maximum-sum-of-two-non-overlapping-subarrays

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Algorithms

Given an integer array <code>nums</code> and two integers <code>firstLen</code> and <code>secondLen</code>, return the maximum sum of elements in two non-overlapping subarrays with lengths <code>firstLen</code> and <code>secondLen</code>.

The array with length <code>firstLen</code> could occur before or after the array with length <code>secondLen</code>, but they have to be non-overlapping.

A subarray is a contiguous part of an array.

&nbsp;
Example 1:

<pre>
Input: nums = [0,6,5,2,2,5,1,9,4], firstLen = 1, secondLen = 2
Output: 20
Explanation: One choice of subarrays is [9] with length 1, and [6,5] with length 2.
</pre>

Example 2:

<pre>
Input: nums = [3,8,1,3,2,1,8,9,0], firstLen = 3, secondLen = 2
Output: 29
Explanation: One choice of subarrays is [3,8,1] with length 3, and [8,9] with length 2.
</pre>

Example 3:

<pre>
Input: nums = [2,1,5,6,0,9,5,0,3,8], firstLen = 4, secondLen = 3
Output: 31
Explanation: One choice of subarrays is [5,6,0,9] with length 4, and [0,3,8] with length 3.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= firstLen, secondLen &lt;= 1000</code></li>
	<li><code>2 &lt;= firstLen + secondLen &lt;= 1000</code></li>
	<li><code>firstLen + secondLen &lt;= nums.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>
</ul>

maxSumTwoNoOverlap

Maximum Sum of Two Non-Overlapping Subarrays

Array

Dynamic Programming

Sliding Window

给你一个整数数组 <code>nums</code> 和两个整数 <code>firstLen</code> 和 <code>secondLen</code>，请你找出并返回两个非重叠 子数组 中元素的最大和，长度分别为 <code>firstLen</code> 和 <code>secondLen</code> 。

长度为 <code>firstLen</code> 的子数组可以出现在长为 <code>secondLen</code> 的子数组之前或之后，但二者必须是不重叠的。

子数组是数组的一个 连续 部分。

&nbsp;

示例 1：

<pre>
输入：nums = [0,6,5,2,2,5,1,9,4], firstLen = 1, secondLen = 2
输出：20
解释：子数组的一种选择中，[9] 长度为 1，[6,5] 长度为 2。
</pre>

示例 2：

<pre>
输入：nums = [3,8,1,3,2,1,8,9,0], firstLen = 3, secondLen = 2
输出：29
解释：子数组的一种选择中，[3,8,1] 长度为 3，[8,9] 长度为 2。
</pre>

示例 3：

<pre>
输入：nums = [2,1,5,6,0,9,5,0,3,8], firstLen = 4, secondLen = 3
输出：31
解释：子数组的一种选择中，[5,6,0,9] 长度为 4，[0,3,8] 长度为 3。
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= firstLen, secondLen &lt;= 1000</code></li>
	<li><code>2 &lt;= firstLen + secondLen &lt;= 1000</code></li>
	<li><code>firstLen + secondLen &lt;= nums.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>
</ul>