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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
    }
}

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