Is Count Equal and Divisible Pairs in an Array easy or hard?

Count Equal and Divisible Pairs in an Array is categorized as an Easy problem. The brute force solution is straightforward with nested loops, while the HashMap optimization introduces a common interview technique for reducing unnecessary comparisons.

Count Equal and Divisible Pairs in an Array Python/Java solution

In Python and Java, the common implementation first builds a dictionary or HashMap mapping values to index lists. Then it iterates through each list and checks pairs (i, j) to verify whether (i * j) % k == 0 while maintaining i < j.

How to solve Count Equal and Divisible Pairs in an Array in O(n)?

An exact O(n) solution is not generally possible because the algorithm must potentially evaluate many index combinations that satisfy the equality condition. The typical optimization is to group indices by value using a HashMap and only evaluate pairs within each group, reducing unnecessary comparisons.

What is the best approach for Count Equal and Divisible Pairs in an Array?

The practical approach groups indices by value using a HashMap, then checks index pairs only within each group. This avoids comparing elements with different values. The worst-case time complexity remains O(n^2), but it performs much better when the array contains many distinct numbers.

Is Count Equal and Divisible Pairs in an Array asked at Google/Amazon/Meta?

Problems combining array iteration, hashing, and index-based conditions appear frequently in interviews at companies like Amazon, Google, and Meta. While this exact question may vary, the underlying pattern of filtering pairs with hash maps and validating conditions is common.

What data structure is used in Count Equal and Divisible Pairs in an Array?

The optimized approach uses a HashMap (hash table) that maps each number to the list of indices where it appears. This structure allows you to process only indices with equal values instead of scanning the entire array repeatedly.

What is the time complexity of Count Equal and Divisible Pairs in an Array?

The straightforward brute force solution runs in O(n^2) time because it checks every pair of indices. The HashMap grouping approach also has O(n^2) worst-case time when all values are identical, with O(n) additional space for storing index groups.

Description Hints 1 Solution Video 30:04

2176. Count Equal and Divisible Pairs in an Array

Q: How to solve Count Equal and Divisible Pairs in an Array in O(n)?

An exact O(n) solution is not generally possible because the algorithm must potentially evaluate many index combinations that satisfy the equality condition. The typical optimization is to group indices by value using a HashMap and only evaluate pairs within each group, reducing unnecessary comparisons.

Q: What is the best approach for Count Equal and Divisible Pairs in an Array?

The practical approach groups indices by value using a HashMap, then checks index pairs only within each group. This avoids comparing elements with different values. The worst-case time complexity remains O(n^2), but it performs much better when the array contains many distinct numbers.

Q: Is Count Equal and Divisible Pairs in an Array asked at Google/Amazon/Meta?

Problems combining array iteration, hashing, and index-based conditions appear frequently in interviews at companies like Amazon, Google, and Meta. While this exact question may vary, the underlying pattern of filtering pairs with hash maps and validating conditions is common.

Q: What data structure is used in Count Equal and Divisible Pairs in an Array?

The optimized approach uses a HashMap (hash table) that maps each number to the list of indices where it appears. This structure allows you to process only indices with equal values instead of scanning the entire array repeatedly.

Q: What is the time complexity of Count Equal and Divisible Pairs in an Array?

The straightforward brute force solution runs in O(n^2) time because it checks every pair of indices. The HashMap grouping approach also has O(n^2) worst-case time when all values are identical, with O(n) additional space for storing index groups.

Easy84.0% acceptance5 companies

Given a 0-indexed integer array nums of length n and an integer k, return the number of pairs (i, j) where 0 <= i < j < n, such that nums[i] == nums[j] and (i * j) is divisible by k.

Example 1:

Input: nums = [3,1,2,2,2,1,3], k = 2
Output: 4
Explanation:
There are 4 pairs that meet all the requirements:
- nums[0] == nums[6], and 0 * 6 == 0, which is divisible by 2.
- nums[2] == nums[3], and 2 * 3 == 6, which is divisible by 2.
- nums[2] == nums[4], and 2 * 4 == 8, which is divisible by 2.
- nums[3] == nums[4], and 3 * 4 == 12, which is divisible by 2.

Example 2:

Input: nums = [1,2,3,4], k = 1
Output: 0
Explanation: Since no value in nums is repeated, there are no pairs (i,j) that meet all the requirements.

Constraints:

1 <= nums.length <= 100
1 <= nums[i], k <= 100

Asked at 5 companies

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Topics

Array

Hints open one at a time

Hint 1

For every possible pair of indices (i, j) where i < j, check if it satisfies the given conditions.

30:04

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