leetcode 1589 - Maximum Sum Obtained of Any Permutation

https://leetcode.com/problems/maximum-sum-obtained-of-any-permutation/

We have an array of integers, nums, and an array of requests where requests[i] = [starti, endi]. The ith request asks for the sum of nums[starti] + nums[starti + 1] + … + nums[endi - 1] + nums[endi]. Both starti and endi are 0-indexed.

Return the maximum total sum of all requests among all permutations of nums.

Since the answer may be too large, return it modulo 109 + 7.

Example 1:

Input: nums = [1,2,3,4,5], requests = [[1,3],[0,1]]
Output: 19
Explanation: One permutation of nums is [2,1,3,4,5] with the following result:
requests[0] -> nums[1] + nums[2] + nums[3] = 1 + 3 + 4 = 8
requests[1] -> nums[0] + nums[1] = 2 + 1 = 3
Total sum: 8 + 3 = 11.
A permutation with a higher total sum is [3,5,4,2,1] with the following result:
requests[0] -> nums[1] + nums[2] + nums[3] = 5 + 4 + 2 = 11
requests[1] -> nums[0] + nums[1] = 3 + 5 = 8
Total sum: 11 + 8 = 19, which is the best that you can do.
Example 2:

Input: nums = [1,2,3,4,5,6], requests = [[0,1]]
Output: 11
Explanation: A permutation with the max total sum is [6,5,4,3,2,1] with request sums [11].
Example 3:

Input: nums = [1,2,3,4,5,10], requests = [[0,2],[1,3],[1,1]]
Output: 47
Explanation: A permutation with the max total sum is [4,10,5,3,2,1] with request sums [19,18,10].

Constraints:

n == nums.length
1 <= n <= 105
0 <= nums[i] <= 105
1 <= requests.length <= 105
requests[i].length == 2
0 <= starti <= endi < n

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class Solution {
public:
int maxSumRangeQuery(vector<int>& nums, vector<vector<int>>& requests) {
int size = nums.size();

vector<int> count(size, 0);

for (auto &v : requests) {
count[v[0]]++;
if (v[1] + 1 < size)
count[v[1] + 1]--;
}

for (int i = 1; i < size; i++)
count[i] += count[i-1];

sort(nums.begin(), nums.end(), greater<int>());
sort(count.begin(), count.end(), greater<int>());

long sum = 0;
int mod = 1e9 + 7;

for (int i = 0; i < size; i++)
sum += (long)nums[i] * count[i];

return sum % mod;
}
};