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Commit 5d33e36

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feat: add solutions to lc problem: No.3015 (doocs#2255)
No.3015.Count the Number of Houses at a Certain Distance I
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7 files changed

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‎solution/3000-3099/3015.Count the Number of Houses at a Certain Distance I/README.md‎

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## 解法
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### 方法一
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### 方法一:枚举
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我们可以枚举每个点对 $(i, j),ドル那么从 $i$ 到 $j$ 的最短距离为 $min(|i - j|, |i - x| + 1 + |j - y|, |i - y| + 1 + |j - x|),ドル我们将该距离的出现次数加 2ドル,ドル因为 $(i, j)$ 和 $(j, i)$ 都是满足要求的点对。
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时间复杂度 $O(n^2),ドル其中 $n$ 是题目给定的 $n$。忽略答案数组的空间消耗,空间复杂度 $O(1)$。
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<!-- tabs:start -->
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```python
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class Solution:
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def countOfPairs(self, n: int, x: int, y: int) -> List[int]:
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x, y = x - 1, y - 1
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ans = [0] * n
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for i in range(n):
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for j in range(i + 1, n):
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a = j - i
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b = abs(i - x) + 1 + abs(j - y)
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c = abs(i - y) + 1 + abs(j - x)
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ans[min(a, b, c) - 1] += 2
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return ans
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```
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```java
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class Solution {
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public int[] countOfPairs(int n, int x, int y) {
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int[] ans = new int[n];
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x--;
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y--;
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for (int i = 0; i < n; ++i) {
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for (int j = i + 1; j < n; ++j) {
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int a = j - i;
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int b = Math.abs(i - x) + 1 + Math.abs(j - y);
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int c = Math.abs(i - y) + 1 + Math.abs(j - x);
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ans[Math.min(a, Math.min(b, c)) - 1] += 2;
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}
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}
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return ans;
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}
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}
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```
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```cpp
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class Solution {
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public:
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vector<int> countOfPairs(int n, int x, int y) {
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vector<int> ans(n);
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x--;
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y--;
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for (int i = 0; i < n; ++i) {
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for (int j = i + 1; j < n; ++j) {
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int a = j - i;
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int b = abs(x - i) + abs(y - j) + 1;
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int c = abs(y - i) + abs(x - j) + 1;
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ans[min({a, b, c}) - 1] += 2;
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}
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}
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return ans;
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}
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};
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```
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```go
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func countOfPairs(n int, x int, y int) []int {
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ans := make([]int, n)
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x, y = x-1, y-1
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for i := 0; i < n; i++ {
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for j := i + 1; j < n; j++ {
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a := j - i
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b := abs(x-i) + abs(y-j) + 1
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c := abs(x-j) + abs(y-i) + 1
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ans[min(a, min(b, c))-1] += 2
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}
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}
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return ans
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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```ts
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function countOfPairs(n: number, x: number, y: number): number[] {
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const ans: number[] = Array(n).fill(0);
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x--;
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y--;
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for (let i = 0; i < n; ++i) {
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for (let j = i + 1; j < n; ++j) {
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const a = j - i;
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const b = Math.abs(x - i) + Math.abs(y - j) + 1;
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const c = Math.abs(y - i) + Math.abs(x - j) + 1;
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ans[Math.min(a, b, c) - 1] += 2;
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}
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}
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return ans;
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}
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```
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<!-- tabs:end -->

‎solution/3000-3099/3015.Count the Number of Houses at a Certain Distance I/README_EN.md‎

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## Solutions
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### Solution 1
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### Solution 1: Enumeration
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We can enumerate each pair of points $(i, j)$. The shortest distance from $i$ to $j$ is $min(|i - j|, |i - x| + 1 + |j - y|, |i - y| + 1 + |j - x|)$. We add 2ドル$ to the count of this distance because both $(i, j)$ and $(j, i)$ are valid pairs of points.
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The time complexity is $O(n^2),ドル where $n$ is the $n$ given in the problem. Ignoring the space consumption of the answer array, the space complexity is $O(1)$.
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<!-- tabs:start -->
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```python
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class Solution:
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def countOfPairs(self, n: int, x: int, y: int) -> List[int]:
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x, y = x - 1, y - 1
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ans = [0] * n
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for i in range(n):
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for j in range(i + 1, n):
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a = j - i
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b = abs(i - x) + 1 + abs(j - y)
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c = abs(i - y) + 1 + abs(j - x)
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ans[min(a, b, c) - 1] += 2
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return ans
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```
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```java
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class Solution {
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public int[] countOfPairs(int n, int x, int y) {
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int[] ans = new int[n];
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x--;
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y--;
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for (int i = 0; i < n; ++i) {
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for (int j = i + 1; j < n; ++j) {
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int a = j - i;
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int b = Math.abs(i - x) + 1 + Math.abs(j - y);
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int c = Math.abs(i - y) + 1 + Math.abs(j - x);
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ans[Math.min(a, Math.min(b, c)) - 1] += 2;
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}
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}
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return ans;
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}
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}
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```
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```cpp
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class Solution {
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public:
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vector<int> countOfPairs(int n, int x, int y) {
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vector<int> ans(n);
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x--;
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y--;
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for (int i = 0; i < n; ++i) {
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for (int j = i + 1; j < n; ++j) {
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int a = j - i;
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int b = abs(x - i) + abs(y - j) + 1;
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int c = abs(y - i) + abs(x - j) + 1;
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ans[min({a, b, c}) - 1] += 2;
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}
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}
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return ans;
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}
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};
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```
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```go
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func countOfPairs(n int, x int, y int) []int {
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ans := make([]int, n)
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x, y = x-1, y-1
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for i := 0; i < n; i++ {
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for j := i + 1; j < n; j++ {
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a := j - i
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b := abs(x-i) + abs(y-j) + 1
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c := abs(x-j) + abs(y-i) + 1
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ans[min(a, min(b, c))-1] += 2
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}
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}
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return ans
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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```ts
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function countOfPairs(n: number, x: number, y: number): number[] {
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const ans: number[] = Array(n).fill(0);
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x--;
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y--;
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for (let i = 0; i < n; ++i) {
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for (let j = i + 1; j < n; ++j) {
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const a = j - i;
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const b = Math.abs(x - i) + Math.abs(y - j) + 1;
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const c = Math.abs(y - i) + Math.abs(x - j) + 1;
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ans[Math.min(a, b, c) - 1] += 2;
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}
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}
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return ans;
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}
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```
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<!-- tabs:end -->
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class Solution {
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public:
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vector<int> countOfPairs(int n, int x, int y) {
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vector<int> ans(n);
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x--;
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y--;
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for (int i = 0; i < n; ++i) {
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for (int j = i + 1; j < n; ++j) {
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int a = j - i;
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int b = abs(x - i) + abs(y - j) + 1;
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int c = abs(y - i) + abs(x - j) + 1;
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ans[min({a, b, c}) - 1] += 2;
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}
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}
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return ans;
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}
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};
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func countOfPairs(n int, x int, y int) []int {
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ans := make([]int, n)
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x, y = x-1, y-1
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for i := 0; i < n; i++ {
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for j := i + 1; j < n; j++ {
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a := j - i
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b := abs(x-i) + abs(y-j) + 1
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c := abs(x-j) + abs(y-i) + 1
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ans[min(a, min(b, c))-1] += 2
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}
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}
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return ans
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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class Solution {
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public int[] countOfPairs(int n, int x, int y) {
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int[] ans = new int[n];
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x--;
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y--;
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for (int i = 0; i < n; ++i) {
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for (int j = i + 1; j < n; ++j) {
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int a = j - i;
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int b = Math.abs(i - x) + 1 + Math.abs(j - y);
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int c = Math.abs(i - y) + 1 + Math.abs(j - x);
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ans[Math.min(a, Math.min(b, c)) - 1] += 2;
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}
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}
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return ans;
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}
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}
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class Solution:
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def countOfPairs(self, n: int, x: int, y: int) -> List[int]:
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x, y = x - 1, y - 1
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ans = [0] * n
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for i in range(n):
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for j in range(i + 1, n):
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a = j - i
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b = abs(i - x) + 1 + abs(j - y)
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c = abs(i - y) + 1 + abs(j - x)
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ans[min(a, b, c) - 1] += 2
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return ans
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Original file line numberDiff line numberDiff line change
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function countOfPairs(n: number, x: number, y: number): number[] {
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const ans: number[] = Array(n).fill(0);
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x--;
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y--;
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for (let i = 0; i < n; ++i) {
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for (let j = i + 1; j < n; ++j) {
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const a = j - i;
8+
const b = Math.abs(x - i) + Math.abs(y - j) + 1;
9+
const c = Math.abs(y - i) + Math.abs(x - j) + 1;
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ans[Math.min(a, b, c) - 1] += 2;
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}
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}
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return ans;
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}

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