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feat: add weekly contest 403 #3178

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feat: add weekly contest 403
yanglbme Jun 30, 2024
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fix: update
yanglbme Jun 30, 2024
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idoocs Jun 30, 2024
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184 changes: 184 additions & 0 deletions solution/3200-3299/3200.Maximum Height of a Triangle/README.md
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---
comments: true
difficulty: 简单
edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3200.Maximum%20Height%20of%20a%20Triangle/README.md
---

<!-- problem:start -->

# [3200. 三角形的最大高度](https://leetcode.cn/problems/maximum-height-of-a-triangle)

[English Version](/solution/3200-3299/3200.Maximum%20Height%20of%20a%20Triangle/README_EN.md)

## 题目描述

<!-- description:start -->

<p>给你两个整数 <code>red</code> 和 <code>blue</code>，分别表示红色球和蓝色球的数量。你需要使用这些球来组成一个三角形，满足第 1 行有 1 个球，第 2 行有 2 个球，第 3 行有 3 个球，依此类推。</p>

<p>每一行的球必须是 <strong>相同 </strong>颜色，且相邻行的颜色必须<strong> 不同</strong>。</p>

<p>返回可以实现的三角形的 <strong>最大 </strong>高度。</p>

<p>&nbsp;</p>

<p><strong class="example">示例 1：</strong></p>

<div class="example-block">
<p><strong>输入：</strong> <span class="example-io">red = 2, blue = 4</span></p>

<p><strong>输出：</strong> 3</p>

<p><strong>解释：</strong></p>

<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3200.Maximum%20Height%20of%20a%20Triangle/images/brb.png" style="width: 300px; height: 240px; padding: 10px;" /></p>

<p>上图显示了唯一可能的排列方式。</p>
</div>

<p><strong class="example">示例 2：</strong></p>

<div class="example-block">
<p><strong>输入：</strong> <span class="example-io">red = 2, blue = 1</span></p>

<p><strong>输出：</strong> <span class="example-io">2</span></p>

<p><strong>解释：</strong></p>

<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3200.Maximum%20Height%20of%20a%20Triangle/images/br.png" style="width: 150px; height: 135px; padding: 10px;" /><br />
上图显示了唯一可能的排列方式。</p>
</div>

<p><strong class="example">示例 3：</strong></p>

<div class="example-block">
<p><strong>输入：</strong> <span class="example-io">red = 1, blue = 1</span></p>

<p><strong>输出：</strong> <span class="example-io">1</span></p>
</div>

<p><strong class="example">示例 4：</strong></p>

<div class="example-block">
<p><strong>输入：</strong> <span class="example-io">red = 10, blue = 1</span></p>

<p><strong>输出：</strong> <span class="example-io">2</span></p>

<p><strong>解释：</strong></p>

<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3200.Maximum%20Height%20of%20a%20Triangle/images/br.png" style="width: 150px; height: 135px; padding: 10px;" /><br />
上图显示了唯一可能的排列方式。</p>
</div>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
<li><code>1 &lt;= red, blue &lt;= 100</code></li>
</ul>

<!-- description:end -->

## 解法

<!-- solution:start -->

### 方法一：模拟

我们可以枚举第一行的颜色，然后模拟构造三角形，计算最大高度。

时间复杂度 $O(\sqrt{n})$，其中 $n$ 为红色球和蓝色球的数量。空间复杂度 $O(1)$。

<!-- tabs:start -->

#### Python3

```python
class Solution:
def maxHeightOfTriangle(self, red: int, blue: int) -> int:
ans = 0
for k in range(2):
c = [red, blue]
i, j = 1, k
while i <= c[j]:
c[j] -= i
j ^= 1
ans = max(ans, i)
i += 1
return ans
```

#### Java

```java
class Solution {
public int maxHeightOfTriangle(int red, int blue) {
int ans = 0;
for (int k = 0; k < 2; ++k) {
int[] c = {red, blue};
for (int i = 1, j = k; i <= c[j]; j ^= 1, ++i) {
c[j] -= i;
ans = Math.max(ans, i);
}
}
return ans;
}
}
```

#### C++

```cpp
class Solution {
public:
int maxHeightOfTriangle(int red, int blue) {
int ans = 0;
for (int k = 0; k < 2; ++k) {
int c[2] = {red, blue};
for (int i = 1, j = k; i <= c[j]; j ^= 1, ++i) {
c[j] -= i;
ans = max(ans, i);
}
}
return ans;
}
};
```

#### Go

```go
func maxHeightOfTriangle(red int, blue int) (ans int) {
for k := 0; k < 2; k++ {
c := [2]int{red, blue}
for i, j := 1, k; i <= c[j]; i, j = i+1, j^1 {
c[j] -= i
ans = max(ans, i)
}
}
return
}
```

#### TypeScript

```ts
function maxHeightOfTriangle(red: number, blue: number): number {
let ans = 0;
for (let k = 0; k < 2; ++k) {
const c: [number, number] = [red, blue];
for (let i = 1, j = k; i <= c[j]; ++i, j ^= 1) {
c[j] -= i;
ans = Math.max(ans, i);
}
}
return ans;
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
182 changes: 182 additions & 0 deletions solution/3200-3299/3200.Maximum Height of a Triangle/README_EN.md
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---
comments: true
difficulty: Easy
edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3200.Maximum%20Height%20of%20a%20Triangle/README_EN.md
---

<!-- problem:start -->

# [3200. Maximum Height of a Triangle](https://leetcode.com/problems/maximum-height-of-a-triangle)

[中文文档](/solution/3200-3299/3200.Maximum%20Height%20of%20a%20Triangle/README.md)

## Description

<!-- description:start -->

<p>You are given two integers <code>red</code> and <code>blue</code> representing the count of red and blue colored balls. You have to arrange these balls to form a triangle such that the 1<sup>st</sup> row will have 1 ball, the 2<sup>nd</sup> row will have 2 balls, the 3<sup>rd</sup> row will have 3 balls, and so on.</p>

<p>All the balls in a particular row should be the <strong>same</strong> color, and adjacent rows should have <strong>different</strong> colors.</p>

<p>Return the <strong>maximum</strong><em> height of the triangle</em> that can be achieved.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">red = 2, blue = 4</span></p>

<p><strong>Output:</strong> 3</p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3200.Maximum%20Height%20of%20a%20Triangle/images/brb.png" style="width: 300px; height: 240px; padding: 10px;" /></p>

<p>The only possible arrangement is shown above.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">red = 2, blue = 1</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3200.Maximum%20Height%20of%20a%20Triangle/images/br.png" style="width: 150px; height: 135px; padding: 10px;" /><br />
The only possible arrangement is shown above.</p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">red = 1, blue = 1</span></p>

<p><strong>Output:</strong> <span class="example-io">1</span></p>
</div>

<p><strong class="example">Example 4:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">red = 10, blue = 1</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3200.Maximum%20Height%20of%20a%20Triangle/images/br.png" style="width: 150px; height: 135px; padding: 10px;" /><br />
The only possible arrangement is shown above.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
<li><code>1 &lt;= red, blue &lt;= 100</code></li>
</ul>

<!-- description:end -->

## Solutions

<!-- solution:start -->

### Solution 1: Simulation

We can enumerate the color of the first row, then simulate the construction of the triangle, calculating the maximum height.

The time complexity is $O(\sqrt{n})$, where $n$ is the number of red and blue balls. The space complexity is $O(1)$.

<!-- tabs:start -->

#### Python3

```python
class Solution:
def maxHeightOfTriangle(self, red: int, blue: int) -> int:
ans = 0
for k in range(2):
c = [red, blue]
i, j = 1, k
while i <= c[j]:
c[j] -= i
j ^= 1
ans = max(ans, i)
i += 1
return ans
```

#### Java

```java
class Solution {
public int maxHeightOfTriangle(int red, int blue) {
int ans = 0;
for (int k = 0; k < 2; ++k) {
int[] c = {red, blue};
for (int i = 1, j = k; i <= c[j]; j ^= 1, ++i) {
c[j] -= i;
ans = Math.max(ans, i);
}
}
return ans;
}
}
```

#### C++

```cpp
class Solution {
public:
int maxHeightOfTriangle(int red, int blue) {
int ans = 0;
for (int k = 0; k < 2; ++k) {
int c[2] = {red, blue};
for (int i = 1, j = k; i <= c[j]; j ^= 1, ++i) {
c[j] -= i;
ans = max(ans, i);
}
}
return ans;
}
};
```

#### Go

```go
func maxHeightOfTriangle(red int, blue int) (ans int) {
for k := 0; k < 2; k++ {
c := [2]int{red, blue}
for i, j := 1, k; i <= c[j]; i, j = i+1, j^1 {
c[j] -= i
ans = max(ans, i)
}
}
return
}
```

#### TypeScript

```ts
function maxHeightOfTriangle(red: number, blue: number): number {
let ans = 0;
for (let k = 0; k < 2; ++k) {
const c: [number, number] = [red, blue];
for (let i = 1, j = k; i <= c[j]; ++i, j ^= 1) {
c[j] -= i;
ans = Math.max(ans, i);
}
}
return ans;
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
14 changes: 14 additions & 0 deletions solution/3200-3299/3200.Maximum Height of a Triangle/Solution.cpp
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Original file line number Diff line number Diff line change
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class Solution {
public:
int maxHeightOfTriangle(int red, int blue) {
int ans = 0;
for (int k = 0; k < 2; ++k) {
int c[2] = {red, blue};
for (int i = 1, j = k; i <= c[j]; j ^= 1, ++i) {
c[j] -= i;
ans = max(ans, i);
}
}
return ans;
}
};
10 changes: 10 additions & 0 deletions solution/3200-3299/3200.Maximum Height of a Triangle/Solution.go
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Original file line number Diff line number Diff line change
@@ -0,0 +1,10 @@
func maxHeightOfTriangle(red int, blue int) (ans int) {
for k := 0; k < 2; k++ {
c := [2]int{red, blue}
for i, j := 1, k; i <= c[j]; i, j = i+1, j^1 {
c[j] -= i
ans = max(ans, i)
}
}
return
}
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