1)
if (check(arr2, a, d))
++ans;
return ans;
}
bool check(vector<int>& arr, int a, int d) {
int idx = lower_bound(arr.begin(), arr.end(), a - d) - arr.begin();
if (idx != arr.size() && arr[idx] <= a + d) return false;
return true;
}
};
func findTheDistanceValue(arr1 []int, arr2 []int, d int) int {
check := func(arr []int, a int) bool {
for _, b := range arr {
if -d <= a-b && a-b <= d {
return false
}
}
return true
}
ans := 0
for _, a := range arr1 {
if check(arr2, a) {
ans++
}
}
return ans
}
func findTheDistanceValue(arr1 []int, arr2 []int, d int) int {
sort.Ints(arr2)
check := func(a int) bool {
left, right := 0, len(arr2)
for left < right {
mid := (left + right) >> 1
if arr2[mid] >= a-d {
right = mid
} else {
left = mid + 1
}
}
if left != len(arr2) && arr2[left] <= a+d {
return false
}
return true
}
ans := 0
for _, a := range arr1 {
if check(a) {
ans++
}
}
return ans
}
function findTheDistanceValue(
arr1: number[],
arr2: number[],
d: number,
): number {
let res = 0;
for (const num of arr1) {
if (arr2.every(v => Math.abs(num - v) > d)) {
res++;
}
}
return res;
}
function findTheDistanceValue(
arr1: number[],
arr2: number[],
d: number,
): number {
arr2.sort((a, b) => a - b);
const n = arr2.length;
let res = 0;
for (const num of arr1) {
let left = 0;
let right = n - 1;
while (left < right) {
const mid = (left + right) >>> 1;
if (arr2[mid] <= num) {
left = mid + 1;
} else {
right = mid;
}
}
if (
Math.abs(num - arr2[left]) <= d ||
(left !== 0 && Math.abs(num - arr2[left - 1]) <= d)
) {
continue;
}
res++;
}
return res;
}
impl Solution {
pub fn find_the_distance_value(arr1: Vec<i32>, arr2: Vec<i32>, d: i32) -> i32 {
let mut res = 0;
for num in arr1.iter() {
if arr2.iter().all(|v| i32::abs(num - v) > d) {
res += 1;
}
}
res
}
}
impl Solution {
pub fn find_the_distance_value(arr1: Vec<i32>, mut arr2: Vec<i32>, d: i32) -> i32 {
arr2.sort();
let n = arr2.len();
let mut res = 0;
for &num in arr1.iter() {
let mut left = 0;
let mut right = n - 1;
while left < right {
let mid = left + (right - left) / 2;
if arr2[mid] <= num {
left = mid + 1;
} else {
right = mid;
}
}
if i32::abs(num - arr2[left]) <= d || (left != 0 && i32::abs(num - arr2[left - 1]) <= d) {
continue;
}
res += 1;
}
res
}
}