Is Most Frequent IDs easy or hard?

Most Frequent IDs is considered a medium-level problem. The main difficulty is managing dynamic updates while always returning the correct maximum frequency. Understanding how to combine hash maps with heaps or ordered structures is the key concept.

Most Frequent IDs Python/Java solution

In Python, the problem is typically solved using a dictionary for frequency storage and the heapq module for a priority queue with lazy deletion. In Java, developers use HashMap for frequencies and PriorityQueue for the max heap. Both implementations achieve O(n log n) time complexity.

How to solve Most Frequent IDs in O(n)?

Achieving strict O(n) time is difficult because every update potentially changes the ordering of frequencies. Data structures like heaps or balanced trees are needed to track the maximum efficiently, leading to O(n log n) complexity. The hash map plus max heap approach is considered optimal for this problem.

What is the best approach for Most Frequent IDs?

The most practical solution uses a hash map to store the current frequency of each ID and a max heap (priority queue) to track the largest frequency. Each update pushes a new frequency into the heap and stale entries are removed lazily. This keeps updates efficient with O(log n) heap operations and O(n log n) total complexity.

Is Most Frequent IDs asked at Google/Amazon/Meta?

Problems involving dynamic frequency tracking and priority queues appear frequently in interviews at companies like Amazon, Google, and Meta. Variants that combine hash maps with heaps to maintain top elements are especially common in system and algorithm interviews.

What data structure is used in Most Frequent IDs?

The core data structures are a hash table for tracking the current frequency of each ID and a max heap (priority queue) for retrieving the highest frequency quickly. Some implementations also use ordered sets or balanced binary search trees to maintain sorted frequencies.

What is the time complexity of Most Frequent IDs?

The optimal solution runs in O(n log n) time. Each update modifies a hash map entry in O(1) and inserts a new value into a max heap in O(log n). Lazy removal of outdated heap entries keeps the total heap operations bounded by the number of updates.

Description Hints 3 Solution Video 31:59

3092. Most Frequent IDs

Medium43.0% acceptance2 companies

The problem involves tracking the frequency of IDs in a collection that changes over time. You have two integer arrays, nums and freq, of equal length n. Each element in nums represents an ID, and the corresponding element in freq indicates how many times that ID should be added to or removed from the collection at each step.

Addition of IDs: If freq[i] is positive, it means freq[i] IDs with the value nums[i] are added to the collection at step i.
Removal of IDs: If freq[i] is negative, it means -freq[i] IDs with the value nums[i] are removed from the collection at step i.

Return an array ans of length n, where ans[i] represents the count of the most frequent ID in the collection after the i^th step. If the collection is empty at any step, ans[i] should be 0 for that step.

Example 1:

Input: nums = [2,3,2,1], freq = [3,2,-3,1]

Output: [3,3,2,2]

Explanation:

After step 0, we have 3 IDs with the value of 2. So ans[0] = 3.
After step 1, we have 3 IDs with the value of 2 and 2 IDs with the value of 3. So ans[1] = 3.
After step 2, we have 2 IDs with the value of 3. So ans[2] = 2.
After step 3, we have 2 IDs with the value of 3 and 1 ID with the value of 1. So ans[3] = 2.

Example 2:

Input: nums = [5,5,3], freq = [2,-2,1]

Output: [2,0,1]

Explanation:

After step 0, we have 2 IDs with the value of 5. So ans[0] = 2.
After step 1, there are no IDs. So ans[1] = 0.
After step 2, we have 1 ID with the value of 3. So ans[2] = 1.

Constraints:

1 <= nums.length == freq.length <= 10⁵
1 <= nums[i] <= 10⁵
-10⁵ <= freq[i] <= 10⁵
freq[i] != 0
The input is generated such that the occurrences of an ID will not be negative in any step.

Asked at 2 companies

Snowflake Amazon

Topics

Hints open one at a time

Hint 1

Use an ordered set for maintaining the occurrences of each ID.

Hint 2

After step i find the occurrences of nums[i].

Hint 3

Change the occurrences of nums[i] in the ordered set.

31:59

Still stuck?Read 3 approaches, or watch a walkthrough.

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1class Solution:
2    def mostFrequentIDs(self, nums: List[int], freq: List[int]) -> List[int]:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

[2,3,2,1]
[3,2,-3,1]

Input

[5,5,3]
[2,-2,1]

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