How do you solve Find the Number of Good Pairs I on LeetCode?

Find the Number of Good Pairs I (LeetCode #3162) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(m) space complexity. Key insight: Understanding divisibility is crucial for this problem.

What is the brute force solution for Find the Number of Good Pairs I?

The brute force approach for Find the Number of Good Pairs I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible pair of elements from nums1 and nums2 to see if the condition for being a good pair is met. This is straightforward and guarantees that we find all good pairs, but it can be slow for larger inputs. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Find the Number of Good Pairs I?

Find the Number of Good Pairs I uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find the Number of Good Pairs I?

Clarify the problem statement and constraints before jumping to code. Think about the efficiency of your solution and whether you can reduce complexity. Practice explaining your thought process while coding to simulate the interview environment.

What are common mistakes when solving Find the Number of Good Pairs I?

Not checking all pairs correctly, especially edge cases. Forgetting to multiply nums2[j] by k before checking divisibility.

#3162

Find the Number of Good Pairs I

Easy

ArrayHash TableHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n * m)Space O(m)

Instead of checking each pair, we can use a HashMap to store the frequency of the values in nums2 multiplied by k. This allows us to quickly check how many times each divisor appears, reducing the need for nested loops.

⚙️

Algorithm

6 steps

1Step 1: Create a HashMap to store the frequency of each value in nums2 multiplied by k.
2Step 2: Loop through nums2 and populate the HashMap with counts.
3Step 3: Initialize a counter to zero for good pairs.
4Step 4: Loop through nums1 and for each element, check how many times it can be divided by the keys in the HashMap.
5Step 5: For each valid division, add the frequency from the HashMap to the counter.
6Step 6: Return the counter.

solution.py9 lines

1def countGoodPairs(nums1, nums2, k):
2    from collections import Counter
3    freq = Counter(num * k for num in nums2)
4    count = 0
5    for num in nums1:
6        for key in freq:
7            if num % key == 0:
8                count += freq[key]
9    return count

ℹ

Complexity note: The time complexity is O(n * m) where n is the length of nums1 and m is the length of nums2, due to the nested loops. The space complexity is O(m) because we store the frequency of elements from nums2 in a HashMap.

1Understanding divisibility is crucial for this problem.
2Using a HashMap can significantly reduce the number of checks needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.