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feat: add solutions to lc problem: No.1385 #3840

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115 changes: 41 additions & 74 deletions solution/1300-1399/1385.Find the Distance Value Between Two Arrays/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -33,16 +33,16 @@ tags:
<strong>输出：</strong>2
<strong>解释：</strong>
对于 arr1[0]=4 我们有：
|4-10|=6 &gt; d=2
|4-9|=5 &gt; d=2
|4-1|=3 &gt; d=2
|4-8|=4 &gt; d=2
|4-10|=6 &gt; d=2
|4-9|=5 &gt; d=2
|4-1|=3 &gt; d=2
|4-8|=4 &gt; d=2
所以 arr1[0]=4 符合距离要求

对于 arr1[1]=5 我们有：
|5-10|=5 &gt; d=2
|5-9|=4 &gt; d=2
|5-1|=4 &gt; d=2
|5-10|=5 &gt; d=2
|5-9|=4 &gt; d=2
|5-1|=4 &gt; d=2
|5-8|=3 &gt; d=2
所以 arr1[1]=5 也符合距离要求

Expand All @@ -51,7 +51,7 @@ tags:
<strong>|8-9|=1 &lt;= d=2</strong>
|8-1|=7 &gt; d=2
<strong>|8-8|=0 &lt;= d=2</strong>
存在距离小于等于 2 的情况，不符合距离要求
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</pre>

Expand Down Expand Up @@ -85,9 +85,9 @@ tags:

### 方法一：排序 + 二分查找

我们可以先对数组 $arr2$ 排序，然后对于数组 $arr1$ 中的每个元素 $a$，使用二分查找，找到数组 $arr2$ 中第一个大于等于 $a-d$ 的元素，如果元素存在，且小于等于 $a+d$，则说明不符合距离要求，否则说明符合距离要求。我们将符合距离要求的元素个数累加，即为答案。
我们可以先对数组 $\textit{arr2}$ 排序，然后对于数组 $\textit{arr1}$ 中的每个元素 $x$，使用二分查找，找到数组 $\textit{arr2}$ 中第一个大于等于 $x - d$ 的元素，如果元素存在，且小于等于 $x + d$，则说明不符合距离要求，否则说明符合距离要求。我们将符合距离要求的元素个数累加，即为答案。

时间复杂度 $O((m + n) \times \log n)$，空间复杂度 $O(\log n)$。其中 $m$ 和 $n$ 分别是数组 $arr1$ 和 $arr2$ 的长度。
时间复杂度 $O((m + n) \times \log n)$，空间复杂度 $O(\log n)$。其中 $m$ 和 $n$ 分别是数组 $\textit{arr1}$ 和 $\textit{arr2}$ 的长度。

<!-- tabs:start -->

Expand All @@ -96,12 +96,12 @@ tags:
```python
class Solution:
def findTheDistanceValue(self, arr1: List[int], arr2: List[int], d: int) -> int:
def check(a: int) -> bool:
i = bisect_left(arr2, a - d)
return i == len(arr2) or arr2[i] > a + d

arr2.sort()
return sum(check(a) for a in arr1)
ans = 0
for x in arr1:
i = bisect_left(arr2, x - d)
ans += i == len(arr2) or arr2[i] > x + d
return ans
```

#### Java
Expand All @@ -111,26 +111,15 @@ 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)) {
for (int x : arr1) {
int i = Arrays.binarySearch(arr2, x - d);
i = i < 0 ? -i - 1 : i;
if (i == arr2.length || arr2[i] > x + d) {
++ans;
}
}
return ans;
}

private boolean check(int[] arr, int a, int d) {
int l = 0, r = arr.length;
while (l < r) {
int mid = (l + r) >> 1;
if (arr[mid] >= a - d) {
r = mid;
} else {
l = mid + 1;
}
}
return l >= arr.length || arr[l] > a + d;
}
}
```

Expand All @@ -140,14 +129,13 @@ class Solution {
class Solution {
public:
int findTheDistanceValue(vector<int>& arr1, vector<int>& arr2, int d) {
auto check = [&](int a) -> bool {
auto it = lower_bound(arr2.begin(), arr2.end(), a - d);
return it == arr2.end() || *it > a + d;
};
sort(arr2.begin(), arr2.end());
ranges::sort(arr2);
int ans = 0;
for (int& a : arr1) {
ans += check(a);
for (int x : arr1) {
auto it = ranges::lower_bound(arr2, x - d);
if (it == arr2.end() || *it > x + d) {
++ans;
}
}
return ans;
}
Expand All @@ -159,9 +147,9 @@ public:
```go
func findTheDistanceValue(arr1 []int, arr2 []int, d int) (ans int) {
sort.Ints(arr2)
for _, a := range arr1 {
i := sort.SearchInts(arr2, a-d)
if i == len(arr2) || arr2[i] > a+d {
for _, x := range arr1 {
i := sort.SearchInts(arr2, x-d)
if i == len(arr2) || arr2[i] > x+d {
ans++
}
}
Expand All @@ -173,23 +161,11 @@ func findTheDistanceValue(arr1 []int, arr2 []int, d int) (ans int) {

```ts
function findTheDistanceValue(arr1: number[], arr2: number[], d: number): number {
const check = (a: number) => {
let l = 0;
let r = arr2.length;
while (l < r) {
const mid = (l + r) >> 1;
if (arr2[mid] >= a - d) {
r = mid;
} else {
l = mid + 1;
}
}
return l === arr2.length || arr2[l] > a + d;
};
arr2.sort((a, b) => a - b);
let ans = 0;
for (const a of arr1) {
if (check(a)) {
let ans: number = 0;
for (const x of arr1) {
const i = _.sortedIndex(arr2, x - d);
if (i === arr2.length || arr2[i] > x + d) {
++ans;
}
}
Expand All @@ -203,26 +179,17 @@ function findTheDistanceValue(arr1: number[], arr2: number[], d: number): number
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;
let mut ans = 0;
for &x in &arr1 {
let i = match arr2.binary_search(&(x - d)) {
Ok(j) => j,
Err(j) => j,
};
if i == arr2.len() || arr2[i] > x + d {
ans += 1;
}
res += 1;
}
res
ans
}
}
```
Expand Down
119 changes: 43 additions & 76 deletions solution/1300-1399/1385.Find the Distance Value Between Two Arrays/README_EN.md
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Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -31,16 +31,16 @@ tags:
<pre>
<strong>Input:</strong> arr1 = [4,5,8], arr2 = [10,9,1,8], d = 2
<strong>Output:</strong> 2
<strong>Explanation:</strong>
For arr1[0]=4 we have:
|4-10|=6 &gt; d=2
|4-9|=5 &gt; d=2
|4-1|=3 &gt; d=2
|4-8|=4 &gt; d=2
For arr1[1]=5 we have:
|5-10|=5 &gt; d=2
|5-9|=4 &gt; d=2
|5-1|=4 &gt; d=2
<strong>Explanation:</strong>
For arr1[0]=4 we have:
|4-10|=6 &gt; d=2
|4-9|=5 &gt; d=2
|4-1|=3 &gt; d=2
|4-8|=4 &gt; d=2
For arr1[1]=5 we have:
|5-10|=5 &gt; d=2
|5-9|=4 &gt; d=2
|5-1|=4 &gt; d=2
|5-8|=3 &gt; d=2
For arr1[2]=8 we have:
<strong>|8-10|=2 &lt;= d=2</strong>
Expand Down Expand Up @@ -80,9 +80,9 @@ For arr1[2]=8 we have:

### Solution 1: Sorting + Binary Search

We can first sort the array $arr2$, then for each element $a$ in array $arr1$, use binary search to find the first element in array $arr2$ that is greater than or equal to $a-d$. If such an element exists and is less than or equal to $a+d$, it means that it does not meet the distance requirement. Otherwise, it meets the distance requirement. We accumulate the number of elements that meet the distance requirement, which is the answer.
We can first sort the array $\textit{arr2}$, and then for each element $x$ in the array $\textit{arr1}$, use binary search to find the first element in the array $\textit{arr2}$ that is greater than or equal to $x - d$. If such an element exists and is less than or equal to $x + d$, it does not meet the distance requirement. Otherwise, it meets the distance requirement. We count the number of elements that meet the distance requirement, which is the answer.

The time complexity is $O((m + n) \times \log n)$, and the space complexity is $O(\log n)$. Where $m$ and $n$ are the lengths of arrays $arr1$ and $arr2$, respectively.
The time complexity is $O((m + n) \times \log n)$, and the space complexity is $O(\log n)$. Here, $m$ and $n$ are the lengths of the arrays $\textit{arr1}$ and $\textit{arr2}$, respectively.

<!-- tabs:start -->

Expand All @@ -91,12 +91,12 @@ The time complexity is $O((m + n) \times \log n)$, and the space complexity is $
```python
class Solution:
def findTheDistanceValue(self, arr1: List[int], arr2: List[int], d: int) -> int:
def check(a: int) -> bool:
i = bisect_left(arr2, a - d)
return i == len(arr2) or arr2[i] > a + d

arr2.sort()
return sum(check(a) for a in arr1)
ans = 0
for x in arr1:
i = bisect_left(arr2, x - d)
ans += i == len(arr2) or arr2[i] > x + d
return ans
```

#### Java
Expand All @@ -106,26 +106,15 @@ 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)) {
for (int x : arr1) {
int i = Arrays.binarySearch(arr2, x - d);
i = i < 0 ? -i - 1 : i;
if (i == arr2.length || arr2[i] > x + d) {
++ans;
}
}
return ans;
}

private boolean check(int[] arr, int a, int d) {
int l = 0, r = arr.length;
while (l < r) {
int mid = (l + r) >> 1;
if (arr[mid] >= a - d) {
r = mid;
} else {
l = mid + 1;
}
}
return l >= arr.length || arr[l] > a + d;
}
}
```

Expand All @@ -135,14 +124,13 @@ class Solution {
class Solution {
public:
int findTheDistanceValue(vector<int>& arr1, vector<int>& arr2, int d) {
auto check = [&](int a) -> bool {
auto it = lower_bound(arr2.begin(), arr2.end(), a - d);
return it == arr2.end() || *it > a + d;
};
sort(arr2.begin(), arr2.end());
ranges::sort(arr2);
int ans = 0;
for (int& a : arr1) {
ans += check(a);
for (int x : arr1) {
auto it = ranges::lower_bound(arr2, x - d);
if (it == arr2.end() || *it > x + d) {
++ans;
}
}
return ans;
}
Expand All @@ -154,9 +142,9 @@ public:
```go
func findTheDistanceValue(arr1 []int, arr2 []int, d int) (ans int) {
sort.Ints(arr2)
for _, a := range arr1 {
i := sort.SearchInts(arr2, a-d)
if i == len(arr2) || arr2[i] > a+d {
for _, x := range arr1 {
i := sort.SearchInts(arr2, x-d)
if i == len(arr2) || arr2[i] > x+d {
ans++
}
}
Expand All @@ -168,23 +156,11 @@ func findTheDistanceValue(arr1 []int, arr2 []int, d int) (ans int) {

```ts
function findTheDistanceValue(arr1: number[], arr2: number[], d: number): number {
const check = (a: number) => {
let l = 0;
let r = arr2.length;
while (l < r) {
const mid = (l + r) >> 1;
if (arr2[mid] >= a - d) {
r = mid;
} else {
l = mid + 1;
}
}
return l === arr2.length || arr2[l] > a + d;
};
arr2.sort((a, b) => a - b);
let ans = 0;
for (const a of arr1) {
if (check(a)) {
let ans: number = 0;
for (const x of arr1) {
const i = _.sortedIndex(arr2, x - d);
if (i === arr2.length || arr2[i] > x + d) {
++ans;
}
}
Expand All @@ -198,26 +174,17 @@ function findTheDistanceValue(arr1: number[], arr2: number[], d: number): number
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;
let mut ans = 0;
for &x in &arr1 {
let i = match arr2.binary_search(&(x - d)) {
Ok(j) => j,
Err(j) => j,
};
if i == arr2.len() || arr2[i] > x + d {
ans += 1;
}
res += 1;
}
res
ans
}
}
```
Expand Down
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