What is the brute force solution for Number of Ways Where Square of Number Is Equal to Product of Two Numbers?

The brute force approach for Number of Ways Where Square of Number Is Equal to Product of Two Numbers is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible triplets in both arrays to see if they satisfy the given conditions. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Ways Where Square of Number Is Equal to Product of Two Numbers?

Number of Ways Where Square of Number Is Equal to Product of Two Numbers uses the following concepts: Array, Hash Table, Math, Two Pointers. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Ways Where Square of Number Is Equal to Product of Two Numbers?

Always start with a brute force approach to understand the problem better. Explain your thought process clearly when transitioning to an optimal solution. Practice recognizing patterns in problems to identify potential optimizations.

What are common mistakes when solving Number of Ways Where Square of Number Is Equal to Product of Two Numbers?

Not considering all pairs in the second array when checking conditions. Failing to account for the uniqueness of squares in frequency maps.

#1577

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

Medium

ArrayHash TableMathTwo PointersHash MapArray

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

By using frequency maps to count occurrences of squares, we can significantly reduce the number of operations needed to find valid triplets. This allows us to check conditions in constant time.

⚙️

Algorithm

3 steps

1Step 1: Create a frequency map for the squares of elements in nums1 and nums2.
2Step 2: For each unique square from nums1, calculate how many pairs from nums2 can form that square using the frequency map.
3Step 3: Repeat the process for squares from nums2 and count pairs from nums1.

solution.py24 lines

1# Full working Python code
2
3def countTriplets(nums1, nums2):
4    from collections import Counter
5    count = 0
6    freq1 = Counter(num ** 2 for num in nums1)
7    freq2 = Counter(num ** 2 for num in nums2)
8
9    for square in freq1:
10        for j in range(len(nums2)):
11            for k in range(j + 1, len(nums2)):
12                if nums2[j] * nums2[k] == square:
13                    count += freq1[square]
14
15    for square in freq2:
16        for j in range(len(nums1)):
17            for k in range(j + 1, len(nums1)):
18                if nums1[j] * nums1[k] == square:
19                    count += freq2[square]
20
21    return count
22
23# Example usage
24print(countTriplets([7,4], [5,2,8,9]))

ℹ

Complexity note: The time complexity is O(n) for creating frequency maps, but we still have to check pairs which can lead to O(n²) in the worst case. However, the use of frequency maps allows us to reduce redundant calculations. The space complexity is O(n) due to the storage of frequency maps.

1Using frequency maps can significantly reduce redundant calculations.
2Understanding the conditions for valid triplets is crucial for optimizing the solution.

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