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feat: add solutions to lc problems: No.2549~2552
* No.2549.Count Distinct Numbers on Board * No.2550.Count Collisions of Monkeys on a Polygon * No.2551.Put Marbles in Bag * No.2552.Count Increasing Quadruplets
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‎solution/0100-0199/0170.Two Sum III - Data structure design/README_EN.md‎

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public TwoSum() {
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}
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public void add(int number) {
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cnt.merge(number, 1, Integer::sum);
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}
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public boolean find(int value) {
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for (var e : cnt.entrySet()) {
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int x = e.getKey(), v = e.getValue();
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TwoSum() {
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}
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void add(int number) {
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++cnt[number];
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}
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bool find(int value) {
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for (auto& [x, v] : cnt) {
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long y = (long) value - x;

‎solution/0500-0599/0599.Minimum Index Sum of Two Lists/README_EN.md‎

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<li><code>list1[i]</code> and <code>list2[i]</code> consist of spaces <code>&#39; &#39;</code> and English letters.</li>
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<li>All the strings of <code>list1</code> are <strong>unique</strong>.</li>
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<li>All the strings of <code>list2</code> are <strong>unique</strong>.</li>
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<li>There is at least a common string between <code>list1</code> and <code>list2</code>.</li>
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</ul>
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## Solutions

‎solution/1600-1699/1671.Minimum Number of Removals to Make Mountain Array/README_EN.md‎

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}
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}
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return n - ans;
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};
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}
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```
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### **...**

‎solution/2300-2399/2317.Maximum XOR After Operations/README.md‎

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**方法一:位运算**
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在一次操作中,我们可以把 $nums[i]$ 更新为 $nums[i] \& (nums[i] \oplus x),ドル其中 $\&$ 表示逐位与运算,$\oplus$ 表示逐位异或运算。由于 $x$ 是任意非负整数,因此 $nums[i] \oplus x$ 的结果是一个任意值,再与 $nums[i]$ 逐位与运算,可以把 $nums[i]$ 的二进制表示中的若干位 1ドル$ 变为 0ドル$。
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在一次操作中,我们可以把 `nums[i]` 更新为 `nums[i] AND (nums[i] XOR x)`。由于 $x$ 是任意非负整数,因此 $nums[i] \oplus x$ 的结果是一个任意值,再与 `nums[i]` 逐位与运算,可以把 `nums[i]` 的二进制表示中的若干位 1ドル$ 变为 0ドル$。
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而题目中要获取的是 `nums` 所有元素的最大逐位异或和,对于一个二进制位,只要在 `nums` 中存在一个元素对应的二进制位为 1ドル,ドル那么这个二进制位对于最大逐位异或和的贡献就是 1ドル$。因此答案就是 `nums` 中所有元素的逐位或运算的结果。
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# [2549. 统计桌面上的不同数字](https://leetcode.cn/problems/count-distinct-numbers-on-board)
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[English Version](/solution/2500-2599/2549.Count%20Distinct%20Numbers%20on%20Board/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>给你一个正整数 <code>n</code> ,开始时,它放在桌面上。在 <code>10<sup>9</sup></code> 天内,每天都要执行下述步骤:</p>
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<ul>
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<li>对于出现在桌面上的每个数字 <code>x</code> ,找出符合 <code>1 &lt;= i &lt;= n</code> 且满足 <code>x % i == 1</code> 的所有数字 <code>i</code> 。</li>
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<li>然后,将这些数字放在桌面上。</li>
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</ul>
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<p>返回在 <code>10<sup>9</sup></code> 天之后,出现在桌面上的 <strong>不同</strong> 整数的数目。</p>
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<p><strong>注意:</strong></p>
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<ul>
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<li>一旦数字放在桌面上,则会一直保留直到结束。</li>
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<li><code>%</code> 表示取余运算。例如,<code>14 % 3</code> 等于 <code>2</code> 。</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>
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<strong>输入:</strong>n = 5
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<strong>输出:</strong>4
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<strong>解释:</strong>最开始,5 在桌面上。
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第二天,2 和 4 也出现在桌面上,因为 5 % 2 == 1 且 5 % 4 == 1 。
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再过一天 3 也出现在桌面上,因为 4 % 3 == 1 。
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在十亿天结束时,桌面上的不同数字有 2 、3 、4 、5 。
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre>
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<strong>输入:</strong>n = 3
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<strong>输出:</strong>2
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<strong>解释:</strong>
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因为 3 % 2 == 1 ,2 也出现在桌面上。
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在十亿天结束时,桌面上的不同数字只有两个:2 和 3 。
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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>1 &lt;= n &lt;= 100</code></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:脑筋急转弯**
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由于每一次对桌面上的数字 $n$ 进行操作,会使得数字 $n-1$ 也出现在桌面上,因此最终桌面上的数字为 $[2,...n],ドル即 $n-1$ 个数字。
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注意到 $n$ 可能为 1ドル,ドル因此需要特判。
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时间复杂度 $O(1),ドル空间复杂度 $O(1)$。
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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 distinctIntegers(self, n: int) -> int:
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return max(1, n - 1)
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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 int distinctIntegers(int n) {
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return Math.max(1, n - 1);
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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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int distinctIntegers(int n) {
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return max(1, n - 1);
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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 distinctIntegers(n int) int {
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if n == 1 {
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return 1
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}
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return n - 1
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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# [2549. Count Distinct Numbers on Board](https://leetcode.com/problems/count-distinct-numbers-on-board)
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[中文文档](/solution/2500-2599/2549.Count%20Distinct%20Numbers%20on%20Board/README.md)
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## Description
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<p>You are given a positive integer <code>n</code>, that is initially placed on a board. Every day, for <code>10<sup>9</sup></code> days, you perform the following procedure:</p>
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<ul>
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<li>For each number <code>x</code> present on the board, find all numbers <code>1 &lt;= i &lt;= n</code> such that <code>x % i == 1</code>.</li>
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<li>Then, place those numbers on the board.</li>
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</ul>
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<p>Return<em> the number of <strong>distinct</strong> integers present on the board after</em> <code>10<sup>9</sup></code> <em>days have elapsed</em>.</p>
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<p><strong>Note:</strong></p>
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<ul>
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<li>Once a number is placed on the board, it will remain on it until the end.</li>
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<li><code>%</code>&nbsp;stands&nbsp;for the modulo operation. For example,&nbsp;<code>14 % 3</code> is <code>2</code>.</li>
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</ul>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> n = 5
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<strong>Output:</strong> 4
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<strong>Explanation:</strong> Initially, 5 is present on the board.
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The next day, 2 and 4 will be added since 5 % 2 == 1 and 5 % 4 == 1.
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After that day, 3 will be added to the board because 4 % 3 == 1.
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At the end of a billion days, the distinct numbers on the board will be 2, 3, 4, and 5.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> n = 3
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<strong>Output:</strong> 2
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<strong>Explanation:</strong>
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Since 3 % 2 == 1, 2 will be added to the board.
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After a billion days, the only two distinct numbers on the board are 2 and 3.
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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>1 &lt;= n &lt;= 100</code></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 distinctIntegers(self, n: int) -> int:
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return max(1, n - 1)
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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 int distinctIntegers(int n) {
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return Math.max(1, n - 1);
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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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int distinctIntegers(int n) {
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return max(1, n - 1);
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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 distinctIntegers(n int) int {
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if n == 1 {
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return 1
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}
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return n - 1
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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class Solution {
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public:
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int distinctIntegers(int n) {
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return max(1, n - 1);
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}
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};
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func distinctIntegers(n int) int {
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if n == 1 {
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return 1
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}
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return n - 1
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}
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class Solution {
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public int distinctIntegers(int n) {
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return Math.max(1, n - 1);
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}
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}
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class Solution:
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def distinctIntegers(self, n: int) -> int:
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return max(1, n - 1)

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