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:
For each request [i,j], we set count[i]++ and count[j + 1]--,
Then we sweep once the whole count, we can find the frequency for count[i].
class Solution {
int mod = (int) 1e9 + 7;
public int maxSumRangeQuery(int[] nums, int[][] requests) {
int n = nums.length;
int[] freq = new int[n];
for (int[] req : requests) {
freq[req[0]] ++;
if (req[1] + 1 < n) {
freq[req[1] + 1] --;
}
}
for (int i = 1; i < n; i ++) {
freq[i] += freq[i - 1];
}
Arrays.sort(nums);
Arrays.sort(freq);
int max = 0;
for (int i = 0; i < nums.length; i ++) {
max = (int) (((long) nums[i] * freq[i] + max) % mod);
}
return max;
}
}