Leetcode 1726: Tuple with Same Product

Input: You are given an array `nums` containing distinct positive integers.

Example: Input: nums = [1, 2, 3, 4, 6]

Constraints:

• 1 <= nums.length <= 1000

• 1 <= nums[i] <= 10^4

• All elements in nums are distinct.

Output: Return the number of valid tuples `(a, b, c, d)` such that `a * b = c * d` and `a != b != c != d`.

Example: Output: 6

Constraints:

• The output is an integer representing the number of valid tuples.

Goal: The goal is to find pairs of tuples `(a, b)` and `(c, d)` such that the products `a * b` and `c * d` are equal, and then count how many such valid tuples exist.

Steps:

• 1. For each pair `(a, b)` in the array, compute the product `a * b`.

• 2. Keep track of how many times each product appears using a map.

• 3. For each new product, if it has appeared before, add the number of times it has appeared to the result.

• 4. Since the tuples `(a, b)` and `(c, d)` are interchangeable, multiply the result by 8 to account for all possible orderings of the tuples.

Goal: The solution must efficiently handle cases where the array length is up to 1000 and the elements are distinct integers up to 10^4.

Steps:

• 1 <= nums.length <= 1000

• 1 <= nums[i] <= 10^4

• All elements in nums are distinct.

Assumptions:

• All elements in the array are distinct positive integers.

• Input: Input: nums = [1, 2, 3, 4, 6]

• Explanation: The valid tuples where `a * b = c * d` are (1,6,2,3), (1,6,3,2), (6,1,2,3), (6,1,3,2), (2,3,1,6), (3,2,1,6). Therefore, the output is 6.

• Input: Input: nums = [2, 4, 8, 16]

• Explanation: The valid tuples where `a * b = c * d` are (2,8,4,4), (8,2,4,4), and there are no other valid tuples. Hence, the output is 2.

Approach: We can approach this problem by iterating over all unique pairs `(a, b)` in the array, computing the product `a * b`, and counting how many times each product appears. If a product appears more than once, we count all valid pairs formed by these occurrences.

Observations:

• The problem essentially asks for counting pairs of pairs that produce the same product.

• Using a map to store the products of pairs is a good way to count and then use this information to compute the valid tuples.

Steps:

• 1. Initialize a map to store the frequency of each product from pairs of numbers in the array.

• 2. Loop through all pairs `(a, b)` in the array, compute the product `a * b`, and update the map.

• 3. For each product that appears more than once, calculate how many tuples can be formed by multiplying the occurrences by 8 (for the orderings).

• 4. Return the total count of valid tuples.

Empty Inputs:

• The array should not be empty as it is stated that the length of `nums` will be at least 1.

Large Inputs:

• The solution should handle the case where `nums.length` is close to 1000 efficiently.

Special Values:

• There will be no cases where `nums[i] == nums[j]` because all elements in `nums` are distinct.

Constraints:

• The constraints suggest that the solution must efficiently compute the result even with the maximum input size.

int tupleSameProduct(vector<int>& nums) {
    map<int, int> mp;
    
    int n = nums.size(), res = 0;
    
    for(int i = 0; i < n; i++)
    for(int j = i + 1; j < n; j++) {
        int x = nums[i] * nums[j];
        if(mp.count(x)) res+=mp[x];
        mp[x]++;
    }
    return res * 8;
}

1 : Function Definition

int tupleSameProduct(vector<int>& nums) {

The function `tupleSameProduct` is defined, taking a vector `nums` as input. The goal is to find how many quadruples (a, b, c, d) satisfy nums[a] * nums[b] == nums[c] * nums[d].

2 : Map Initialization

    map<int, int> mp;

A map `mp` is initialized to store the frequency of each product that can be formed by multiplying two numbers from the input list `nums`.

3 : Variable Declaration

    int n = nums.size(), res = 0;

The variable `n` is initialized to the size of the input vector `nums`, and `res` is initialized to 0. `res` will store the total number of valid quadruples.

4 : Outer Loop

    for(int i = 0; i < n; i++)

An outer loop is initiated to iterate over each element `nums[i]` in the input vector.

5 : Inner Loop

    for(int j = i + 1; j < n; j++) {

An inner loop is initiated to iterate over all elements `nums[j]` that come after `nums[i]` to form pairs of products.

6 : Product Calculation

        int x = nums[i] * nums[j];

The product `x` is calculated by multiplying the values of `nums[i]` and `nums[j]`.

7 : Count Check

        if(mp.count(x)) res+=mp[x];

The function checks if the product `x` already exists in the map `mp`. If it does, the value in `mp[x]` is added to `res`, indicating the number of valid quadruples that can be formed using this product.

8 : Map Update

        mp[x]++;

The map `mp` is updated by incrementing the count of the product `x`. This tracks how many times the product `x` has been encountered.

9 : Return Result

    return res * 8;

The result `res` is multiplied by 8 because each pair of products represents a combination of four distinct indices, and there are 8 permutations of four distinct indices that can form the same product.

Best Case: O(n^2), since we need to check every pair of numbers in the array.

Average Case: O(n^2), as each pair is checked independently.

Worst Case: O(n^2), the worst case is when we check all pairs.

Description: The time complexity is quadratic due to the nested loops over all pairs in the array.

Best Case: O(n^2), as we store the products of all pairs.

Worst Case: O(n^2), to store the products of all pairs.

Description: The space complexity is quadratic, as we store the frequency of products for each pair.

Leetcode 1726: Tuple with Same Product

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