1589. Maximum Sum Obtained of Any Permutation

Description

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

 

Solutions

Solution: Greedy

• Time complexity: O(nlogn)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} nums
 * @param {number[][]} requests
 * @return {number}
 */
const maxSumRangeQuery = function (nums, requests) {
  const MODULO = 10 ** 9 + 7;
  const size = nums.length;
  const prefixRequest = new Array(size).fill(0);

  requests.forEach(([start, end]) => {
    prefixRequest[start] += 1;
    end + 1 < size && (prefixRequest[end + 1] -= 1);
  });

  for (let index = 1; index < size; index++) prefixRequest[index] += prefixRequest[index - 1];

  nums.sort((a, b) => a - b);
  prefixRequest.sort((a, b) => a - b);

  return prefixRequest.reduce((result, count, index) => {
    return (result + count * nums[index]) % MODULO;
  }, 0);
};