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feat: add solutions to lc problem: No.2423~2430
* No.2423.Remove Letter To Equalize Frequency * No.2424.Longest Uploaded Prefix * No.2425.Bitwise XOR of All Pairings * No.2426.Number of Pairs Satisfying Inequality * No.2427.Number of Common Factors * No.2428.Maximum Sum of an Hourglass * No.2429.Minimize XOR * No.2430.Maximum Deletions on a String
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‎solution/2400-2499/2423.Remove Letter To Equalize Frequency/README.md

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# [2423. 删除字符使频率相同](https://leetcode.cn/problems/remove-letter-to-equalize-frequency)
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[English Version](/solution/2400-2499/2423.Remove%20Letter%20To%20Equalize%20Frequency/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给你一个下标从 <strong>0</strong>&nbsp;开始的字符串&nbsp;<code>word</code>&nbsp;，字符串只包含小写英文字母。你需要选择 <strong>一个</strong>&nbsp;下标并 <strong>删除</strong>&nbsp;下标处的字符，使得 <code>word</code>&nbsp;中剩余每个字母出现 <strong>频率</strong>&nbsp;相同。</p>
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<p>如果删除一个字母后，<code>word</code>&nbsp;中剩余所有字母的出现频率都相同，那么返回 <code>true</code>&nbsp;，否则返回 <code>false</code>&nbsp;。</p>
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<p><strong>注意：</strong></p>
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<ul>
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<li>字母&nbsp;<code>x</code>&nbsp;的 <strong>频率</strong><strong>&nbsp;</strong>是这个字母在字符串中出现的次数。</li>
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<li>你 <strong>必须</strong>&nbsp;恰好删除一个字母，不能一个字母都不删除。</li>
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</ul>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<pre><b>输入：</b>word = "abcc"
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<b>输出：</b>true
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<b>解释：</b>选择下标 3 并删除该字母，word 变成 "abc" 且每个字母出现频率都为 1 。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre><b>输入：</b>word = "aazz"
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<b>输出：</b>false
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<b>解释：</b>我们必须删除一个字母，所以要么 "a" 的频率变为 1 且 "z" 的频率为 2 ，要么两个字母频率反过来。所以不可能让剩余所有字母出现频率相同。
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>2 &lt;= word.length &lt;= 100</code></li>
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<li><code>word</code>&nbsp;只包含小写英文字母。</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：直接模拟**
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遍历字符串中的每个字符，删除该字符后，判断剩余字符串中每个字符出现的频率是否相同。如果相同，返回 `true`，否则遍历结束，返回 `false`。
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时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。其中 $n$ 为字符串的长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def equalFrequency(self, word: str) -> bool:
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for i in range(len(word)):
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cnt = Counter(word[:i] + word[i + 1:])
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if len(set(cnt.values())) == 1:
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return True
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return False
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public boolean equalFrequency(String word) {
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for (int i = 0; i < word.length(); ++i) {
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int[] cnt = new int[26];
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for (int j = 0; j < word.length(); ++j) {
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if (j != i) {
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++cnt[word.charAt(j) - 'a'];
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}
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}
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Set<Integer> vis = new HashSet<>();
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for (int v : cnt) {
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if (v > 0) {
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vis.add(v);
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}
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}
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if (vis.size() == 1) {
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return true;
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}
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}
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return false;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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bool equalFrequency(string word) {
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for (int i = 0; i < word.size(); ++i) {
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int cnt[26] = {0};
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for (int j = 0; j < word.size(); ++j) {
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if (j != i) {
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++cnt[word[j] - 'a'];
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}
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}
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unordered_set<int> vis;
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for (int v : cnt) {
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if (v) {
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vis.insert(v);
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}
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}
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if (vis.size() == 1) {
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return true;
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}
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}
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return false;
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}
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};
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```
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### **Go**
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```go
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func equalFrequency(word string) bool {
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for i := range word {
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cnt := make([]int, 26)
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for j, c := range word {
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if j != i {
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cnt[c-'a']++
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}
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}
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vis := map[int]bool{}
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for _, v := range cnt {
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if v > 0 {
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vis[v] = true
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}
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}
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if len(vis) == 1 {
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return true
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}
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}
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return false
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}
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2400-2499/2423.Remove Letter To Equalize Frequency/README_EN.md

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# [2423. Remove Letter To Equalize Frequency](https://leetcode.com/problems/remove-letter-to-equalize-frequency)
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[中文文档](/solution/2400-2499/2423.Remove%20Letter%20To%20Equalize%20Frequency/README.md)
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## Description
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<p>You are given a <strong>0-indexed</strong> string <code>word</code>, consisting of lowercase English letters. You need to select <strong>one</strong> index and <strong>remove</strong> the letter at that index from <code>word</code> so that the <strong>frequency</strong> of every letter present in <code>word</code> is equal.</p>
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<p>Return<em> </em><code>true</code><em> if it is possible to remove one letter so that the frequency of all letters in </em><code>word</code><em> are equal, and </em><code>false</code><em> otherwise</em>.</p>
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<p><strong>Note:</strong></p>
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<ul>
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<li>The <b>frequency</b> of a letter <code>x</code> is the number of times it occurs in the string.</li>
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<li>You <strong>must</strong> remove exactly one letter and cannot chose to do nothing.</li>
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</ul>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> word = &quot;abcc&quot;
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<strong>Output:</strong> true
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<strong>Explanation:</strong> Select index 3 and delete it: word becomes &quot;abc&quot; and each character has a frequency of 1.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> word = &quot;aazz&quot;
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<strong>Output:</strong> false
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<strong>Explanation:</strong> We must delete a character, so either the frequency of &quot;a&quot; is 1 and the frequency of &quot;z&quot; is 2, or vice versa. It is impossible to make all present letters have equal frequency.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>2 &lt;= word.length &lt;= 100</code></li>
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<li><code>word</code> consists of lowercase English letters only.</li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def equalFrequency(self, word: str) -> bool:
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for i in range(len(word)):
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cnt = Counter(word[:i] + word[i + 1:])
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if len(set(cnt.values())) == 1:
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return True
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return False
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```
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### **Java**
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```java
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class Solution {
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public boolean equalFrequency(String word) {
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for (int i = 0; i < word.length(); ++i) {
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int[] cnt = new int[26];
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for (int j = 0; j < word.length(); ++j) {
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if (j != i) {
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++cnt[word.charAt(j) - 'a'];
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}
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}
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Set<Integer> vis = new HashSet<>();
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for (int v : cnt) {
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if (v > 0) {
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vis.add(v);
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}
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}
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if (vis.size() == 1) {
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return true;
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}
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}
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return false;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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bool equalFrequency(string word) {
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for (int i = 0; i < word.size(); ++i) {
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int cnt[26] = {0};
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for (int j = 0; j < word.size(); ++j) {
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if (j != i) {
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++cnt[word[j] - 'a'];
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}
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}
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unordered_set<int> vis;
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for (int v : cnt) {
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if (v) {
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vis.insert(v);
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}
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}
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if (vis.size() == 1) {
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return true;
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}
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}
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return false;
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}
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};
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```
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### **Go**
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```go
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func equalFrequency(word string) bool {
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for i := range word {
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cnt := make([]int, 26)
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for j, c := range word {
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if j != i {
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cnt[c-'a']++
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}
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}
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vis := map[int]bool{}
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for _, v := range cnt {
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if v > 0 {
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vis[v] = true
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}
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}
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if len(vis) == 1 {
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return true
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}
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}
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return false
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}
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2400-2499/2423.Remove Letter To Equalize Frequency/Solution.cpp

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class Solution {
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public:
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bool equalFrequency(string word) {
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for (int i = 0; i < word.size(); ++i) {
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int cnt[26] = {0};
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for (int j = 0; j < word.size(); ++j) {
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if (j != i) {
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++cnt[word[j] - 'a'];
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}
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}
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unordered_set<int> vis;
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for (int v : cnt) {
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if (v) {
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vis.insert(v);
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}
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}
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if (vis.size() == 1) {
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return true;
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}
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}
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return false;
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}
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};

‎solution/2400-2499/2423.Remove Letter To Equalize Frequency/Solution.go

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func equalFrequency(word string) bool {
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for i := range word {
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cnt := make([]int, 26)
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for j, c := range word {
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if j != i {
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cnt[c-'a']++
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}
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}
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vis := map[int]bool{}
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for _, v := range cnt {
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if v > 0 {
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vis[v] = true
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}
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}
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if len(vis) == 1 {
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return true
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}
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}
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return false
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}

‎solution/2400-2499/2423.Remove Letter To Equalize Frequency/Solution.java

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class Solution {
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public boolean equalFrequency(String word) {
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for (int i = 0; i < word.length(); ++i) {
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int[] cnt = new int[26];
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for (int j = 0; j < word.length(); ++j) {
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if (j != i) {
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++cnt[word.charAt(j) - 'a'];
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}
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}
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Set<Integer> vis = new HashSet<>();
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for (int v : cnt) {
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if (v > 0) {
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vis.add(v);
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}
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}
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if (vis.size() == 1) {
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return true;
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}
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}
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return false;
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}
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}

‎solution/2400-2499/2423.Remove Letter To Equalize Frequency/Solution.py

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class Solution:
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def equalFrequency(self, word: str) -> bool:
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for i in range(len(word)):
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cnt = Counter(word[:i] + word[i + 1 :])
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if len(set(cnt.values())) == 1:
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return True
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return False

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