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feat: add solutions to lc problems: No.3025,3027 (#2311)
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‎solution/3000-3099/3025.Find the Number of Ways to Place People I/README.md

+100-4
Original file line numberDiff line numberDiff line change
@@ -73,24 +73,120 @@
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## 解法
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### 方法一
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### 方法一：排序 + 枚举
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我们不妨考虑枚举矩形左上角的点 $(x_1, y_1)$，那么根据题目，右下角的点 $(x_2, y_2)$ 随着 $x$ 的增大，纵坐标 $y$ 也会要严格增大，才符合题意。
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因此，我们对所有点按照 $x$ 坐标升序排序，如果 $x$ 坐标相同，按照 $y$ 坐标降序排序。
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然后我们枚举左上角的点 $(x_1, y_1)$，并且维护一个最大的 $y_2$，记为 $maxY$，表示所有右下角的点的纵坐标的最大值。然后我们枚举右下角的点 $(x_2, y_2)$，如果 $y_2$ 大于 $maxY$ 并且小于等于 $y_1$，那么我们就找到了一个合法的方案，将答案加一，然后更新 $maxY$ 为 $y_2$。
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枚举完所有的点对后，我们就得到了答案。
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时间复杂度 $O(n^2)$，空间复杂度 $O(\log n)$。其中 $n$ 是点的数量。
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<!-- tabs:start -->
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8090
```python
81-
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class Solution:
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def numberOfPairs(self, points: List[List[int]]) -> int:
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points.sort(key=lambda x: (x[0], -x[1]))
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ans = 0
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for i, (_, y1) in enumerate(points):
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max_y = -inf
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for _, y2 in points[i + 1 :]:
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if max_y < y2 <= y1:
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max_y = y2
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ans += 1
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return ans
82102
```
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84104
```java
85-
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class Solution {
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public int numberOfPairs(int[][] points) {
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Arrays.sort(points, (a, b) -> a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
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int ans = 0;
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int n = points.length;
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final int inf = 1 << 30;
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for (int i = 0; i < n; ++i) {
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int y1 = points[i][1];
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int maxY = -inf;
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for (int j = i + 1; j < n; ++j) {
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int y2 = points[j][1];
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if (maxY < y2 && y2 <= y1) {
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maxY = y2;
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++ans;
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}
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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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88127
```cpp
89-
128+
class Solution {
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public:
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int numberOfPairs(vector<vector<int>>& points) {
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sort(points.begin(), points.end(), [](const vector<int>& a, const vector<int>& b) {
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return a[0] < b[0] || (a[0] == b[0] && b[1] < a[1]);
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});
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int n = points.size();
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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int y1 = points[i][1];
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int maxY = INT_MIN;
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for (int j = i + 1; j < n; ++j) {
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int y2 = points[j][1];
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if (maxY < y2 && y2 <= y1) {
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maxY = y2;
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++ans;
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}
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}
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}
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return ans;
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}
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};
90150
```
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```go
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func numberOfPairs(points [][]int) (ans int) {
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sort.Slice(points, func(i, j int) bool {
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return points[i][0] < points[j][0] || points[i][0] == points[j][0] && points[j][1] < points[i][1]
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})
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for i, p1 := range points {
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y1 := p1[1]
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maxY := math.MinInt32
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for _, p2 := range points[i+1:] {
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y2 := p2[1]
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if maxY < y2 && y2 <= y1 {
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maxY = y2
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ans++
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}
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}
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}
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return
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}
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```
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```ts
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function numberOfPairs(points: number[][]): number {
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points.sort((a, b) => (a[0] === b[0] ? b[1] - a[1] : a[0] - b[0]));
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const n = points.length;
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let ans = 0;
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for (let i = 0; i < n; ++i) {
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const [_, y1] = points[i];
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let maxY = -Infinity;
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for (let j = i + 1; j < n; ++j) {
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const [_, y2] = points[j];
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if (maxY < y2 && y2 <= y1) {
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maxY = y2;
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++ans;
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}
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}
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}
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return ans;
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}
94190
```
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<!-- tabs:end -->

‎solution/3000-3099/3025.Find the Number of Ways to Place People I/README_EN.md

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@@ -63,24 +63,119 @@ Note that it does not matter if the fence encloses any area, the first and secon
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## Solutions
6565

66-
### Solution 1
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### Solution 1: Sorting and Classification
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First, we sort the array. Then, we can classify the results based on the properties of a triangle.
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- If the sum of the two smaller numbers is less than or equal to the largest number, it cannot form a triangle. Return "Invalid".
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- If the three numbers are equal, it is an equilateral triangle. Return "Equilateral".
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- If two numbers are equal, it is an isosceles triangle. Return "Isosceles".
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- If none of the above conditions are met, it is a scalene triangle. Return "Scalene".
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The time complexity is $O(1)$, and the space complexity is $O(1)$.
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6877
<!-- tabs:start -->
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7079
```python
71-
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class Solution:
81+
def numberOfPairs(self, points: List[List[int]]) -> int:
82+
points.sort(key=lambda x: (x[0], -x[1]))
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ans = 0
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for i, (_, y1) in enumerate(points):
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max_y = -inf
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for _, y2 in points[i + 1 :]:
87+
if max_y < y2 <= y1:
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max_y = y2
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ans += 1
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return ans
7291
```
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7493
```java
75-
94+
class Solution {
95+
public int numberOfPairs(int[][] points) {
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Arrays.sort(points, (a, b) -> a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
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int ans = 0;
98+
int n = points.length;
99+
final int inf = 1 << 30;
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for (int i = 0; i < n; ++i) {
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int y1 = points[i][1];
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int maxY = -inf;
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for (int j = i + 1; j < n; ++j) {
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int y2 = points[j][1];
105+
if (maxY < y2 && y2 <= y1) {
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maxY = y2;
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++ans;
108+
}
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}
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}
111+
return ans;
112+
}
113+
}
76114
```
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78116
```cpp
79-
117+
class Solution {
118+
public:
119+
int numberOfPairs(vector<vector<int>>& points) {
120+
sort(points.begin(), points.end(), [](const vector<int>& a, const vector<int>& b) {
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return a[0] < b[0] || (a[0] == b[0] && b[1] < a[1]);
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});
123+
int n = points.size();
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int ans = 0;
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for (int i = 0; i < n; ++i) {
126+
int y1 = points[i][1];
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int maxY = INT_MIN;
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for (int j = i + 1; j < n; ++j) {
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int y2 = points[j][1];
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if (maxY < y2 && y2 <= y1) {
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maxY = y2;
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++ans;
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}
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}
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}
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return ans;
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}
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};
80139
```
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82141
```go
142+
func numberOfPairs(points [][]int) (ans int) {
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sort.Slice(points, func(i, j int) bool {
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return points[i][0] < points[j][0] || points[i][0] == points[j][0] && points[j][1] < points[i][1]
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})
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for i, p1 := range points {
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y1 := p1[1]
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maxY := math.MinInt32
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for _, p2 := range points[i+1:] {
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y2 := p2[1]
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if maxY < y2 && y2 <= y1 {
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maxY = y2
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ans++
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}
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}
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}
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return
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}
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```
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161+
```ts
162+
function numberOfPairs(points: number[][]): number {
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points.sort((a, b) => (a[0] === b[0] ? b[1] - a[1] : a[0] - b[0]));
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const n = points.length;
165+
let ans = 0;
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for (let i = 0; i < n; ++i) {
167+
const [_, y1] = points[i];
168+
let maxY = -Infinity;
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for (let j = i + 1; j < n; ++j) {
170+
const [_, y2] = points[j];
171+
if (maxY < y2 && y2 <= y1) {
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maxY = y2;
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++ans;
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}
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}
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}
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return ans;
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}
84179
```
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86181
<!-- tabs:end -->

‎solution/3000-3099/3025.Find the Number of Ways to Place People I/Solution.cpp

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1+
class Solution {
2+
public:
3+
int numberOfPairs(vector<vector<int>>& points) {
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sort(points.begin(), points.end(), [](const vector<int>& a, const vector<int>& b) {
5+
return a[0] < b[0] || (a[0] == b[0] && b[1] < a[1]);
6+
});
7+
int n = points.size();
8+
int ans = 0;
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for (int i = 0; i < n; ++i) {
10+
int y1 = points[i][1];
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int maxY = INT_MIN;
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for (int j = i + 1; j < n; ++j) {
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int y2 = points[j][1];
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if (maxY < y2 && y2 <= y1) {
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maxY = y2;
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++ans;
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}
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}
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}
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return ans;
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}
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};

‎solution/3000-3099/3025.Find the Number of Ways to Place People I/Solution.go

+17
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,17 @@
1+
func numberOfPairs(points [][]int) (ans int) {
2+
sort.Slice(points, func(i, j int) bool {
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return points[i][0] < points[j][0] || points[i][0] == points[j][0] && points[j][1] < points[i][1]
4+
})
5+
for i, p1 := range points {
6+
y1 := p1[1]
7+
maxY := math.MinInt32
8+
for _, p2 := range points[i+1:] {
9+
y2 := p2[1]
10+
if maxY < y2 && y2 <= y1 {
11+
maxY = y2
12+
ans++
13+
}
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}
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}
16+
return
17+
}

‎solution/3000-3099/3025.Find the Number of Ways to Place People I/Solution.java

+20
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,20 @@
1+
class Solution {
2+
public int numberOfPairs(int[][] points) {
3+
Arrays.sort(points, (a, b) -> a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
4+
int ans = 0;
5+
int n = points.length;
6+
final int inf = 1 << 30;
7+
for (int i = 0; i < n; ++i) {
8+
int y1 = points[i][1];
9+
int maxY = -inf;
10+
for (int j = i + 1; j < n; ++j) {
11+
int y2 = points[j][1];
12+
if (maxY < y2 && y2 <= y1) {
13+
maxY = y2;
14+
++ans;
15+
}
16+
}
17+
}
18+
return ans;
19+
}
20+
}

‎solution/3000-3099/3025.Find the Number of Ways to Place People I/Solution.py

+11
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,11 @@
1+
class Solution:
2+
def numberOfPairs(self, points: List[List[int]]) -> int:
3+
points.sort(key=lambda x: (x[0], -x[1]))
4+
ans = 0
5+
for i, (_, y1) in enumerate(points):
6+
max_y = -inf
7+
for _, y2 in points[i + 1 :]:
8+
if max_y < y2 <= y1:
9+
max_y = y2
10+
ans += 1
11+
return ans

‎solution/3000-3099/3025.Find the Number of Ways to Place People I/Solution.ts

+17
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,17 @@
1+
function numberOfPairs(points: number[][]): number {
2+
points.sort((a, b) => (a[0] === b[0] ? b[1] - a[1] : a[0] - b[0]));
3+
const n = points.length;
4+
let ans = 0;
5+
for (let i = 0; i < n; ++i) {
6+
const [_, y1] = points[i];
7+
let maxY = -Infinity;
8+
for (let j = i + 1; j < n; ++j) {
9+
const [_, y2] = points[j];
10+
if (maxY < y2 && y2 <= y1) {
11+
maxY = y2;
12+
++ans;
13+
}
14+
}
15+
}
16+
return ans;
17+
}

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