Count Good Meals LeetCode solution in Python

1
class Solution:
2
    def countPairs(self, deliciousness: List[int]) -> int:
3
        pows = [2**i for i in range(0, 22)]  # form our list of powers of 2
4
        dp_seen = (
5
            {}
6
        )  # dict to store what we've seen - dynamic programming solution for time requirement
7
        count = 0  # to store the answer
8

9
        for j in range(0, len(deliciousness)):
10
            for i in range(0, len(pows)):
11
                if (
12
                    pows[i] - deliciousness[j] in dp_seen
13
                ):  # "if we find a previous deliciousness[j] as pows[i] - deliciousness[j], then we will add dp_seen[deliciousness[j]] to count"
14
                    count += dp_seen[pows[i] - deliciousness[j]]
15
            if deliciousness[j] in dp_seen:
16
                dp_seen[deliciousness[j]] += 1
17
            else:
18
                dp_seen[deliciousness[j]] = 1
19

20
        return count % (
21
            10**9 + 7
22
        )  # the arbitrary modulo, presumably to reduce the answer size