How do you solve Count Equal and Divisible Pairs in an Array on LeetCode?

Count Equal and Divisible Pairs in an Array (LeetCode #2176) 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: Using a hashmap allows us to efficiently group indices by value, reducing the number of comparisons needed.

What is the brute force solution for Count Equal and Divisible Pairs in an Array?

The brute force approach for Count Equal and Divisible Pairs in an Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of indices (i, j) in the array to see if they meet the conditions. This is straightforward but inefficient for larger arrays due to its quadratic time complexity. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Equal and Divisible Pairs in an Array?

Count Equal and Divisible Pairs in an Array uses the following concepts: Array. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Equal and Divisible Pairs in an Array?

Always clarify the constraints and edge cases before diving into coding. Discuss your thought process and potential optimizations with the interviewer. Practice writing clean and efficient code, as readability is key in interviews.

What are common mistakes when solving Count Equal and Divisible Pairs in an Array?

Failing to check the condition i < j, leading to incorrect pairs being counted. Not considering the case where k = 1, which can lead to confusion in divisibility checks.

#2176

Count Equal and Divisible Pairs in an Array

Easy

ArrayHash 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 a hashmap to count occurrences of each number and their indices. This allows us to efficiently check pairs without needing to iterate through all combinations, significantly reducing the time complexity.

⚙️

Algorithm

5 steps

1Step 1: Create a hashmap to store lists of indices for each unique number in nums.
2Step 2: Populate the hashmap with indices for each number in nums.
3Step 3: For each unique number, retrieve its list of indices and check all pairs of indices (i, j) where i < j.
4Step 4: For each pair, check if (i * j) is divisible by k and increment the count if true.
5Step 5: Return the final count.

solution.py21 lines

1# Full working Python code
2
3def countPairs(nums, k):
4    from collections import defaultdict
5    index_map = defaultdict(list)
6    count = 0
7    n = len(nums)
8
9    for i in range(n):
10        index_map[nums[i]].append(i)
11
12    for indices in index_map.values():
13        m = len(indices)
14        for i in range(m):
15            for j in range(i + 1, m):
16                if (indices[i] * indices[j]) % k == 0:
17                    count += 1
18    return count
19
20# Example usage
21print(countPairs([3,1,2,2,2,1,3], 2))  # Output: 4

ℹ

Complexity note: The time complexity is O(n) because we traverse the array once to build the hashmap and then process each unique number's indices. The space complexity is O(n) due to the hashmap storing indices for each unique number.

1Using a hashmap allows us to efficiently group indices by value, reducing the number of comparisons needed.
2Divisibility checks can be performed directly on the indices, avoiding unnecessary calculations.

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