3Sum With Multiplicity Python Solution

lcpy gen -n 923   # by problem number
lcpy gen -s three_sum_multiplicity   # by problem name

Input: arr = [1,1,2,2,3,3,4,4,5,5], target = 8
Output: 20
Explanation:
Enumerating by the values (arr[i], arr[j], arr[k]):
(1, 2, 5) occurs 8 times;
(1, 3, 4) occurs 8 times;
(2, 2, 4) occurs 2 times;
(2, 3, 3) occurs 2 times.

Input: arr = [1,1,2,2,2,2], target = 5
Output: 12
Explanation:
arr[i] = 1, arr[j] = arr[k] = 2 occurs 12 times:
We choose one 1 from [1,1] in 2 ways,
and two 2s from [2,2,2,2] in 6 ways.

Input: arr = [2,1,3], target = 6
Output: 1
Explanation: (1, 2, 3) occured one time in the array so we return 1.

from collections import Counter
from math import comb


class Solution:
    # Time: O(n + V^2) where V = 101 distinct values
    # Space: O(V)
    def three_sum_multiplicity(self, arr: list[int], target: int) -> int:
        mod = 10**9 + 7
        count = Counter(arr)
        values = sorted(count)
        total = 0
        for i, x in enumerate(values):
            for y in values[i:]:
                z = target - x - y
                if z < y or z not in count:
                    continue
                if x == y == z:
                    total += comb(count[x], 3)
                elif x == y:
                    total += comb(count[x], 2) * count[z]
                elif y == z:
                    total += comb(count[y], 2) * count[x]
                else:
                    total += count[x] * count[y] * count[z]
        return total % mod