1726. Tuple with Same Product

Problem

Given an array nums of distinct positive integers, return the number of tuples (a, b, c, d) such that a * b = c * d where a, b, c, and d are elements of nums, and a != b != c != d.

https://leetcode.com/problems/tuple-with-same-product/

Example 1:

Input: nums = [2,3,4,6]
Output: 8
Explanation: There are 8 valid tuples:

1 2	(2,6,3,4) , (2,6,4,3) , (6,2,3,4) , (6,2,4,3) (3,4,2,6) , (4,3,2,6) , (3,4,6,2) , (4,3,6,2)

Example 2:

Input: nums = [1,2,4,5,10]
Output: 16
Explanation: There are 16 valid tuples:

1 2 3 4

(1,10,2,5) , (1,10,5,2) , (10,1,2,5) , (10,1,5,2) (2,5,1,10) , (2,5,10,1) , (5,2,1,10) , (5,2,10,1) (2,10,4,5) , (2,10,5,4) , (10,2,4,5) , (10,2,5,4) (4,5,2,10) , (4,5,10,2) , (5,4,2,10) , (5,4,10,2)

Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 10⁴
• All elements in nums are distinct.

Test Cases

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class Solution:
    def tupleSameProduct(self, nums: List[int]) -> int:

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import pytest

from solution import Solution


@pytest.mark.parametrize('nums, expected', [
    ([2,3,4,6], 8),
    ([1,2,4,5,10], 16),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, nums, expected):
    assert sol.tupleSameProduct(nums) == expected

Thoughts

枚举 nums 中所有可能的数字对，计算每个数字对之积，并统计每个可能的乘积能对应多少个数字对。

设某个乘积共有 m 个可能的数字对，因为 nums 中的数字是各不相同的正整数，可知这 m 个数字对是由各不相同的共 2m 个数字构成的。从 m 个数字对中任选两个做排列，有 m * (m - 1) 种可能，选中的每对数字都可以有两种顺序，所以总共有 4 * m * (m - 1) 种可能。

时间复杂度 O(n²)，空间复杂度 O(n²)。

Code

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from collections import defaultdict
from itertools import combinations


class Solution:
    def tupleSameProduct(self, nums: list[int]) -> int:
        prods: dict[int, int] = defaultdict(int) # product -> count of pairs
        for a, b in combinations(nums, 2):
            prods[a * b] += 1

        return sum(m * (m - 1) << 2 for m in prods.values() if m > 1)

许可协议

本文除了题目描述（Problem）部分之外，其他内容由 Calf 原创，采用 CC BY-NC-SA 4.0 许可协议，转载请注明出处。

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