1508. Range Sum of Sorted Subarray Sums
/*
* @lc app=leetcode id=1508 lang=cpp
*
* [1508] Range Sum of Sorted Subarray Sums
*
* https://leetcode.com/problems/range-sum-of-sorted-subarray-sums/description/
*
* algorithms
* Medium (63.14%)
* Likes: 1557
* Dislikes: 262
* Total Accepted: 179.2K
* Total Submissions: 284K
* Testcase Example: '[1,2,3,4]\n4\n1\n5'
*
* You are given the array nums consisting of n 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 n * (n + 1) / 2
* numbers.
*
* Return the sum of the numbers from index left to index right (indexed from
* 1), inclusive, in the new array. Since the answer can be a huge number
* return it modulo 10^9 + 7.
*
*
* Example 1:
*
*
* 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.
*
*
* Example 2:
*
*
* 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.
*
*
* Example 3:
*
*
* Input: nums = [1,2,3,4], n = 4, left = 1, right = 10
* Output: 50
*
*
*
* Constraints:
*
*
* n == nums.length
* 1 <= nums.length <= 1000
* 1 <= nums[i] <= 100
* 1 <= left <= right <= n * (n + 1) / 2
*
*
*/
// @lc code=start
class Solution {
public:
int rangeSum(vector<int>& nums, int n, int left, int right) {
vector<int> rangeSum;
for (int i = 0; i < nums.size(); i++) {
int sum = 0;
for (int j = i; j < nums.size(); j++) {
sum += nums[j];
rangeSum.push_back(sum);
}
}
sort(rangeSum.begin(), rangeSum.end());
int res = 0;
const int MOD = 1e9 + 7;
for (int i = left - 1; i < right; i++) {
res = (res + rangeSum[i]) % MOD;
}
return res;
}
};
// @lc code=end
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