transform-array-by-parity

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Algorithms

You are given an integer array <code>nums</code>. Transform <code>nums</code> by performing the following operations in the exact order specified:

<ol>
	<li>Replace each even number with 0.</li>
	<li>Replace each odd numbers with 1.</li>
	<li>Sort the modified array in non-decreasing order.</li>
</ol>

Return the resulting array after performing these operations.

&nbsp;
Example 1:

<div class="example-block">
Input: nums = [4,3,2,1]

Output: [0,0,1,1]

Explanation:

<ul>
	<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>
	<li>After sorting <code>nums</code> in non-descending order, <code>nums = [0, 0, 1, 1]</code>.</li>
</ul>
</div>

Example 2:

<div class="example-block">
Input: nums = [1,5,1,4,2]

Output: [0,0,1,1,1]

Explanation:

<ul>
	<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>
	<li>After sorting <code>nums</code> in non-descending order, <code>nums = [0, 0, 1, 1, 1]</code>.</li>
</ul>
</div>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 1000</code></li>
</ul>

第 151 场双周赛

biweekly-contest-151

transformArray

中级工程师

Transform Array by Parity

Odd Even Linked List

odd-even-linked-list

奇偶链表

Array

Counting

Sorting

给你一个整数数组 <code>nums</code>。请你按照以下顺序 依次&nbsp;执行操作，转换 <code>nums</code>：

<ol>
	<li>将每个偶数替换为 0。</li>
	<li>将每个奇数替换为 1。</li>
	<li>按&nbsp;非递减&nbsp;顺序排序修改后的数组。</li>
</ol>

执行完这些操作后，返回结果数组。

&nbsp;

示例 1:

<div class="example-block">
输入：nums = [4,3,2,1]

输出：[0,0,1,1]

解释：

<ul>
	<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>
	<li>按非递减顺序排序 <code>nums</code>，得到 <code>nums = [0, 0, 1, 1]</code>。</li>
</ul>
</div>

示例 2:

<div class="example-block">
输入：nums = [1,5,1,4,2]

输出：[0,0,1,1,1]

解释：

<ul>
	<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>
	<li>按非递减顺序排序&nbsp;<code>nums</code>，得到 <code>nums = [0, 0, 1, 1, 1]</code>。</li>
</ul>
</div>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 1000</code></li>
</ul>