# 1726. Tuple with Same Product

## 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 $a \times b = c \times d$, there are a total of $\mathrm{C}_n^2 = \frac{n \times (n-1)}{2}$ 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 $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the length of the array.

• 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;
}
}

• 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;
}
};

• 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


• 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
}

• 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;
}


• 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
}
}