Number of Single Divisor Triplets — LeetCode 2198 Python Solution

The problem

You are given a 0-indexed array of positive integers nums. A triplet of three distinct indices (i, j, k) is called a single divisor triplet of nums if nums[i] + nums[j] + nums[k] is divisible by exactly one of nums[i], nums[j], or nums[k].

Example

Input

nums = [4,6,7,3,2]

Output

12

Explanation

The triplets (0, 3, 4), (0, 4, 3), (3, 0, 4), (3, 4, 0), (4, 0, 3), and (4, 3, 0) have the values of [4, 3, 2] (or a permutation of [4, 3, 2]).

Python solution

class Solution:
    def singleDivisorTriplet(self, nums: List[int]) -> int:
        cnt = Counter(nums)
        ans = 0
        for a, x in cnt.items():
            for b, y in cnt.items():
                for c, z in cnt.items():
                    s = a + b + c
                    if sum(s % v == 0 for v in (a, b, c)) == 1:
                        if a == b:
                            ans += x * (x - 1) * z
                        elif a == c:
                            ans += x * (x - 1) * y
                        elif b == c:
                            ans += x * y * (y - 1)
                        else:
                            ans += x * y * z
        return ans

Complexity

Pattern: Math and Number Theory

Find the closed form, the invariant, or the modular identity — and skip the loop entirely. LeetCode 2198. Number of Single Divisor Triplets is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Math.

The math and number theory guide has the Python template for the pattern and the 485 LeetCode problems that use it.

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