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feat: add solutions to lc problem: No.3199 (doocs#3168)
No.3199.Count Triplets with Even XOR Set Bits I
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‎solution/1000-1099/1048.Longest String Chain/README.md‎

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- 双指针
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- 字符串
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- 动态规划
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- 排序
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---
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<!-- problem:start -->

‎solution/1000-1099/1048.Longest String Chain/README_EN.md‎

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- Two Pointers
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- String
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- Dynamic Programming
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- Sorting
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---
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‎solution/2900-2999/2966.Divide Array Into Arrays With Max Difference/README.md‎

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<p>给你一个长度为 <code>n</code> 的整数数组 <code>nums</code>,以及一个正整数 <code>k</code> 。</p>
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<p>将这个数组划分为一个或多个长度为 <code>3</code> 的子数组,并满足以下条件:</p>
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<p>将这个数组划分为&nbsp;<code>n / 3</code>&nbsp;个长度为 <code>3</code> 的子数组,并满足以下条件:</p>
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<ul>
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<li><code>nums</code> 中的 <strong>每个 </strong>元素都必须 <strong>恰好 </strong>存在于某个子数组中。</li>
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<li>子数组中<strong> 任意 </strong>两个元素的差必须小于或等于 <code>k</code> 。</li>
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<li>子数组中<strong> 任意 </strong>两个元素的差必须 <strong>小于或等于</strong> <code>k</code> 。</li>
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</ul>
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<p>返回一个<em> </em><strong>二维数组 </strong>,包含所有的子数组。如果不可能满足条件,就返回一个空数组。如果有多个答案,返回 <strong>任意一个</strong> 即可。</p>
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<p><strong class="example">示例 1:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [1,3,4,8,7,9,3,5,1], k = 2
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<strong>输出:</strong>[[1,1,3],[3,4,5],[7,8,9]]
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<strong>解释:</strong>可以将数组划分为以下子数组:[1,1,3],[3,4,5] 和 [7,8,9] 。
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每个子数组中任意两个元素的差都小于或等于 2 。
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注意,元素的顺序并不重要。
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</pre>
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<div class="example-block">
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<p><span class="example-io"><b>输入:</b>nums = [1,3,4,8,7,9,3,5,1], k = 2</span></p>
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<p><span class="example-io"><b>输出:</b>[[1,1,3],[3,4,5],[7,8,9]]</span></p>
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<p><strong>解释:</strong></p>
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<p>每个数组中任何两个元素之间的差小于或等于 2。</p>
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</div>
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<p><strong class="example">示例 2:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [1,3,3,2,7,3], k = 3
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<strong>输出:</strong>[]
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<strong>解释:</strong>无法划分数组满足所有条件。
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</pre>
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<div class="example-block">
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<p><span class="example-io"><b>输入:</b></span><span class="example-io">nums = [2,4,2,2,5,2], k = 2</span></p>
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<p><span class="example-io"><b>输出:</b></span><span class="example-io">[]</span></p>
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<p><strong>解释:</strong></p>
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<p>将&nbsp;<code>nums</code>&nbsp;划分为 2 个长度为 3 的数组的不同方式有:</p>
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<ul>
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<li>[[2,2,2],[2,4,5]] (及其排列)</li>
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<li>[[2,2,4],[2,2,5]] (及其排列)</li>
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</ul>
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<p>因为有四个 2,所以无论我们如何划分,都会有一个包含元素 2 和 5 的数组。因为&nbsp;<code>5 - 2 = 3 &gt; k</code>,条件无法被满足,所以没有合法的划分。</p>
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</div>
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<p><strong class="example">示例 3:</strong></p>
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<div class="example-block">
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<p><span class="example-io"><b>输入:</b></span><span class="example-io">nums = [4,2,9,8,2,12,7,12,10,5,8,5,5,7,9,2,5,11], k = 14</span></p>
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<p><span class="example-io"><b>输出:</b></span><span class="example-io">[[2,2,12],[4,8,5],[5,9,7],[7,8,5],[5,9,10],[11,12,2]]</span></p>
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<p><strong>解释:</strong></p>
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<p>每个数组中任何两个元素之间的差小于或等于 14。</p>
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</div>
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<p>&nbsp;</p>
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‎solution/3100-3199/3195.Find the Minimum Area to Cover All Ones I/README.md‎

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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/3100-3199/3195.Find%20the%20Minimum%20Area%20to%20Cover%20All%20Ones%20I/README.md
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tags:
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- 数组
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- 矩阵
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‎solution/3100-3199/3195.Find the Minimum Area to Cover All Ones I/README_EN.md‎

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difficulty: Medium
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3195.Find%20the%20Minimum%20Area%20to%20Cover%20All%20Ones%20I/README_EN.md
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tags:
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- Array
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- Matrix
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‎solution/3100-3199/3196.Maximize Total Cost of Alternating Subarrays/README.md‎

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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3196.Maximize%20Total%20Cost%20of%20Alternating%20Subarrays/README.md
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tags:
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‎solution/3100-3199/3196.Maximize Total Cost of Alternating Subarrays/README_EN.md‎

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tags:
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‎solution/3100-3199/3197.Find the Minimum Area to Cover All Ones II/README.md‎

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tags:
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- 数组
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- 枚举
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‎solution/3100-3199/3197.Find the Minimum Area to Cover All Ones II/README_EN.md‎

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difficulty: 简单
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3199.Count%20Triplets%20with%20Even%20XOR%20Set%20Bits%20I/README.md
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---
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<!-- problem:start -->
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# [3199. Count Triplets with Even XOR Set Bits I 🔒](https://leetcode.cn/problems/count-triplets-with-even-xor-set-bits-i)
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[English Version](/solution/3100-3199/3199.Count%20Triplets%20with%20Even%20XOR%20Set%20Bits%20I/README_EN.md)
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## 题目描述
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<!-- description:start -->
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Given three integer arrays <code>a</code>, <code>b</code>, and <code>c</code>, return the number of triplets <code>(a[i], b[j], c[k])</code>, such that the bitwise <code>XOR</code> of the elements of each triplet has an <strong>even</strong> number of <span data-keyword="set-bit">set bits</span>.
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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">a = [1], b = [2], c = [3]</span></p>
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<p><strong>Output:</strong> <span class="example-io">1</span></p>
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<p><strong>Explanation:</strong></p>
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<p>The only triplet is <code>(a[0], b[0], c[0])</code> and their <code>XOR</code> is: <code>1 XOR 2 XOR 3 = 00<sub>2</sub></code>.</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">a = [1,1], b = [2,3], c = [1,5]</span></p>
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<p><strong>Output:</strong> <span class="example-io">4</span></p>
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<p><strong>Explanation:</strong></p>
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<p>Consider these four triplets:</p>
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<ul>
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<li><code>(a[0], b[1], c[0])</code>: <code>1 XOR 3 XOR 1 = 011<sub>2</sub></code></li>
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<li><code>(a[1], b[1], c[0])</code>: <code>1 XOR 3 XOR 1 = 011<sub>2</sub></code></li>
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<li><code>(a[0], b[0], c[1])</code>: <code>1 XOR 2 XOR 5 = 110<sub>2</sub></code></li>
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<li><code>(a[1], b[0], c[1])</code>: <code>1 XOR 2 XOR 5 = 110<sub>2</sub></code></li>
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</ul>
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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>1 &lt;= a.length, b.length, c.length &lt;= 100</code></li>
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<li><code>0 &lt;= a[i], b[i], c[i] &lt;= 100</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:位运算
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对于两个整数,异或结果中 1ドル$ 的个数的奇偶性,取决于两个整数的二进制表示中 1ドル$ 的个数的奇偶性。
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我们可以用三个数组 `cnt1`、`cnt2`、`cnt3` 分别记录数组 `a`、`b`、`c` 中每个数的二进制表示中 1ドル$ 的个数的奇偶性。
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然后我们在 $[0, 1]$ 的范围内枚举三个数组中的每个数的二进制表示中 1ドル$ 的个数的奇偶性,如果三个数的二进制表示中 1ドル$ 的个数的奇偶性之和为偶数,那么这三个数的异或结果中 1ドル$ 的个数也为偶数,此时我们将这三个数的组合数相乘累加到答案中。
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最后返回答案即可。
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时间复杂度 $O(n),ドル其中 $n$ 为数组 `a`、`b`、`c` 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 tripletCount(self, a: List[int], b: List[int], c: List[int]) -> int:
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cnt1 = Counter(x.bit_count() & 1 for x in a)
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cnt2 = Counter(x.bit_count() & 1 for x in b)
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cnt3 = Counter(x.bit_count() & 1 for x in c)
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ans = 0
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for i in range(2):
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for j in range(2):
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for k in range(2):
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if (i + j + k) & 1 ^ 1:
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ans += cnt1[i] * cnt2[j] * cnt3[k]
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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 tripletCount(int[] a, int[] b, int[] c) {
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int[] cnt1 = new int[2];
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int[] cnt2 = new int[2];
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int[] cnt3 = new int[2];
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for (int x : a) {
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++cnt1[Integer.bitCount(x) & 1];
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}
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for (int x : b) {
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++cnt2[Integer.bitCount(x) & 1];
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}
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for (int x : c) {
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++cnt3[Integer.bitCount(x) & 1];
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}
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int ans = 0;
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for (int i = 0; i < 2; ++i) {
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for (int j = 0; j < 2; ++j) {
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for (int k = 0; k < 2; ++k) {
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if ((i + j + k) % 2 == 0) {
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ans += cnt1[i] * cnt2[j] * cnt3[k];
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}
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}
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}
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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 tripletCount(vector<int>& a, vector<int>& b, vector<int>& c) {
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int cnt1[2]{};
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int cnt2[2]{};
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int cnt3[2]{};
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for (int x : a) {
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++cnt1[__builtin_popcount(x) & 1];
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}
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for (int x : b) {
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++cnt2[__builtin_popcount(x) & 1];
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}
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for (int x : c) {
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++cnt3[__builtin_popcount(x) & 1];
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}
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int ans = 0;
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for (int i = 0; i < 2; ++i) {
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for (int j = 0; j < 2; ++j) {
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for (int k = 0; k < 2; ++k) {
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if ((i + j + k) % 2 == 0) {
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ans += cnt1[i] * cnt2[j] * cnt3[k];
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}
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}
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}
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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 tripletCount(a []int, b []int, c []int) (ans int) {
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cnt1 := [2]int{}
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cnt2 := [2]int{}
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cnt3 := [2]int{}
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for _, x := range a {
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cnt1[bits.OnesCount(uint(x))%2]++
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}
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for _, x := range b {
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cnt2[bits.OnesCount(uint(x))%2]++
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}
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for _, x := range c {
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cnt3[bits.OnesCount(uint(x))%2]++
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}
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for i := 0; i < 2; i++ {
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for j := 0; j < 2; j++ {
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for k := 0; k < 2; k++ {
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if (i+j+k)%2 == 0 {
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ans += cnt1[i] * cnt2[j] * cnt3[k]
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}
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}
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}
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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 tripletCount(a: number[], b: number[], c: number[]): number {
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const cnt1: [number, number] = [0, 0];
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const cnt2: [number, number] = [0, 0];
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const cnt3: [number, number] = [0, 0];
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for (const x of a) {
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++cnt1[bitCount(x) & 1];
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}
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for (const x of b) {
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++cnt2[bitCount(x) & 1];
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}
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for (const x of c) {
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++cnt3[bitCount(x) & 1];
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}
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let ans = 0;
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for (let i = 0; i < 2; ++i) {
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for (let j = 0; j < 2; ++j) {
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for (let k = 0; k < 2; ++k) {
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if ((i + j + k) % 2 === 0) {
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ans += cnt1[i] * cnt2[j] * cnt3[k];
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}
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}
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}
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}
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return ans;
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}
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function bitCount(i: number): number {
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i = i - ((i >>> 1) & 0x55555555);
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i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
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i = (i + (i >>> 4)) & 0x0f0f0f0f;
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i = i + (i >>> 8);
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i = i + (i >>> 16);
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return i & 0x3f;
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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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