LeetCode 1577: Number of Ways Where Square of Number Is Equal to Product of Two Numbers Solution

Master LeetCode problem 1577 (Number of Ways Where Square of Number Is Equal to Product of Two Numbers), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1577. Number of Ways Where Square of Number Is Equal to Product of Two Numbers

Medium

Array Hash Table Math Two Pointers

Problem Explanation

Explanation:

To solve this problem, we can iterate through all possible combinations of (i, j, k) for both Type 1 and Type 2 triplets, and check if the conditions specified in the problem statement are satisfied. We will maintain two hash maps to store the frequencies of squares in nums1 and products in nums2 respectively, to make the checking process more efficient.

For Type 1 triplets, we iterate through all pairs (j, k) in nums2 and calculate the square of nums2[j] * nums2[k]. We then check if this square exists in the map for squares of nums1.
For Type 2 triplets, we iterate through all pairs (j, k) in nums1 and calculate the square of nums1[j] * nums1[k]. We then check if this square exists in the map for products in nums2.

Finally, we return the total count of valid triplets found.

Time Complexity: O(N^2), where N is the size of the larger array between nums1 and nums2.

Space Complexity: O(N), to store the frequencies of squares and products in hash maps.

Solution Code

class Solution {
    public int numTriplets(int[] nums1, int[] nums2) {
        return countTriplets(nums1, nums2) + countTriplets(nums2, nums1);
    }
    
    private int countTriplets(int[] nums1, int[] nums2) {
        int count = 0;
        Map<Long, Integer> squares = new HashMap<>();
        
        for (int i = 0; i < nums1.length; i++) {
            long square = (long) nums1[i] * nums1[i];
            squares.put(square, squares.getOrDefault(square, 0) + 1);
        }
        
        for (int j = 0; j < nums2.length; j++) {
            for (int k = j + 1; k < nums2.length; k++) {
                long product = (long) nums2[j] * nums2[k];
                if (squares.containsKey(product)) {
                    count += squares.get(product);
                }
            }
        }
        
        return count;
    }
}

Try It Yourself

Loading code editor...

Frequently Asked Questions

How to solve LeetCode 1577 (Number of Ways Where Square of Number Is Equal to Product of Two Numbers)?

This page provides optimized solutions for LeetCode problem 1577 (Number of Ways Where Square of Number Is Equal to Product of Two Numbers) in Java, C++, and Python, along with a detailed explanation and an interactive code editor to test your code.

What is the time complexity of LeetCode 1577 (Number of Ways Where Square of Number Is Equal to Product of Two Numbers)?

The time complexity for LeetCode 1577 (Number of Ways Where Square of Number Is Equal to Product of Two Numbers) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

Can I run code for LeetCode 1577 on DevExCode?

Yes, DevExCode provides an interactive code editor where you can write, test, and run your code for LeetCode 1577 in Java, C++, or Python.

Back to LeetCode Solutions