Program to find number of unique four indices where they can generate sum less than target from four lists in python

Suppose we have four list of numbers A, B, C and D and also have another number target. We have to find the number of different unique indices i, j, k, l such that A[i] + B[j] + C[k] + D[l] ≤ target.

So, if the input is like A = [3, 2] B = [5, 3] C = [1] D = [2, 3] target = 9, then the output will be 3, as We can pick the following combinations: [3, 3, 1, 2] [3, 3, 1, 2] [2, 3, 1, 3]

To solve this, we will follow these steps:

• temp_list := a new list
• for i in range 0 to size of A, do
  • for j in range 0 to size of B, do
    • insert (A[i] + B[j]) at the end of temp_list
• sort the list temp_list
• ans := 0
• for i in range 0 to size of C, do
  • for j in range 0 to size of D, do
    • sum_cd := C[i] + D[j]
    • sum_ab := target - sum_cd
    • ans := ans + number of elements in temp_list whose sum <= sum_ab
• return ans

Let us see the following implementation to get better understanding:

Example

Live Demo

from bisect import bisect_right

class Solution:
   def solve(self, A, B, C, D, target):
      temp_list = []
      for i in range(len(A)):
         for j in range(len(B)):
            temp_list.append(A[i] + B[j])

      temp_list.sort()

      ans = 0
      for i in range(len(C)):
         for j in range(len(D)):
            sum_cd = C[i] + D[j]
            sum_ab = target - sum_cd

            ans += bisect_right(temp_list, sum_ab)

      return ans

ob = Solution()
A = [3, 2]
B = [5, 3]
C = [1]
D = [2, 3]
target = 9
print(ob.solve(A, B, C, D, target))

Input

[3, 2], [5, 3], [1], [2, 3], 9

Output

3

Arnab Chakraborty

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Updated on: 26-Nov-2020

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