Given the array nums consisting of n positive integers. You computed the sum of all non-empty continous subarrays from the array and then sort them in non-decreasing order, creating a new array of n * (n + 1) / 2 numbers.
Return the sum of the numbers from index left to index right (indexed from 1), inclusive, in the new array. Since the answer can be a huge number return it modulo 10^9 + 7.
Example 1:
Input: nums = [1,2,3,4], n = 4, left = 1, right = 5
Output: 13
Explanation: All subarray sums are 1, 3, 6, 10, 2, 5, 9, 3, 7, 4. After sorting them in non-decreasing order we have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 1 to ri = 5 is 1 + 2 + 3 + 3 + 4 = 13.
Example 2:
Input: nums = [1,2,3,4], n = 4, left = 3, right = 4
Output: 6
Explanation: The given array is the same as example 1. We have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 3 to ri = 4 is 3 + 3 = 6.
Example 3:
Input: nums = [1,2,3,4], n = 4, left = 1, right = 10
Output: 50
Constraints:
1 <= nums.length <= 10^3
nums.length == n
1 <= nums[i] <= 100
1 <= left <= right <= n * (n + 1) / 2
class Solution {
static class Pair {
long sum;
int index;
Pair(long sum, int index) {
this.sum = sum;
this.index = index;
}
}
// The benefit of using PQ is, that instead of using N*(N+1)/2 space
// We will only use O(N) space
public int rangeSum(int[] nums, int n, int left, int right) {
PriorityQueue<Pair> PQ = new PriorityQueue<>((a, b) -> (int) a.sum - (int) b.sum);
for (int i = 0; i < n; i++)
PQ.add(new Pair(nums[i], i + 1));
long ans = 0, MOD = 1_000_000_007;
for (int i = 1; i <= right; i++) {
Pair smallSum = PQ.poll();
if (i >= left)
ans = (ans + smallSum.sum) % MOD;
if (smallSum.index < n) {
smallSum.sum = (smallSum.sum + nums[smallSum.index++]) % MOD;
PQ.add(smallSum);
}
}
return (int) ans;
}
}