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feat: add weekly contest 439 and biweekly contest 151 (#4121)
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‎solution/3400-3499/3467.Transform Array by Parity/README.md

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---
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comments: true
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difficulty: 简单
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3467.Transform%20Array%20by%20Parity/README.md
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---
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<!-- problem:start -->
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# [3467. 将数组按照奇偶性转化](https://leetcode.cn/problems/transform-array-by-parity)
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[English Version](/solution/3400-3499/3467.Transform%20Array%20by%20Parity/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>给你一个整数数组 <code>nums</code>。请你按照以下顺序 <strong>依次</strong>&nbsp;执行操作，转换 <code>nums</code>：</p>
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<ol>
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<li>将每个偶数替换为 0。</li>
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<li>将每个奇数替换为 1。</li>
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<li>按&nbsp;<strong>非递减&nbsp;</strong>顺序排序修改后的数组。</li>
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</ol>
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<p>执行完这些操作后，返回结果数组。</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">nums = [4,3,2,1]</span></p>
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<p><strong>输出：</strong><span class="example-io">[0,0,1,1]</span></p>
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<p><strong>解释：</strong></p>
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<ul>
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<li>将偶数（4 和 2）替换为 0，将奇数（3 和 1）替换为 1。现在，<code>nums = [0, 1, 0, 1]</code>。</li>
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<li>按非递减顺序排序 <code>nums</code>，得到 <code>nums = [0, 0, 1, 1]</code>。</li>
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</ul>
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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><strong>输入：</strong><span class="example-io">nums = [1,5,1,4,2]</span></p>
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<p><strong>输出：</strong><span class="example-io">[0,0,1,1,1]</span></p>
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<p><strong>解释：</strong></p>
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<ul>
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<li>将偶数（4 和 2）替换为 0，将奇数（1, 5 和 1）替换为 1。现在，<code>nums = [1, 1, 1, 0, 0]</code>。</li>
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<li>按非递减顺序排序&nbsp;<code>nums</code>，得到 <code>nums = [0, 0, 1, 1, 1]</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>提示：</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 100</code></li>
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<li><code>1 &lt;= nums[i] &lt;= 1000</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：计数
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我们可以遍历数组 $\textit{nums}$，统计其中偶数的个数 $\textit{even}$。然后我们将数组的前 $\textit{even}$ 个元素置为 $0$，剩余的元素置为 $1$。
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时间复杂度 $O(n)$，其中 $n$ 是数组 $\textit{nums}$ 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 transformArray(self, nums: List[int]) -> List[int]:
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even = sum(x % 2 == 0 for x in nums)
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for i in range(even):
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nums[i] = 0
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for i in range(even, len(nums)):
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nums[i] = 1
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return nums
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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[] transformArray(int[] nums) {
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int even = 0;
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for (int x : nums) {
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even += (x & 1 ^ 1);
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}
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for (int i = 0; i < even; ++i) {
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nums[i] = 0;
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}
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for (int i = even; i < nums.length; ++i) {
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nums[i] = 1;
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}
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return nums;
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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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vector<int> transformArray(vector<int>& nums) {
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int even = 0;
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for (int x : nums) {
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even += (x & 1 ^ 1);
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}
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for (int i = 0; i < even; ++i) {
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nums[i] = 0;
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}
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for (int i = even; i < nums.size(); ++i) {
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nums[i] = 1;
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}
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return nums;
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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 transformArray(nums []int) []int {
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even := 0
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for _, x := range nums {
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even += x&1 ^ 1
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}
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for i := 0; i < even; i++ {
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nums[i] = 0
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}
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for i := even; i < len(nums); i++ {
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nums[i] = 1
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}
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return nums
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}
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```
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#### TypeScript
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```ts
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function transformArray(nums: number[]): number[] {
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const even = nums.filter(x => x % 2 === 0).length;
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for (let i = 0; i < even; ++i) {
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nums[i] = 0;
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}
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for (let i = even; i < nums.length; ++i) {
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nums[i] = 1;
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}
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return nums;
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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 -->

‎solution/3400-3499/3467.Transform Array by Parity/README_EN.md

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---
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comments: true
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difficulty: Easy
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3467.Transform%20Array%20by%20Parity/README_EN.md
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---
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<!-- problem:start -->
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# [3467. Transform Array by Parity](https://leetcode.com/problems/transform-array-by-parity)
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[中文文档](/solution/3400-3499/3467.Transform%20Array%20by%20Parity/README.md)
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## Description
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<!-- description:start -->
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<p>You are given an integer array <code>nums</code>. Transform <code>nums</code> by performing the following operations in the <strong>exact</strong> order specified:</p>
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<ol>
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<li>Replace each even number with 0.</li>
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<li>Replace each odd numbers with 1.</li>
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<li>Sort the modified array in <strong>non-decreasing</strong> order.</li>
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</ol>
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<p>Return the resulting array after performing these operations.</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">nums = [4,3,2,1]</span></p>
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<p><strong>Output:</strong> <span class="example-io">[0,0,1,1]</span></p>
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<p><strong>Explanation:</strong></p>
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<ul>
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<li>Replace the even numbers (4 and 2) with 0 and the odd numbers (3 and 1) with 1. Now, <code>nums = [0, 1, 0, 1]</code>.</li>
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<li>After sorting <code>nums</code> in non-descending order, <code>nums = [0, 0, 1, 1]</code>.</li>
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</ul>
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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">nums = [1,5,1,4,2]</span></p>
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<p><strong>Output:</strong> <span class="example-io">[0,0,1,1,1]</span></p>
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<p><strong>Explanation:</strong></p>
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<ul>
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<li>Replace the even numbers (4 and 2) with 0 and the odd numbers (1, 5 and 1) with 1. Now, <code>nums = [1, 1, 1, 0, 0]</code>.</li>
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<li>After sorting <code>nums</code> in non-descending order, <code>nums = [0, 0, 1, 1, 1]</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;= nums.length &lt;= 100</code></li>
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<li><code>1 &lt;= nums[i] &lt;= 1000</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: Counting
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We can traverse the array $\textit{nums}$ and count the number of even elements $\textit{even}$. Then, we set the first $\textit{even}$ elements of the array to $0$ and the remaining elements to $1$.
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The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. 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 transformArray(self, nums: List[int]) -> List[int]:
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even = sum(x % 2 == 0 for x in nums)
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for i in range(even):
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nums[i] = 0
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for i in range(even, len(nums)):
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nums[i] = 1
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return nums
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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[] transformArray(int[] nums) {
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int even = 0;
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for (int x : nums) {
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even += (x & 1 ^ 1);
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}
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for (int i = 0; i < even; ++i) {
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nums[i] = 0;
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}
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for (int i = even; i < nums.length; ++i) {
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nums[i] = 1;
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}
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return nums;
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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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vector<int> transformArray(vector<int>& nums) {
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int even = 0;
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for (int x : nums) {
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even += (x & 1 ^ 1);
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}
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for (int i = 0; i < even; ++i) {
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nums[i] = 0;
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}
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for (int i = even; i < nums.size(); ++i) {
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nums[i] = 1;
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}
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return nums;
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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 transformArray(nums []int) []int {
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even := 0
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for _, x := range nums {
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even += x&1 ^ 1
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}
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for i := 0; i < even; i++ {
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nums[i] = 0
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}
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for i := even; i < len(nums); i++ {
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nums[i] = 1
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}
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return nums
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}
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```
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#### TypeScript
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```ts
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function transformArray(nums: number[]): number[] {
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const even = nums.filter(x => x % 2 === 0).length;
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for (let i = 0; i < even; ++i) {
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nums[i] = 0;
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}
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for (let i = even; i < nums.length; ++i) {
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nums[i] = 1;
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}
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return nums;
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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 -->

‎solution/3400-3499/3467.Transform Array by Parity/Solution.cpp

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class Solution {
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public:
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vector<int> transformArray(vector<int>& nums) {
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int even = 0;
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for (int x : nums) {
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even += (x & 1 ^ 1);
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}
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for (int i = 0; i < even; ++i) {
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nums[i] = 0;
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}
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for (int i = even; i < nums.size(); ++i) {
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nums[i] = 1;
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}
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return nums;
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}
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};

‎solution/3400-3499/3467.Transform Array by Parity/Solution.go

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func transformArray(nums []int) []int {
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even := 0
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for _, x := range nums {
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even += x&1 ^ 1
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}
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for i := 0; i < even; i++ {
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nums[i] = 0
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}
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for i := even; i < len(nums); i++ {
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nums[i] = 1
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}
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return nums
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}

‎solution/3400-3499/3467.Transform Array by Parity/Solution.java

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class Solution {
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public int[] transformArray(int[] nums) {
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int even = 0;
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for (int x : nums) {
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even += (x & 1 ^ 1);
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}
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for (int i = 0; i < even; ++i) {
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nums[i] = 0;
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}
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for (int i = even; i < nums.length; ++i) {
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nums[i] = 1;
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}
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return nums;
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}
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}

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