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Commit 312ee51

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feat: add solutions to lc problem: No.3400 (#3890)
No.3400.Maximum Number of Matching Indices After Right Shifts
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‎solution/1500-1599/1565.Unique Orders and Customers Per Month/README_EN.md‎

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@@ -43,7 +43,7 @@ This table contains information about the orders made by customer_id.
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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<strong>Input:</strong>
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Orders table:
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+----------+------------+-------------+------------+
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| order_id | order_date | customer_id | invoice |
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| 9 | 2021年01月07日 | 3 | 31 |
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| 10 | 2021年01月15日 | 2 | 20 |
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+----------+------------+-------------+------------+
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<strong>Output:</strong>
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<strong>Output:</strong>
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+---------+-------------+----------------+
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| month | order_count | customer_count |
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+---------+-------------+----------------+
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| 2020-12 | 2 | 1 |
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| 2021-01 | 1 | 1 |
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+---------+-------------+----------------+
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<strong>Explanation:</strong>
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<strong>Explanation:</strong>
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In September 2020 we have two orders from 2 different customers with invoices &gt; 20ドル.
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In October 2020 we have two orders from 1 customer, and only one of the two orders has invoice &gt; 20ドル.
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In November 2020 we have two orders from 2 different customers but invoices &lt; 20,ドル so we don&#39;t include that month.

‎solution/1500-1599/1577.Number of Ways Where Square of Number Is Equal to Product of Two Numbers/README_EN.md‎

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<pre>
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<strong>Input:</strong> nums1 = [7,4], nums2 = [5,2,8,9]
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<strong>Output:</strong> 1
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<strong>Explanation:</strong> Type 1: (1, 1, 2), nums1[1]<sup>2</sup> = nums2[1] * nums2[2]. (4<sup>2</sup> = 2 * 8).
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<strong>Explanation:</strong> Type 1: (1, 1, 2), nums1[1]<sup>2</sup> = nums2[1] * nums2[2]. (4<sup>2</sup> = 2 * 8).
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</pre>
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<p><strong class="example">Example 2:</strong></p>

‎solution/1500-1599/1589.Maximum Sum Obtained of Any Permutation/README_EN.md‎

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<pre>
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<strong>Input:</strong> nums = [1,2,3,4,5], requests = [[1,3],[0,1]]
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<strong>Output:</strong> 19
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<strong>Explanation:</strong> One permutation of nums is [2,1,3,4,5] with the following result:
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<strong>Explanation:</strong> One permutation of nums is [2,1,3,4,5] with the following result:
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requests[0] -&gt; nums[1] + nums[2] + nums[3] = 1 + 3 + 4 = 8
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requests[1] -&gt; nums[0] + nums[1] = 2 + 1 = 3
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Total sum: 8 + 3 = 11.
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---
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comments: true
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difficulty: 中等
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3400.Maximum%20Number%20of%20Matching%20Indices%20After%20Right%20Shifts/README.md
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---
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<!-- problem:start -->
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# [3400. 右移后的最大匹配索引数 🔒](https://leetcode.cn/problems/maximum-number-of-matching-indices-after-right-shifts)
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[English Version](/solution/3400-3499/3400.Maximum%20Number%20of%20Matching%20Indices%20After%20Right%20Shifts/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>给定两个长度相同的整数数组&nbsp;<code>nums1</code> 和&nbsp;<code>nums2</code>。</p>
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<p>如果&nbsp;<code>nums1[i] == nums2[i]</code>&nbsp;则认为下标&nbsp;<code>i</code> 是 <strong>匹配</strong> 的。</p>
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<p>返回在&nbsp;<code>nums1</code>&nbsp;上进行任意次数 <strong>右移</strong>&nbsp;后 <strong>最大</strong>&nbsp;的 <strong>匹配&nbsp;</strong>下标数量。</p>
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<p><strong>右移&nbsp;</strong>是对于所有下标,将位于下标&nbsp;<code>i</code>&nbsp;的元素移动到&nbsp;<code>(i + 1) % n</code>。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1:</strong></p>
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<div class="example-block">
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<p><strong>输入:</strong><span class="example-io">nums1 = [3,1,2,3,1,2], nums2 = [1,2,3,1,2,3]</span></p>
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<p><span class="example-io"><b>输出:</b>6</span></p>
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<p><strong>解释:</strong></p>
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<p>如果我们右移&nbsp;<code>nums1</code> 2 次,它变为&nbsp;<code>[1, 2, 3, 1, 2, 3]</code>。每个下标都匹配,所以输出为 6。</p>
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</div>
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<p><strong class="example">示例 2:</strong></p>
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<div class="example-block">
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<p><span class="example-io"><b>输入:</b>nums1 = [1,4,2,5,3,1], nums2 = [2,3,1,2,4,6]</span></p>
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<p><span class="example-io"><b>输出:</b>3</span></p>
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<p><strong>解释:</strong></p>
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<p>如果我们右移&nbsp;<code>nums1</code> 3 次,它变为&nbsp;<code>[5, 3, 1, 1, 4, 2]</code>。下标 1,2,4 匹配,所以输出为 3。</p>
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</div>
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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>nums1.length == nums2.length</code></li>
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<li><code>1 &lt;= nums1.length, nums2.length &lt;= 3000</code></li>
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<li><code>1 &lt;= nums1[i], nums2[i] &lt;= 10<sup>9</sup></code></li>
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</ul>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一:枚举
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我们可以枚举右移的次数 $k,ドル其中 0ドル \leq k \lt n$。对于每一个 $k,ドル我们可以计算出右移 $k$ 次后的数组 $\textit{nums1}$ 和 $\textit{nums2}$ 的匹配下标数量,取最大值作为答案即可。
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时间复杂度 $O(n^2),ドル其中 $n$ 为数组 $\textit{nums1}$ 的长度。空间复杂度 $O(1)$。
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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 maximumMatchingIndices(self, nums1: List[int], nums2: List[int]) -> int:
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n = len(nums1)
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ans = 0
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for k in range(n):
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t = sum(nums1[(i + k) % n] == x for i, x in enumerate(nums2))
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ans = max(ans, t)
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return ans
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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 maximumMatchingIndices(int[] nums1, int[] nums2) {
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int n = nums1.length;
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int ans = 0;
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for (int k = 0; k < n; ++k) {
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int t = 0;
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for (int i = 0; i < n; ++i) {
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if (nums1[(i + k) % n] == nums2[i]) {
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++t;
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}
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}
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ans = Math.max(ans, t);
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}
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return ans;
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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 maximumMatchingIndices(vector<int>& nums1, vector<int>& nums2) {
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int n = nums1.size();
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int ans = 0;
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for (int k = 0; k < n; ++k) {
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int t = 0;
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for (int i = 0; i < n; ++i) {
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if (nums1[(i + k) % n] == nums2[i]) {
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++t;
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}
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}
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ans = max(ans, t);
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}
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return ans;
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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 maximumMatchingIndices(nums1 []int, nums2 []int) (ans int) {
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n := len(nums1)
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for k := range nums1 {
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t := 0
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for i, x := range nums2 {
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if nums1[(i+k)%n] == x {
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t++
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}
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}
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ans = max(ans, t)
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}
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return
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}
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```
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#### TypeScript
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```ts
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function maximumMatchingIndices(nums1: number[], nums2: number[]): number {
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const n = nums1.length;
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let ans: number = 0;
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for (let k = 0; k < n; ++k) {
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let t: number = 0;
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for (let i = 0; i < n; ++i) {
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if (nums1[(i + k) % n] === nums2[i]) {
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++t;
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}
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}
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ans = Math.max(ans, t);
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}
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return ans;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->
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---
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comments: true
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difficulty: Medium
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3400.Maximum%20Number%20of%20Matching%20Indices%20After%20Right%20Shifts/README_EN.md
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---
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<!-- problem:start -->
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# [3400. Maximum Number of Matching Indices After Right Shifts 🔒](https://leetcode.com/problems/maximum-number-of-matching-indices-after-right-shifts)
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[中文文档](/solution/3400-3499/3400.Maximum%20Number%20of%20Matching%20Indices%20After%20Right%20Shifts/README.md)
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## Description
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<!-- description:start -->
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<p>You are given two integer arrays, <code>nums1</code> and <code>nums2</code>, of the same length.</p>
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<p>An index <code>i</code> is considered <strong>matching</strong> if <code>nums1[i] == nums2[i]</code>.</p>
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<p>Return the <strong>maximum</strong> number of <strong>matching</strong> indices after performing any number of <strong>right shifts</strong> on <code>nums1</code>.</p>
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<p>A <strong>right shift</strong> is defined as shifting the element at index <code>i</code> to index <code>(i + 1) % n</code>, for all indices.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">nums1 = [3,1,2,3,1,2], nums2 = [1,2,3,1,2,3]</span></p>
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<p><strong>Output:</strong> <span class="example-io">6</span></p>
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<p><strong>Explanation:</strong></p>
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<p>If we right shift <code>nums1</code> 2 times, it becomes <code>[1, 2, 3, 1, 2, 3]</code>. Every index matches, so the output is 6.</p>
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</div>
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<p><strong class="example">Example 2:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">nums1 = [1,4,2,5,3,1], nums2 = [2,3,1,2,4,6]</span></p>
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<p><strong>Output:</strong> <span class="example-io">3</span></p>
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<p><strong>Explanation:</strong></p>
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<p>If we right shift <code>nums1</code> 3 times, it becomes <code>[5, 3, 1, 1, 4, 2]</code>. Indices 1, 2, and 4 match, so the output is 3.</p>
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</div>
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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>nums1.length == nums2.length</code></li>
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<li><code>1 &lt;= nums1.length, nums2.length &lt;= 3000</code></li>
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<li><code>1 &lt;= nums1[i], nums2[i] &lt;= 10<sup>9</sup></code></li>
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</ul>
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<!-- description:end -->
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## Solutions
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<!-- solution:start -->
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### Solution 1: Enumeration
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We can enumerate the number of right shifts $k,ドル where 0ドル \leq k < n$. For each $k,ドル we can calculate the number of matching indices between the array $\textit{nums1}$ after right shifting $k$ times and $\textit{nums2}$. The maximum value is taken as the answer.
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The time complexity is $O(n^2),ドル where $n$ is the length of the array $\textit{nums1}$. The space complexity is $O(1)$.
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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 maximumMatchingIndices(self, nums1: List[int], nums2: List[int]) -> int:
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n = len(nums1)
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ans = 0
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for k in range(n):
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t = sum(nums1[(i + k) % n] == x for i, x in enumerate(nums2))
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ans = max(ans, t)
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return ans
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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 maximumMatchingIndices(int[] nums1, int[] nums2) {
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int n = nums1.length;
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int ans = 0;
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for (int k = 0; k < n; ++k) {
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int t = 0;
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for (int i = 0; i < n; ++i) {
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if (nums1[(i + k) % n] == nums2[i]) {
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++t;
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}
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}
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ans = Math.max(ans, t);
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}
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return ans;
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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 maximumMatchingIndices(vector<int>& nums1, vector<int>& nums2) {
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int n = nums1.size();
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int ans = 0;
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for (int k = 0; k < n; ++k) {
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int t = 0;
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for (int i = 0; i < n; ++i) {
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if (nums1[(i + k) % n] == nums2[i]) {
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++t;
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}
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}
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ans = max(ans, t);
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}
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return ans;
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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 maximumMatchingIndices(nums1 []int, nums2 []int) (ans int) {
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n := len(nums1)
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for k := range nums1 {
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t := 0
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for i, x := range nums2 {
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if nums1[(i+k)%n] == x {
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t++
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}
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}
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ans = max(ans, t)
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}
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return
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}
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```
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#### TypeScript
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```ts
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function maximumMatchingIndices(nums1: number[], nums2: number[]): number {
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const n = nums1.length;
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let ans: number = 0;
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for (let k = 0; k < n; ++k) {
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let t: number = 0;
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for (let i = 0; i < n; ++i) {
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if (nums1[(i + k) % n] === nums2[i]) {
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++t;
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}
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}
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ans = Math.max(ans, t);
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}
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return ans;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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