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feat: add solutions to lc problem: No.2417
No.2417.Closest Fair Integer
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‎solution/1600-1699/1686.Stone Game VI/Solution.py

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class Solution:
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def stoneGameVI(self, aliceValues: List[int], bobValues: List[int]) -> int:
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arr = [(a + b, i)
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for i, (a, b) in enumerate(zip(aliceValues, bobValues))]
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arr = [(a + b, i) for i, (a, b) in enumerate(zip(aliceValues, bobValues))]
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arr.sort(reverse=True)
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a = sum(aliceValues[v[1]] for i, v in enumerate(arr) if i % 2 == 0)
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b = sum(bobValues[v[1]] for i, v in enumerate(arr) if i % 2 == 1)

‎solution/2400-2499/2417.Closest Fair Integer/README.md

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# [2417. Closest Fair Integer](https://leetcode.cn/problems/closest-fair-integer)
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[English Version](/solution/2400-2499/2417.Closest%20Fair%20Integer/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>You are given a <strong>positive</strong> integer <code>n</code>.</p>
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<p>We call an integer <code>k</code> fair if the number of <strong>even</strong> digits in <code>k</code> is equal to the number of <strong>odd</strong> digits in it.</p>
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<p>Return <em>the <strong>smallest</strong> fair integer that is <strong>greater than or equal</strong> to </em><code>n</code>.</p>
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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 = 2
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<strong>Output:</strong> 10
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<strong>Explanation:</strong> The smallest fair integer that is greater than or equal to 2 is 10.
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10 is fair because it has an equal number of even and odd digits (one odd digit and one even digit).</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 = 403
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<strong>Output:</strong> 1001
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<strong>Explanation:</strong> The smallest fair integer that is greater than or equal to 403 is 1001.
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1001 is fair because it has an equal number of even and odd digits (two odd digits and two even digits).
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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;= 10<sup>9</sup></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：分类讨论**
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我们记 $n$ 的位数为 $k$，奇数位数、偶数位数分别为 $a$ 和 $b$。
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- 若 $a=b$，则 $n$ 本身就是 `fair` 的，直接返回 $n$ 即可；
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- 否则，若 $k$ 为奇数，那么我们找到 $k+1$ 位的最小 `fair` 数即可，形如 `10000111`；若 $k$ 为偶数，我们直接暴力递归 `closestFair(n+1)` 即可。
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时间复杂度 $O(\sqrt{n} \times \log_{10} n)$。
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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 closestFair(self, n: int) -> int:
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a = b = k = 0
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t = n
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while t:
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if (t % 10) & 1:
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a += 1
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else:
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b += 1
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t //= 10
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k += 1
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if k & 1:
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x = 10**k
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y = int('1' * (k >> 1) or '0')
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return x + y
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if a == b:
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return n
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return self.closestFair(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 closestFair(int n) {
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int a = 0, b = 0;
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int k = 0, t = n;
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while (t > 0) {
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if ((t % 10) % 2 == 1) {
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++a;
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} else {
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++b;
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}
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t /= 10;
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++k;
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}
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if (k % 2 == 1) {
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int x = (int) Math.pow(10, k);
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int y = 0;
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for (int i = 0; i < k >> 1; ++i) {
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y = y * 10 + 1;
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}
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return x + y;
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}
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if (a == b) {
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return n;
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}
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return closestFair(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 closestFair(int n) {
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int a = 0, b = 0;
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int t = n, k = 0;
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while (t) {
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if ((t % 10) & 1) {
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++a;
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} else {
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++b;
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}
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++k;
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t /= 10;
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}
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if (a == b) {
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return n;
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}
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if (k % 2 == 1) {
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int x = pow(10, k);
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int y = 0;
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for (int i = 0; i < k >> 1; ++i) {
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y = y * 10 + 1;
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}
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return x + y;
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}
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return closestFair(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 closestFair(n int) int {
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a, b := 0, 0
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t, k := n, 0
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for t > 0 {
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if (t%10)%2 == 1 {
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a++
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} else {
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b++
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}
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k++
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t /= 10
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}
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if a == b {
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return n
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}
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if k%2 == 1 {
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x := int(math.Pow(10, float64(k)))
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y := 0
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for i := 0; i < k>>1; i++ {
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y = y*10 + 1
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}
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return x + y
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}
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return closestFair(n + 1)
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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/2400-2499/2417.Closest Fair Integer/README_EN.md

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# [2417. Closest Fair Integer](https://leetcode.com/problems/closest-fair-integer)
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[中文文档](/solution/2400-2499/2417.Closest%20Fair%20Integer/README.md)
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## Description
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<p>You are given a <strong>positive</strong> integer <code>n</code>.</p>
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<p>We call an integer <code>k</code> fair if the number of <strong>even</strong> digits in <code>k</code> is equal to the number of <strong>odd</strong> digits in it.</p>
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<p>Return <em>the <strong>smallest</strong> fair integer that is <strong>greater than or equal</strong> to </em><code>n</code>.</p>
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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 = 2
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<strong>Output:</strong> 10
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<strong>Explanation:</strong> The smallest fair integer that is greater than or equal to 2 is 10.
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10 is fair because it has an equal number of even and odd digits (one odd digit and one even digit).</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 = 403
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<strong>Output:</strong> 1001
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<strong>Explanation:</strong> The smallest fair integer that is greater than or equal to 403 is 1001.
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1001 is fair because it has an equal number of even and odd digits (two odd digits and two even digits).
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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;= 10<sup>9</sup></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 closestFair(self, n: int) -> int:
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a = b = k = 0
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t = n
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while t:
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if (t % 10) & 1:
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a += 1
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else:
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b += 1
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t //= 10
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k += 1
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if k & 1:
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x = 10**k
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y = int('1' * (k >> 1) or '0')
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return x + y
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if a == b:
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return n
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return self.closestFair(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 closestFair(int n) {
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int a = 0, b = 0;
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int k = 0, t = n;
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while (t > 0) {
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if ((t % 10) % 2 == 1) {
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++a;
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} else {
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++b;
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}
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t /= 10;
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++k;
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}
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if (k % 2 == 1) {
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int x = (int) Math.pow(10, k);
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int y = 0;
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for (int i = 0; i < k >> 1; ++i) {
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y = y * 10 + 1;
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}
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return x + y;
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}
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if (a == b) {
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return n;
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}
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return closestFair(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 closestFair(int n) {
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int a = 0, b = 0;
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int t = n, k = 0;
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while (t) {
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if ((t % 10) & 1) {
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++a;
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} else {
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++b;
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}
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++k;
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t /= 10;
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}
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if (a == b) {
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return n;
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}
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if (k % 2 == 1) {
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int x = pow(10, k);
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int y = 0;
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for (int i = 0; i < k >> 1; ++i) {
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y = y * 10 + 1;
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}
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return x + y;
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}
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return closestFair(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 closestFair(n int) int {
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a, b := 0, 0
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t, k := n, 0
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for t > 0 {
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if (t%10)%2 == 1 {
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a++
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} else {
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b++
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}
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k++
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t /= 10
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}
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if a == b {
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return n
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}
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if k%2 == 1 {
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x := int(math.Pow(10, float64(k)))
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y := 0
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for i := 0; i < k>>1; i++ {
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y = y*10 + 1
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}
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return x + y
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}
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return closestFair(n + 1)
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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 -->

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