Program to find maximum sum obtained of any permutation in Python

Suppose we have an array nums, and another array called requests where requests[i] = [start_i, end_i], this represents the ith request asks for the sum of nums[start_i] + nums[start_i+1] + ... + nums[end_i-1] + nums[end_i]. We have to find the maximum total sum of all requests among all permutations of nums. The answer may be very large, so return it modulo 10^9+7.

So, if the input is like nums = [10,20,30,40,50] requests = [[1,3],[0,1]], then the output will be 190, because if we arrange like [30,50,40,20,10] we get: from requests[0] : nums[1] + nums[2] + nums[3] = 50 + 40 + 20 = 110, and from requests[1] : nums[0] + nums[1] = 30 + 50 = 80, so total sum 110+80 = 190.

To solve this, we will follow these steps:

• A := a new list
• for each request (s, e) in requests, do
  • insert pair (s, 0) at the end of A
  • insert pair (e, 1) at the end of A
• sort the list A
• fr := an empty map
• cnt := 0
• n := size of A
• i := 0
• while i
• r := 1
• while i
• r := r + 1
• i := i + 1

Example

Let us see the following implementation to get better understanding −

from collections import defaultdict
def solve(nums, requests):
   A = []
   for s, e in requests:
      A.append((s, 0))
      A.append((e, 1))
   A.sort()
   fr = defaultdict(list)
   cnt = 0

   n = len(A)
   i = 0
   while i  0:
            fr[cnt-r].append((pre, p-1))
         pre = p
      elif flag == 1:
         cnt -= r
         fr[cnt+r].append((pre, p))
         pre = p+1
      i += 1

   nums.sort(reverse=True)
   ks = list(fr.keys())
   ks.sort(reverse=True)
   ans = 0
   m = 10**9 + 7
   i = 0
   for k in ks:
      for s, e in fr[k]:
         d = e - s + 1
         ans += sum(nums[i:i+d]) * k
         ans %= m
         i += d
   return ans

nums = [10,20,30,40,50]
requests = [[1,3],[0,1]]
print(solve(nums, requests))

Input

[10,20,30,40,50],[[1,3],[0,1]]

Output

190