Count Values With Equally Spaced Occurrences I Python solution?

In Python, use `defaultdict(list)` to store indices. After building the map, for each value check that the list is empty or all adjacent differences are same. For example: `d[v].append(i)` during the loop, then `all(l[i] - l[i-1] == l[1] - l[0] for i in range(2, len(l)))`. This is O(n) time.

How to solve Count Values With Equally Spaced Occurrences I in O(n)?

Use a hash map that maps each integer to a list of indices. During the first pass, append the current index to the list for that value. After that, iterate the map and confirm the adjacent index differences are constant. Each index is processed twice, so the total time is O(n).

What is the best approach for Count Values With Equally Spaced Occurrences I?

The best approach uses a hash map to store each value's list of indices and then checks all adjacent gaps equal the same. This runs in O(n) time and O(n) space. A one-pass gap-validation variant stores only the last index and expected gap, reducing per-value overhead while keeping O(n).

Is Count Values With Equally Spaced Occurrences I asked at Google/Amazon/Meta?

It is not a specific FAANG interview question, but the same hash map grouped index-pattern appears in many interview tasks. Interviewers use related variants to test data structure understanding and edge case handling. Therefore, solving this problem is useful practice for big tech interviews.

What data structure is used in Count Values With Equally Spaced Occurrences I?

A hash map (dictionary) is the core data structure. It maps each value to a list of index positions, or to a compact record with last index and gap. Hash maps provide average O(1) lookups, making the solution linear in the array size.

What is the time complexity of Count Values With Equally Spaced Occurrences I?

The optimal solution is O(n) time and O(n) space. The brute-force version that rescans the whole array for each distinct value is O(n²) time. Hash table lookups in the optimal solution average O(1).

Description Hints 2 Solution Video 3:44

4048. Count Values With Equally Spaced Occurrences I

Q: Is Count Values With Equally Spaced Occurrences I easy or hard?

It is classified as an easy problem. The main logic is index grouping and comparing constant gaps, which requires only a hash map and a few loops. The only subtle part is deciding how to handle values that appear once or exactly twice.

Easy75.3% acceptance

You are given an integer array nums.

An integer x is called special if:

x appears exactly three times in nums.
All three occurrences of x are equally spaced in nums. In other words, if all occurrences of x are at indices i₁ < i₂ < i₃, then i₂ - i₁ = i₃ - i₂.

Return the number of distinct special integers in nums.

Example 1:

Input: nums = [1,8,1,5,1,5,8,5]

Output: 2

Explanation:

1 is special because it occurs exactly three times at equally spaced indices 0, 2, and 4.
5 is special because it occurs exactly three times at equally spaced indices 3, 5, and 7.
8 is not special because it occurs only twice.

Therefore, the answer is 2.

Example 2:

Input: nums = [8,8,8,8]

Output: 0

Explanation:

8 is not special because it does not occur exactly three times. Therefore, the answer is 0.

Example 3:

Input: nums = [8,6,6,8,8]

Output: 0

Explanation:

8 occurs at indices 0, 3, and 4, which are not equally spaced. 6 occurs only twice. Therefore, no integer is special.

Constraints:

3 <= nums.length <= 100
1 <= nums[i] <= 100

Topics

Array Hash Table

Hints open one at a time

Hint 1

1a (Brute Force). Count each value's occurrences, then enumerate all triples of indices i < j < k. Count a triple if nums[i] == nums[j] == nums[k], j - i == k - j, and the value appears exactly three times in the array.

Hint 2

2a (Grouping Indices). Group occurrence indices by value. Count the groups containing exactly three indices whose two consecutive gaps are equal.

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4048. Count Values With Equally Spaced Occurrences I

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