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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

  • class Solution {
    public int maxSumRangeQuery(int[] nums, int[][] requests) {
        int n = nums.length;
        int[] d = new int[n];
        for (var req : requests) {
            int l = req[0], r = req[1];
            d[l]++;
            if (r + 1 < n) {
                d[r + 1]--;
            }
        }
        for (int i = 1; i < n; ++i) {
            d[i] += d[i - 1];
        }
        Arrays.sort(nums);
        Arrays.sort(d);
        final int mod = (int) 1e9 + 7;
        long ans = 0;
        for (int i = 0; i < n; ++i) {
            ans = (ans + 1L * nums[i] * d[i]) % mod;
        }
        return (int) ans;
    }
}

  • class Solution {
public:
    int maxSumRangeQuery(vector<int>& nums, vector<vector<int>>& requests) {
        int n = nums.size();
        int d[n];
        memset(d, 0, sizeof(d));
        for (auto& req : requests) {
            int l = req[0], r = req[1];
            d[l]++;
            if (r + 1 < n) {
                d[r + 1]--;
            }
        }
        for (int i = 1; i < n; ++i) {
            d[i] += d[i - 1];
        }
        sort(nums.begin(), nums.end());
        sort(d, d + n);
        long long ans = 0;
        const int mod = 1e9 + 7;
        for (int i = 0; i < n; ++i) {
            ans = (ans + 1LL * nums[i] * d[i]) % mod;
        }
        return ans;
    }
};

  • class Solution:
    def maxSumRangeQuery(self, nums: List[int], requests: List[List[int]]) -> int:
        n = len(nums)
        d = [0] * n
        for l, r in requests:
            d[l] += 1
            if r + 1 < n:
                d[r + 1] -= 1
        for i in range(1, n):
            d[i] += d[i - 1]
        nums.sort()
        d.sort()
        mod = 10**9 + 7
        return sum(a * b for a, b in zip(nums, d)) % mod


  • func maxSumRangeQuery(nums []int, requests [][]int) (ans int) {
	n := len(nums)
	d := make([]int, n)
	for _, req := range requests {
		l, r := req[0], req[1]
		d[l]++
		if r+1 < n {
			d[r+1]--
		}
	}
	for i := 1; i < n; i++ {
		d[i] += d[i-1]
	}
	sort.Ints(nums)
	sort.Ints(d)
	const mod = 1e9 + 7
	for i, a := range nums {
		b := d[i]
		ans = (ans + a*b) % mod
	}
	return

  • function maxSumRangeQuery(nums: number[], requests: number[][]): number {
    const n = nums.length;
    const d = new Array(n).fill(0);
    for (const [l, r] of requests) {
        d[l]++;
        if (r + 1 < n) {
            d[r + 1]--;
        }
    }
    for (let i = 1; i < n; ++i) {
        d[i] += d[i - 1];
    }
    nums.sort((a, b) => a - b);
    d.sort((a, b) => a - b);
    let ans = 0;
    const mod = 10 ** 9 + 7;
    for (let i = 0; i < n; ++i) {
        ans = (ans + nums[i] * d[i]) % mod;
    }
    return ans;
}

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