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
class Solution:
    def maxSumRangeQuery(self, nums: List[int], requests: List[List[int]]) -> int:
        requestCount = defaultdict(int) #Number of request per index
        
        #Create a min-heap of the start and end intervals.
        begHeap = []
        endHeap = []
        for req in requests:
            heapq.heappush(begHeap, req[0])
            heapq.heappush(endHeap, req[1]+1)
        
        poppedCount = 0 #Keeps track of how many intervals we have entered without exiting. (Currently active)
        for i in range(len(nums)):
            
            while begHeap and begHeap[0] <= i:
                heapq.heappop(begHeap)
                poppedCount += 1
                
            while endHeap and endHeap[0] <= i:
                heapq.heappop(endHeap)
                poppedCount -= 1
                continue
            requestCount[i] = poppedCount
        
        #Then we simply get the highest interval frequency and match with the highest nums
        sortedRequest = sorted(list(requestCount.items()), key = lambda x: -x[1])
        nums.sort(key = lambda x: -x)
        
        out = 0
        for n, s in zip(nums, sortedRequest):
            out += (n*s[1])
            
        return out % (pow(10, 9)+7)

class Solution:
    def maxSumRangeQuery(self, nums: List[int], requests: List[List[int]]) -> int:
        requestCount = defaultdict(int) #Number of request per index
        
        #Create a min-heap of the start and end intervals.
        begHeap = []
        endHeap = []
        for req in requests:
            heapq.heappush(begHeap, req[0])
            heapq.heappush(endHeap, req[1]+1)
        
        poppedCount = 0 #Keeps track of how many intervals we have entered without exiting. (Currently active)
        for i in range(len(nums)):
            
            while begHeap and begHeap[0] <= i:
                heapq.heappop(begHeap)
                poppedCount += 1
                
            while endHeap and endHeap[0] <= i:
                heapq.heappop(endHeap)
                poppedCount -= 1
                continue
            requestCount[i] = poppedCount
        
        #Then we simply get the highest interval frequency and match with the highest nums
        sortedRequest = sorted(list(requestCount.items()), key = lambda x: -x[1])
        nums.sort(key = lambda x: -x)
        
        out = 0
        for n, s in zip(nums, sortedRequest):
            out += (n*s[1])
            
        return out % (pow(10, 9)+7)
