Find Two Pairs of Numbers with Minimal Sum Difference in Python

Suppose we have a list of numbers called nums and we want to select two pairs of numbers from it such that the absolute difference between the sum of these two pairs is minimized.

So, if the input is like nums = [3, 4, 5, 10, 7], then the output will be 1, as we can select these pairs (3 + 7) - (4 + 5) = 1.

To solve this, we will follow these steps:

• distances := a new list
• for i in range 0 to size of nums - 2, do
  • for j in range i + 1 to size of nums - 1, do
    • insert a list [|nums[i] - nums[j]| , i, j] at the end of distances
  • sort the list distances
  • ans := 1^9
  • for i in range 0 to size of distances - 2, do
    • [dist, i1, i2] := distances[i]
    • j := i + 1
    • [dist2, i3, i4] := distances[j]
    • while j < size of distances and elements in (i1, i2, i3, i4) are not unique, do
      • [dist2, i3, i4] := distances[j]
      • j := j + 1
    • if elements in (i1, i2, i3, i4) are unique, then
      • ans := minimum of ans and (dist2 - dist)
    • return ans

Let us see the following implementation to get better understanding:

Example Code

Live Demo

class Solution:
   def solve(self, nums):
      distances = []
      for i in range(len(nums) - 1):
         for j in range(i + 1, len(nums)):
            distances.append((abs(nums[i] - nums[j]), i, j))
      distances.sort()
      ans = 1e9
      for i in range(len(distances) - 1):
         dist, i1, i2 = distances[i]
         j = i + 1
         dist2, i3, i4 = distances[j]
         while j < len(distances) and len({i1, i2, i3, i4}) != 4:
            dist2, i3, i4 = distances[j]
            j += 1
         if len({i1, i2, i3, i4}) == 4:
            ans = min(ans, dist2 - dist)
      return ans

ob = Solution()
nums = [3, 4, 5, 10, 7]
print(ob.solve(nums))

Input

[3, 4, 5, 10, 7]

Output

1