LeetCode 1726. Tuple with Same Product Solution in Java, C++, Python & More | Explanation + Code

Description

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.

 

Example 1:

Input: nums = [2,3,4,6]
Output: 8
Explanation: There are 8 valid tuples:
(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:
(1,10,2,5) , (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.

Solutions

Solution 1: Combination + Hash Table

Assuming there are n pairs of numbers, for any two pairs of numbers a, b and c, d that satisfy the condition a × b = c × d, there are a total of C_n² = ^{n × (n-1)}⁄₂ such combinations.

According to the problem description, each combination that satisfies the above condition can form 8 tuples that satisfy the problem requirements. Therefore, we can multiply the number of combinations with the same product by 8 (equivalent to left shifting by 3 bits) and add them up to get the result.

The time complexity is O(n²), and the space complexity is O(n²). Here, n is the length of the array.

PythonJavaC++GoTypeScriptRust

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(code-box)

class Solution { public int tupleSameProduct(int[] nums) { Map<Integer, Integer> cnt = new HashMap<>(); for (int i = 1; i < nums.length; ++i) { for (int j = 0; j < i; ++j) { int x = nums[i] * nums[j]; cnt.merge(x, 1, Integer::sum); } } int ans = 0; for (int v : cnt.values()) { ans += v * (v - 1) / 2; } return ans << 3; } }(code-box)

class Solution { public: int tupleSameProduct(vector<int>& nums) { unordered_map<int, int> cnt; for (int i = 1; i < nums.size(); ++i) { for (int j = 0; j < i; ++j) { int x = nums[i] * nums[j]; ++cnt[x]; } } int ans = 0; for (auto& [_, v] : cnt) { ans += v * (v - 1) / 2; } return ans << 3; } };(code-box)

func tupleSameProduct(nums []int) int { cnt := map[int]int{} for i := 1; i < len(nums); i++ { for j := 0; j < i; j++ { x := nums[i] * nums[j] cnt[x]++ } } ans := 0 for _, v := range cnt { ans += v * (v - 1) / 2 } return ans << 3 }(code-box)

function tupleSameProduct(nums: number[]): number { const cnt: Map<number, number> = new Map(); for (let i = 1; i < nums.length; ++i) { for (let j = 0; j < i; ++j) { const x = nums[i] * nums[j]; cnt.set(x, (cnt.get(x) ?? 0) + 1); } } let ans = 0; for (const [_, v] of cnt) { ans += (v * (v - 1)) / 2; } return ans << 3; }(code-box)

use std::collections::HashMap; impl Solution { pub fn tuple_same_product(nums: Vec<i32>) -> i32 { let mut cnt: HashMap<i32, i32> = HashMap::new(); let mut ans = 0; for i in 1..nums.len() { for j in 0..i { let x = nums[i] * nums[j]; *cnt.entry(x).or_insert(0) += 1; } } for v in cnt.values() { ans += (v * (v - 1)) / 2; } ans << 3 } }(code-box)