How do you solve Count Array Pairs Divisible by K on LeetCode?

Count Array Pairs Divisible by K (LeetCode #2183) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the relationship between GCD and divisibility is crucial for optimizing the solution.

What is the brute force solution for Count Array Pairs Divisible by K?

The brute force approach for Count Array Pairs Divisible by K is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of indices in the array to see if their product is divisible by k. This is straightforward but inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Array Pairs Divisible by K?

Count Array Pairs Divisible by K uses the following concepts: Array, Hash Table, Math, Counting, Number Theory. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Array Pairs Divisible by K?

Always start with a brute force approach to understand the problem before optimizing. Make sure to explain your thought process clearly, especially when transitioning to a more complex solution. Practice identifying patterns in problems, as they often lead to more efficient solutions.

What are common mistakes when solving Count Array Pairs Divisible by K?

Forgetting to check the product's divisibility correctly. Not considering the GCD and how it affects the required complement for divisibility.

#2183

Count Array Pairs Divisible by K

Hard

ArrayHash TableMathCountingNumber TheoryHash MapArray

LeetCode ↗

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)

The optimal solution leverages the properties of divisibility and the greatest common divisor (GCD) to efficiently count valid pairs without checking each pair explicitly. This reduces the time complexity significantly.

⚙️

Algorithm

4 steps

1Step 1: Create a frequency array to count occurrences of each remainder when nums[i] is divided by k.
2Step 2: For each number in nums, calculate the required complement that would make the product divisible by k using the formula k / gcd(k, nums[i]).
3Step 3: Use the frequency array to find how many numbers can pair with the current number to form a valid product.
4Step 4: Update the frequency array as you process each number to avoid double counting.

solution.py15 lines

1# Full working Python code
2import math
3from collections import defaultdict
4
5def countPairs(nums, k):
6    count = 0
7    freq = defaultdict(int)
8    for num in nums:
9        required = k // math.gcd(k, num)
10        count += freq[required]
11        freq[num % k] += 1
12    return count
13
14# Example usage
15print(countPairs([1, 2, 3, 4, 5], 2))  # Output: 7

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once, and the space complexity is O(n) due to the frequency map storing counts of remainders.

1Understanding the relationship between GCD and divisibility is crucial for optimizing the solution.
2Using a frequency map allows us to efficiently count potential pairs without nested loops.

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