range-sum-of-sorted-subarray-sums

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Algorithms

You are given the array <code>nums</code> consisting of <code>n</code> positive integers. You computed the sum of all non-empty continuous subarrays from the array and then sorted them in non-decreasing order, creating a new array of <code>n * (n + 1) / 2</code> numbers.

Return the sum of the numbers from index <code>left</code> to index <code>right</code> (indexed from 1), inclusive, in the new array. Since the answer can be a huge number return it modulo <code>109 + 7</code>.

&nbsp;
Example 1:

<pre>
Input: nums = [1,2,3,4], n = 4, left = 1, right = 5
Output: 13 
Explanation: All subarray sums are 1, 3, 6, 10, 2, 5, 9, 3, 7, 4. After sorting them in non-decreasing order we have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 1 to ri = 5 is 1 + 2 + 3 + 3 + 4 = 13. 
</pre>

Example 2:

<pre>
Input: nums = [1,2,3,4], n = 4, left = 3, right = 4
Output: 6
Explanation: The given array is the same as example 1. We have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 3 to ri = 4 is 3 + 3 = 6.
</pre>

Example 3:

<pre>
Input: nums = [1,2,3,4], n = 4, left = 1, right = 10
Output: 50
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
	<li><code>1 &lt;= left &lt;= right &lt;= n * (n + 1) / 2</code></li>
</ul>

rangeSum

Range Sum of Sorted Subarray Sums

Array

Two Pointers

Binary Search

Prefix Sum

Sorting

给你一个数组&nbsp;<code>nums</code>&nbsp;，它包含&nbsp;<code>n</code>&nbsp;个正整数。你需要计算所有非空连续子数组的和，并将它们按升序排序，得到一个新的包含&nbsp;<code>n * (n + 1) / 2</code>&nbsp;个数字的数组。

请你返回在新数组中下标为&nbsp;<code>left</code>&nbsp;到&nbsp;<code>right</code> （下标从 1 开始）的所有数字和（包括左右端点）。由于答案可能很大，请你将它对 10^9 + 7 取模后返回。

&nbsp;

示例 1：

<pre>
输入：nums = [1,2,3,4], n = 4, left = 1, right = 5
输出：13 
解释：所有的子数组和为 1, 3, 6, 10, 2, 5, 9, 3, 7, 4 。将它们升序排序后，我们得到新的数组 [1, 2, 3, 3, 4, 5, 6, 7, 9, 10] 。下标从 le = 1 到 ri = 5 的和为 1 + 2 + 3 + 3 + 4 = 13 。
</pre>

示例 2：

<pre>
输入：nums = [1,2,3,4], n = 4, left = 3, right = 4
输出：6
解释：给定数组与示例 1 一样，所以新数组为 [1, 2, 3, 3, 4, 5, 6, 7, 9, 10] 。下标从 le = 3 到 ri = 4 的和为 3 + 3 = 6 。
</pre>

示例 3：

<pre>
输入：nums = [1,2,3,4], n = 4, left = 1, right = 10
输出：50
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10^3</code></li>
	<li><code>nums.length == n</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
	<li><code>1 &lt;= left &lt;= right&nbsp;&lt;= n * (n + 1) / 2</code></li>
</ul>