Skip to content

1695. Maximum Sum Obtained Of Any Permutation

Difficulty: Medium

LeetCode Problem View on GitHub

1695. Maximum Sum Obtained of Any Permutation

Medium

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

Solution

class Solution {
    private final int mod = (int)(1e9 + 7);
    private int freq[];
    static class Pair {
        int node, idx;
        public Pair(int node, int idx) {
            this.node = node;
            this.idx = idx;
        }
        @Override
        public String toString() {
            return "(" + node + " " + idx + ")";
        }
    }
    static class customSort implements Comparator<Pair> {
        @Override
        public int compare(Pair first, Pair second) {
            return Integer.compare(second.node, first.node);
        }
    }
    public int maxSumRangeQuery(int[] nums, int[][] requests) {
        int n = nums.length;
        freq = new int[n];

        for (int query[] : requests) {
            int l = query[0], r = query[1];
            freq[l]++; if (r + 1 < n) freq[r + 1]--;
        }

        for (int i = 1; i < n; i++) 
            freq[i] += freq[i - 1];

        ArrayList<Pair> maxFreq = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            maxFreq.add(new Pair(freq[i], i));
        }
        Collections.sort(maxFreq, new customSort()); 

        ArrayList<Integer> values = new ArrayList<>();
        for (int ele : nums)
            values.add(ele);
        Collections.sort(values);
        int idx = values.size() - 1;

        int arr[] = new int[n];
        for (int i = 0; i < n; i++)
            arr[maxFreq.get(i).idx] = values.get(idx--);

        int pref[] = new int[n];
        pref[0] = arr[0];
        for (int i = 1; i < n; i++) 
            pref[i] += pref[i - 1] + arr[i];

        long res = 0;
        for (int query[] : requests) {
            int l = query[0], r = query[1];
            long total = pref[r] * 1L;
            if (l - 1 >= 0) {
                total -= pref[l - 1] * 1L;
            }
            res = (res + total) % mod;
        }
        return (int)(res);    
    }
}

Complexity Analysis

  • Time Complexity: O(?)
  • Space Complexity: O(?)

Approach

Detailed explanation of the approach will be added here