What is the key pattern in the Tuple with Same Product problem?

The key pattern is array scanning combined with hash lookups to count distinct pairs of products efficiently.

How do I handle large inputs in the Tuple with Same Product problem?

To handle large inputs, use hash maps to store pair products and avoid brute-force solutions that would take too long.

Can I optimize the space complexity of this problem?

The space complexity is O(n^2) due to the hash map storing pair products. While it's hard to reduce, you can consider pruning unnecessary entries to optimize.

What is the time complexity of the Tuple with Same Product problem?

The time complexity is O(n^2), as we compute products for each pair of distinct integers and perform hash map operations.

How does hashing help in solving the Tuple with Same Product problem?

Hashing allows efficient storage and lookup of pair products, enabling the counting of valid tuples without redundant calculations.

#1726

Medium

auto_awesome数组·哈希·扫描

LeetCode 题解工作台

同积元组

给你一个由不同正整数组成的数组 nums ，请你返回满足 a * b = c * d 的元组 (a, b, c, d) 的数量。其中 a 、 b 、 c 和 d 都是 nums 中的元素，且 a != b != c != d 。示例 1：输入： nums = [2,3,4,6] 输出： 8 …

数组哈希表计数

题目描述

给你一个由不同正整数组成的数组 nums ，请你返回满足 a * b = c * d 的元组 (a, b, c, d) 的数量。其中 a、b、c 和 d 都是 nums 中的元素，且 a != b != c != d 。

示例 1：

输入：nums = [2,3,4,6]
输出：8
解释：存在 8 个满足题意的元组：
(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)

示例 2：

输入：nums = [1,2,4,5,10]
输出：16
解释：存在 16 个满足题意的元组：
(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)

提示：

1 <= nums.length <= 1000
1 <= nums[i] <= 10⁴
nums 中的所有元素 互不相同

lightbulb

解题思路

方法一：组合数 + 哈希表

假设存在 $n$ 组数，对于其中任意两组数 $a, b$ 和 $c, d$ ，均满足 $a \times b = c \times d$ 的条件，则这样的组合一共有 $\mathrm{C}_n^2 = \frac{n \times (n-1)}{2}$ 个。

根据题意每一组满足上述条件的组合可以构成 $8$ 个满足题意的元组，故将各个相同乘积的组合数乘以 $8$ 相加（等价于：左移 $3$ 位）即可得到结果。

时间复杂度 $O(n^2)$ ，空间复杂度 $O(n^2)$ 。其中 $n$ 为数组长度。

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class Solution:
    def tupleSameProduct(self, nums: List[int]) -> int:
        cnt = defaultdict(int)
        for i in range(1, len(nums)):
            for j in range(i):
                x = nums[i] * nums[j]
                cnt[x] += 1
        return sum(v * (v - 1) // 2 for v in cnt.values()) << 3

speed

复杂度分析

指标	值
时间	O(n^2)
空间	O(n^2)

psychology

面试官常问的追问

外企场景

question_mark
The candidate demonstrates understanding of hashing techniques for counting distinct pairs.
question_mark
The candidate proposes an efficient solution with O(n^2) time complexity, recognizing the inefficiency of brute force.
question_mark
The candidate properly identifies the trade-off between time complexity and space complexity in the solution.

warning

常见陷阱

外企场景

error
Forgetting to handle distinctness of the elements in the pairs, which could lead to incorrect tuple counts.
error
Overlooking the need to use hash maps efficiently, potentially causing unnecessary recomputation.
error
Not considering space complexity when dealing with large input sizes, which may lead to inefficient solutions.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Consider implementing the same problem but allowing repeated elements in the array.
arrow_right_alt
Optimize the solution by using a more advanced data structure, like a balanced binary search tree, instead of a hash map.
arrow_right_alt
Expand the problem to include tuples of more than four elements with the same product.

help

常见问题

外企场景

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