find-all-good-indices

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Algorithms

You are given a 0-indexed integer array <code>nums</code> of size <code>n</code> and a positive integer <code>k</code>.

We call an index <code>i</code> in the range <code>k &lt;= i &lt; n - k</code> good if the following conditions are satisfied:

<ul>
	<li>The <code>k</code> elements that are just before the index <code>i</code> are in non-increasing order.</li>
	<li>The <code>k</code> elements that are just after the index <code>i</code> are in non-decreasing order.</li>
</ul>

Return an array of all good indices sorted in increasing order.

&nbsp;
Example 1:

<pre>
Input: nums = [2,1,1,1,3,4,1], k = 2
Output: [2,3]
Explanation: There are two good indices in the array:
- Index 2. The subarray [2,1] is in non-increasing order, and the subarray [1,3] is in non-decreasing order.
- Index 3. The subarray [1,1] is in non-increasing order, and the subarray [3,4] is in non-decreasing order.
Note that the index 4 is not good because [4,1] is not non-decreasing.</pre>

Example 2:

<pre>
Input: nums = [2,1,1,2], k = 2
Output: []
Explanation: There are no good indices in this array.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>3 &lt;= n &lt;= 105</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 106</code></li>
	<li><code>1 &lt;= k &lt;= n / 2</code></li>
</ul>

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Dynamic Programming

Prefix Sum

给你一个大小为 <code>n</code>&nbsp;下标从 0&nbsp;开始的整数数组&nbsp;<code>nums</code>&nbsp;和一个正整数&nbsp;<code>k</code>&nbsp;。

对于&nbsp;<code>k &lt;= i &lt; n - k</code>&nbsp;之间的一个下标&nbsp;<code>i</code>&nbsp;，如果它满足以下条件，我们就称它为一个&nbsp;好&nbsp;下标：

<ul>
	<li>下标 <code>i</code> 之前 的 <code>k</code>&nbsp;个元素是 非递增的&nbsp;。</li>
	<li>下标 <code>i</code> 之后&nbsp;的 <code>k</code>&nbsp;个元素是 非递减的&nbsp;。</li>
</ul>

按 升序&nbsp;返回所有好下标。

&nbsp;

示例 1：

<pre>
输入：nums = [2,1,1,1,3,4,1], k = 2
输出：[2,3]
解释：数组中有两个好下标：
- 下标 2 。子数组 [2,1] 是非递增的，子数组 [1,3] 是非递减的。
- 下标 3 。子数组 [1,1] 是非递增的，子数组 [3,4] 是非递减的。
注意，下标 4 不是好下标，因为 [4,1] 不是非递减的。</pre>

示例 2：

<pre>
输入：nums = [2,1,1,2], k = 2
输出：[]
解释：数组中没有好下标。
</pre>

&nbsp;

提示：

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>3 &lt;= n &lt;= 105</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 106</code></li>
	<li><code>1 &lt;= k &lt;= n / 2</code></li>
</ul>