Question

1726. Tuple with Same Product

中文文档

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:
  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

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;
  }
};

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
  }
}