feat: add weekly contest 435 (#4015)

Author: yanglbme

solution/3400-3499/3442.Maximum Difference Between Even and Odd Frequency I/README.md

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+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3442.Maximum%20Difference%20Between%20Even%20and%20Odd%20Frequency%20I/README.md
+---
+
+<!-- problem:start -->
+
+# [3442. 奇偶频次间的最大差值 I](https://leetcode.cn/problems/maximum-difference-between-even-and-odd-frequency-i)
+
+[English Version](/solution/3400-3499/3442.Maximum%20Difference%20Between%20Even%20and%20Odd%20Frequency%20I/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个由小写英文字母组成的字符串&nbsp;<code>s</code> 。请你找出字符串中两个字符的出现频次之间的 <strong>最大</strong> 差值，这两个字符需要满足：</p>
+
+<ul>
+	<li>一个字符在字符串中出现 <strong>偶数次</strong> 。</li>
+	<li>另一个字符在字符串中出现 <strong>奇数次</strong>&nbsp;。</li>
+</ul>
+
+<p>返回 <strong>最大</strong> 差值，计算方法是出现 <strong>奇数次</strong> 字符的次数 <strong>减去</strong> 出现 <strong>偶数次</strong> 字符的次数。</p>
+
+<p>&nbsp;</p>
+
+<p><b>示例 1：</b></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>s = "aaaaabbc"</span></p>
+
+<p><b>输出：</b>3</p>
+
+<p><b>解释：</b></p>
+
+<ul>
+	<li>字符&nbsp;<code>'a'</code>&nbsp;出现 <strong>奇数次</strong>&nbsp;，次数为&nbsp;<code><font face="monospace">5</font></code><font face="monospace"> ；字符</font>&nbsp;<code>'b'</code>&nbsp;出现 <strong>偶数次</strong> ，次数为&nbsp;<code><font face="monospace">2</font></code>&nbsp;。</li>
+	<li>最大差值为&nbsp;<code>5 - 2 = 3</code>&nbsp;。</li>
+</ul>
+</div>
+
+<p><b>示例 2：</b></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>s = "abcabcab"</span></p>
+
+<p><b>输出：</b>1</p>
+
+<p><b>解释：</b></p>
+
+<ul>
+	<li>字符&nbsp;<code>'a'</code>&nbsp;出现 <strong>奇数次</strong>&nbsp;，次数为&nbsp;<code><font face="monospace">3</font></code><font face="monospace"> ；字符</font>&nbsp;<code>'c'</code>&nbsp;出现 <strong>偶数次</strong>&nbsp;，次数为&nbsp;<font face="monospace">2 。</font></li>
+	<li>最大差值为&nbsp;<code>3 - 2 = 1</code> 。</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+<p><b>提示：</b></p>
+
+<ul>
+	<li><code>3 &lt;= s.length &lt;= 100</code></li>
+	<li><code>s</code>&nbsp;仅由小写英文字母组成。</li>
+	<li><code>s</code>&nbsp;至少由一个出现奇数次的字符和一个出现偶数次的字符组成。</li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：计数
+
+我们可以用一个哈希表或数组 $\textit{cnt}$ 记录字符串 $s$ 中每个字符的出现次数。然后遍历 $\textit{cnt}$，找出出现奇数次的字符的最大频次 $a$ 和出现偶数次的字符的最小频次 $b$，最后返回 $a - b$ 即可。
+
+时间复杂度 $O(n)$，其中 $n$ 是字符串 $s$ 的长度。空间复杂度 $O(|\Sigma|)$，其中 $\Sigma$ 是字符集，本题中 $|\Sigma| = 26$。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def maxDifference(self, s: str) -> int:
+        cnt = Counter(s)
+        a, b = 0, inf
+        for v in cnt.values():
+            if v % 2:
+                a = max(a, v)
+            else:
+                b = min(b, v)
+        return a - b
+```
+
+#### Java
+
+```java
+class Solution {
+    public int maxDifference(String s) {
+        int[] cnt = new int[26];
+        for (char c : s.toCharArray()) {
+            ++cnt[c - 'a'];
+        }
+        int a = 0, b = 1 << 30;
+        for (int v : cnt) {
+            if (v % 2 == 1) {
+                a = Math.max(a, v);
+            } else if (v > 0) {
+                b = Math.min(b, v);
+            }
+        }
+        return a - b;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int maxDifference(string s) {
+        int cnt[26]{};
+        for (char c : s) {
+            ++cnt[c - 'a'];
+        }
+        int a = 0, b = 1 << 30;
+        for (int v : cnt) {
+            if (v % 2 == 1) {
+                a = max(a, v);
+            } else if (v > 0) {
+                b = min(b, v);
+            }
+        }
+        return a - b;
+    }
+};
+```
+
+#### Go
+
+```go
+func maxDifference(s string) int {
+	cnt := [26]int{}
+	for _, c := range s {
+		cnt[c-'a']++
+	}
+	a, b := 0, 1<<30
+	for _, v := range cnt {
+		if v%2 == 1 {
+			a = max(a, v)
+		} else if v > 0 {
+			b = min(b, v)
+		}
+	}
+	return a - b
+}
+```
+
+#### TypeScript
+
+```ts
+function maxDifference(s: string): number {
+    const cnt: Record<string, number> = {};
+    for (const c of s) {
+        cnt[c] = (cnt[c] || 0) + 1;
+    }
+    let [a, b] = [0, Infinity];
+    for (const [_, v] of Object.entries(cnt)) {
+        if (v % 2 === 1) {
+            a = Math.max(a, v);
+        } else {
+            b = Math.min(b, v);
+        }
+    }
+    return a - b;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3400-3499/3442.Maximum Difference Between Even and Odd Frequency I/README_EN.md

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+---
+comments: true
+difficulty: Easy
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3442.Maximum%20Difference%20Between%20Even%20and%20Odd%20Frequency%20I/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3442. Maximum Difference Between Even and Odd Frequency I](https://leetcode.com/problems/maximum-difference-between-even-and-odd-frequency-i)
+
+[中文文档](/solution/3400-3499/3442.Maximum%20Difference%20Between%20Even%20and%20Odd%20Frequency%20I/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>You are given a string <code>s</code> consisting of lowercase English letters. Your task is to find the <strong>maximum</strong> difference between the frequency of <strong>two</strong> characters in the string such that:</p>
+
+<ul>
+	<li>One of the characters has an <strong>even frequency</strong> in the string.</li>
+	<li>The other character has an <strong>odd frequency</strong> in the string.</li>
+</ul>
+
+<p>Return the <strong>maximum</strong> difference, calculated as the frequency of the character with an <b>odd</b> frequency <strong>minus</strong> the frequency of the character with an <b>even</b> frequency.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot;aaaaabbc&quot;</span></p>
+
+<p><strong>Output:</strong> 3</p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>The character <code>&#39;a&#39;</code> has an <strong>odd frequency</strong> of <code><font face="monospace">5</font></code><font face="monospace">,</font> and <code>&#39;b&#39;</code> has an <strong>even frequency</strong> of <code><font face="monospace">2</font></code>.</li>
+	<li>The maximum difference is <code>5 - 2 = 3</code>.</li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot;abcabcab&quot;</span></p>
+
+<p><strong>Output:</strong> 1</p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>The character <code>&#39;a&#39;</code> has an <strong>odd frequency</strong> of <code><font face="monospace">3</font></code><font face="monospace">,</font> and <code>&#39;c&#39;</code> has an <strong>even frequency</strong> of <font face="monospace">2</font>.</li>
+	<li>The maximum difference is <code>3 - 2 = 1</code>.</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>3 &lt;= s.length &lt;= 100</code></li>
+	<li><code>s</code> consists only of lowercase English letters.</li>
+	<li><code>s</code> contains at least one character with an odd frequency and one with an even frequency.</li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1: Counting
+
+We can use a hash table or an array $\textit{cnt}$ to record the occurrences of each character in the string $s$. Then, we traverse $\textit{cnt}$ to find the maximum frequency $a$ of characters that appear an odd number of times and the minimum frequency $b$ of characters that appear an even number of times. Finally, we return $a - b$.
+
+The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(|\Sigma|)$, where $\Sigma$ is the character set. In this problem, $|\Sigma| = 26$.
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def maxDifference(self, s: str) -> int:
+        cnt = Counter(s)
+        a, b = 0, inf
+        for v in cnt.values():
+            if v % 2:
+                a = max(a, v)
+            else:
+                b = min(b, v)
+        return a - b
+```
+
+#### Java
+
+```java
+class Solution {
+    public int maxDifference(String s) {
+        int[] cnt = new int[26];
+        for (char c : s.toCharArray()) {
+            ++cnt[c - 'a'];
+        }
+        int a = 0, b = 1 << 30;
+        for (int v : cnt) {
+            if (v % 2 == 1) {
+                a = Math.max(a, v);
+            } else if (v > 0) {
+                b = Math.min(b, v);
+            }
+        }
+        return a - b;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int maxDifference(string s) {
+        int cnt[26]{};
+        for (char c : s) {
+            ++cnt[c - 'a'];
+        }
+        int a = 0, b = 1 << 30;
+        for (int v : cnt) {
+            if (v % 2 == 1) {
+                a = max(a, v);
+            } else if (v > 0) {
+                b = min(b, v);
+            }
+        }
+        return a - b;
+    }
+};
+```
+
+#### Go
+
+```go
+func maxDifference(s string) int {
+	cnt := [26]int{}
+	for _, c := range s {
+		cnt[c-'a']++
+	}
+	a, b := 0, 1<<30
+	for _, v := range cnt {
+		if v%2 == 1 {
+			a = max(a, v)
+		} else if v > 0 {
+			b = min(b, v)
+		}
+	}
+	return a - b
+}
+```
+
+#### TypeScript
+
+```ts
+function maxDifference(s: string): number {
+    const cnt: Record<string, number> = {};
+    for (const c of s) {
+        cnt[c] = (cnt[c] || 0) + 1;
+    }
+    let [a, b] = [0, Infinity];
+    for (const [_, v] of Object.entries(cnt)) {
+        if (v % 2 === 1) {
+            a = Math.max(a, v);
+        } else {
+            b = Math.min(b, v);
+        }
+    }
+    return a - b;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3400-3499/3442.Maximum Difference Between Even and Odd Frequency I/Solution.cpp

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+class Solution {
+public:
+    int maxDifference(string s) {
+        int cnt[26]{};
+        for (char c : s) {
+            ++cnt[c - 'a'];
+        }
+        int a = 0, b = 1 << 30;
+        for (int v : cnt) {
+            if (v % 2 == 1) {
+                a = max(a, v);
+            } else if (v > 0) {
+                b = min(b, v);
+            }
+        }
+        return a - b;
+    }
+};