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feat: add solutions to lc problems: No.3024~3027 #2314

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Feb 4, 2024
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feat: add solutions to lc problems: No.3024~3027 #2314

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18 changes: 18 additions & 0 deletions solution/3000-3099/3024.Type of Triangle II/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -145,6 +145,24 @@ function triangleType(nums: number[]): string {
}
```

```cs
public class Solution {
public string TriangleType(int[] nums) {
Array.Sort(nums);
if (nums[0] + nums[1] <= nums[2]) {
return "none";
}
if (nums[0] == nums[2]) {
return "equilateral";
}
if (nums[0] == nums[1] || nums[1] == nums[2]) {
return "isosceles";
}
return "scalene";
}
}
```

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<!-- end -->
18 changes: 18 additions & 0 deletions solution/3000-3099/3024.Type of Triangle II/README_EN.md
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Expand Up @@ -141,6 +141,24 @@ function triangleType(nums: number[]): string {
}
```

```cs
public class Solution {
public string TriangleType(int[] nums) {
Array.Sort(nums);
if (nums[0] + nums[1] <= nums[2]) {
return "none";
}
if (nums[0] == nums[2]) {
return "equilateral";
}
if (nums[0] == nums[1] || nums[1] == nums[2]) {
return "isosceles";
}
return "scalene";
}
}
```

<!-- tabs:end -->

<!-- end -->
15 changes: 15 additions & 0 deletions solution/3000-3099/3024.Type of Triangle II/Solution.cs
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Original file line number Diff line number Diff line change
@@ -0,0 +1,15 @@
public class Solution {
public string TriangleType(int[] nums) {
Array.Sort(nums);
if (nums[0] + nums[1] <= nums[2]) {
return "none";
}
if (nums[0] == nums[2]) {
return "equilateral";
}
if (nums[0] == nums[1] || nums[1] == nums[2]) {
return "isosceles";
}
return "scalene";
}
}
23 changes: 23 additions & 0 deletions solution/3000-3099/3025.Find the Number of Ways to Place People I/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -189,6 +189,29 @@ function numberOfPairs(points: number[][]): number {
}
```

```cs
public class Solution {
public int NumberOfPairs(int[][] points) {
Array.Sort(points, (a, b) => a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
int ans = 0;
int n = points.Length;
int inf = 1 << 30;
for (int i = 0; i < n; ++i) {
int y1 = points[i][1];
int maxY = -inf;
for (int j = i + 1; j < n; ++j) {
int y2 = points[j][1];
if (maxY < y2 && y2 <= y1) {
maxY = y2;
++ans;
}
}
}
return ans;
}
}
```

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<!-- end -->
23 changes: 23 additions & 0 deletions solution/3000-3099/3025.Find the Number of Ways to Place People I/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -178,6 +178,29 @@ function numberOfPairs(points: number[][]): number {
}
```

```cs
public class Solution {
public int NumberOfPairs(int[][] points) {
Array.Sort(points, (a, b) => a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
int ans = 0;
int n = points.Length;
int inf = 1 << 30;
for (int i = 0; i < n; ++i) {
int y1 = points[i][1];
int maxY = -inf;
for (int j = i + 1; j < n; ++j) {
int y2 = points[j][1];
if (maxY < y2 && y2 <= y1) {
maxY = y2;
++ans;
}
}
}
return ans;
}
}
```

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<!-- end -->
20 changes: 20 additions & 0 deletions solution/3000-3099/3025.Find the Number of Ways to Place People I/Solution.cs
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@@ -0,0 +1,20 @@
public class Solution {
public int NumberOfPairs(int[][] points) {
Array.Sort(points, (a, b) => a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
int ans = 0;
int n = points.Length;
int inf = 1 << 30;
for (int i = 0; i < n; ++i) {
int y1 = points[i][1];
int maxY = -inf;
for (int j = i + 1; j < n; ++j) {
int y2 = points[j][1];
if (maxY < y2 && y2 <= y1) {
maxY = y2;
++ans;
}
}
}
return ans;
}
}
168 changes: 96 additions & 72 deletions solution/3000-3099/3026.Maximum Good Subarray Sum/README.md
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Expand Up @@ -50,53 +50,56 @@

## 解法

### 方法一
### 方法一：前缀和 + 哈希表

我们用一个哈希表 $p$ 记录 $nums[i]$ 的前缀数组 $nums[0..i-1]$ 的和 $s$，如果有多个相同的 $nums[i]$，我们只保留最小的 $s$。初始时，我们将 $p[nums[0]]$ 设为 $0$。另外，我们用一个变量 $s$ 记录当前的前缀和，初始时 $s = 0$。初始化答案 $ans$ 为 $-\infty$。

接下来，我们枚举 $nums[i]$，并且维护一个变量 $s$ 表示 $nums[0..i]$ 的和。如果 $nums[i] - k$ 在 $p$ 中，那么我们就找到了一个好子数组，将答案更新为 $ans = \max(ans, s - p[nums[i] - k])$。同理，如果 $nums[i] + k$ 在 $p$ 中，那么我们也找到了一个好子数组，将答案更新为 $ans = \max(ans, s - p[nums[i] + k])$。然后，如果 $i + 1 \lt n$ 并且 $nums[i + 1]$ 不在 $p$ 中，或者 $p[nums[i + 1]] \gt s$，我们就将 $p[nums[i + 1]]$ 设为 $s$。

最后，如果 $ans = -\infty$，那么我们返回 $0$，否则返回 $ans$。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组的长度。

<!-- tabs:start -->

```python
class Solution:
def maximumSubarraySum(self, nums: List[int], k: int) -> int:
p = {}
r = float('-inf')
p[nums[0]] = 0
s = 0
n = len(nums)
for i in range(n):
s += nums[i]
if nums[i] - k in p:
r = max(r, s - p[nums[i] - k])
if nums[i] + k in p:
r = max(r, s - p[nums[i] + k])
if i + 1 == n:
break
if nums[i + 1] not in p or p[nums[i + 1]] > s:
ans = -inf
p = {nums[0]: 0}
s, n = 0, len(nums)
for i, x in enumerate(nums):
s += x
if x - k in p:
ans = max(ans, s - p[x - k])
if x + k in p:
ans = max(ans, s - p[x + k])
if i + 1 < n and (nums[i + 1] not in p or p[nums[i + 1]] > s):
p[nums[i + 1]] = s
return r if r != float('-inf') else 0
return 0 if ans == -inf else ans
```

```java
class Solution {
public long maximumSubarraySum(int[] nums, int k) {
HashMap<Integer, Long> p = new HashMap<>();
long r = Long.MIN_VALUE;
Map<Integer, Long> p = new HashMap<>();
p.put(nums[0], 0L);
long s = 0;
int n = nums.length;
for (int i = 0;; ++i) {
long ans = Long.MIN_VALUE;
for (int i = 0; i < n; ++i) {
s += nums[i];
if (p.containsKey(nums[i] - k)) {
r = Math.max(r, s - p.get(nums[i] - k));
ans = Math.max(ans, s - p.get(nums[i] - k));
}
if (p.containsKey(nums[i] + k)) {
r = Math.max(r, s - p.get(nums[i] + k));
ans = Math.max(ans, s - p.get(nums[i] + k));
}
if (i + 1 == n) break;
if (!p.containsKey(nums[i + 1]) || p.get(nums[i + 1]) > s) {
if (i + 1 < n && (!p.containsKey(nums[i + 1]) || p.get(nums[i + 1]) > s)) {
p.put(nums[i + 1], s);
}
}
return r == Long.MIN_VALUE ? 0 : r;
return ans == Long.MIN_VALUE ? 0 : ans;
}
}
```
Expand All @@ -106,85 +109,106 @@ class Solution {
public:
long long maximumSubarraySum(vector<int>& nums, int k) {
unordered_map<int, long long> p;
long long r = LONG_LONG_MIN;
p[nums[0]] = 0;
long long s = 0;
const int n = nums.size();
long long ans = LONG_LONG_MIN;
for (int i = 0;; ++i) {
s += nums[i];
auto t = p.find(nums[i] - k);
if (t != p.end()) {
r = max(r, s - t->second);
auto it = p.find(nums[i] - k);
if (it != p.end()) {
ans = max(ans, s - it->second);
}
it = p.find(nums[i] + k);
if (it != p.end()) {
ans = max(ans, s - it->second);
}
t = p.find(nums[i] + k);
if (t != p.end()) {
r = max(r, s - t->second);
if (i + 1 == n) {
break;
}
if (i + 1 == n)
break;
t = p.find(nums[i + 1]);
if (t == p.end() || t->second > s) {
it = p.find(nums[i + 1]);
if (it == p.end() || it->second > s) {
p[nums[i + 1]] = s;
}
}
return r == LONG_LONG_MIN ? 0 : r;
return ans == LONG_LONG_MIN ? 0 : ans;
}
};
```

```go
func maximumSubarraySum(nums []int, k int) int64 {
p := make(map[int]int64)
var r int64 = math.MinInt64
p[nums[0]] = 0
var s int64 = 0
n := len(nums)
for i := 0; ; i++ {
s += int64(nums[i])
if t, ok := p[nums[i]-k]; ok {
r = max(r, s-t)
}
if t, ok := p[nums[i]+k]; ok {
r = max(r, s-t)
}
if i+1 == n {
break
}
if t, ok := p[nums[i+1]]; !ok || t > s {
p[nums[i+1]] = s
}
}
if r == math.MinInt64 {
return 0
}
return r
p := map[int]int64{nums[0]: 0}
var s int64 = 0
n := len(nums)
var ans int64 = math.MinInt64
for i, x := range nums {
s += int64(x)
if t, ok := p[nums[i]-k]; ok {
ans = max(ans, s-t)
}
if t, ok := p[nums[i]+k]; ok {
ans = max(ans, s-t)
}
if i+1 == n {
break
}
if t, ok := p[nums[i+1]]; !ok || s < t {
p[nums[i+1]] = s
}
}
if ans == math.MinInt64 {
return 0
}
return ans
}
```

```ts
function maximumSubarraySum(nums: number[], k: number): number {
const p: Map<number, number> = new Map();
let r: number = Number.MIN_SAFE_INTEGER;
p.set(nums[0], 0);
let ans: number = -Infinity;
let s: number = 0;
const n: number = nums.length;
for (let i = 0; ; ++i) {
for (let i = 0; i < n; ++i) {
s += nums[i];
let t: number | undefined = p.get(nums[i] - k);
if (t !== undefined) {
r = Math.max(r, s - t);
if (p.has(nums[i] - k)) {
ans = Math.max(ans, s - p.get(nums[i] - k)!);
}
t = p.get(nums[i] + k);
if (t !== undefined) {
r = Math.max(r, s - t);
if (p.has(nums[i] + k)) {
ans = Math.max(ans, s - p.get(nums[i] + k)!);
}
if (i + 1 === n) break;
t = p.get(nums[i + 1]);
if (t === undefined || t > s) {
if (i + 1 < n && (!p.has(nums[i + 1]) || p.get(nums[i + 1])! > s)) {
p.set(nums[i + 1], s);
}
}
return r === Number.MIN_SAFE_INTEGER ? 0 : r;
return ans === -Infinity ? 0 : ans;
}
```

```cs
public class Solution {
public long MaximumSubarraySum(int[] nums, int k) {
Dictionary<int, long> p = new Dictionary<int, long>();
p[nums[0]] = 0L;
long s = 0;
int n = nums.Length;
long ans = long.MinValue;
for (int i = 0; i < n; ++i) {
s += nums[i];
if (p.ContainsKey(nums[i] - k)) {
ans = Math.Max(ans, s - p[nums[i] - k]);
}
if (p.ContainsKey(nums[i] + k)) {
ans = Math.Max(ans, s - p[nums[i] + k]);
}
if (i + 1 < n && (!p.ContainsKey(nums[i + 1]) || p[nums[i + 1]] > s)) {
p[nums[i + 1]] = s;
}
}
return ans == long.MinValue ? 0 : ans;
}
}
```

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