Sum of Concatenated Pairs of Elements in a List in Python

Suppose we have a list of numbers called nums. We have to find the sum of every concatenation of every pair of numbers in nums. Here the pair (i, j) and pair (j, i) are considered different.

So, if the input is like nums = [5, 3], then the output will be 176, as We have the following concatenations: (nums[0] + nums[0]) = (5 concat 5) = 55, (nums[0] + nums[1]) = (5 concat 3) = 53, (nums[1] + nums[0]) = (3 concat 5) = 35, (nums[0] + nums[0]) = (3 concat 3) = 33, then the sum is 55 + 53 + 35 + 33 = 176

To solve this, we will follow these steps:

memo := a new map
nums1 := nums
temp := 0
c := sum of all elements in nums1
a := size of nums
for i in range 0 to a, do
   if nums[i] is same as 0, then
      temp := temp + c
   otherwise,
      if nums[i] is present in memo, then
         temp := temp + memo[nums[i]]
      otherwise,
         b := 0
         for j in range 0 to a, do
            b := b + integer of (nums[i] concatenate nums1[j])
         memo[nums[i]] := b
         temp := temp + memo[nums[i]]
return temp

Let us see the following implementation to get better understanding:

Example

 Live Demo

class Solution:
   def solve(self, nums):
      memo = {}
      nums1 = nums
      temp = 0
      c = sum(nums1)
      a = len(nums)
      for i in range(a):
         if nums[i] == 0:
            temp += c
         else:
            if nums[i] in memo:
               temp += memo[nums[i]]
            else:
               b = 0
               for j in range(a):
                  b += int(str(nums[i]) + str(nums1[j]))
               memo[nums[i]] = b
               temp += memo[nums[i]]
      return temp

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

Input

[5, 3]

Output

176