#3092

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Most Frequent IDs

Track the most frequent ID after each update in a dynamic collection of IDs with changing frequencies.

array hash table heap priority queue ordered set

Problem Statement

You are given two integer arrays, nums and freq, both of length n. nums[i] represents an ID, and freq[i] indicates how many times this ID is added or removed from the collection at step i. If freq[i] is positive, the ID is added to the collection, and if freq[i] is negative, it is removed.

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.

Examples

Example 1

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

Output: [3,3,2,2]

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]

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.

Solution Approach

Array Scanning with Hash Lookup

The main idea is to traverse the nums and freq arrays while updating the frequency of each ID in a hash map. This allows quick updates and lookups of ID counts as changes are made. After each update, we can use the hash map to find the most frequent ID efficiently.

Ordered Set for Frequency Tracking

Using an ordered set or a priority queue can optimize the tracking of the most frequent ID. By maintaining the frequency counts in a sorted structure, you can efficiently get the maximum count after each operation. This will help with identifying the most frequent ID without needing to scan the entire hash map every time.

Efficient Handling of Empty Collection

Special attention needs to be given to cases where the collection becomes empty after a step. In such cases, the answer should be 0. Managing the collection’s emptiness requires careful handling during frequency updates to ensure accuracy in the result.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach used for tracking the frequencies. Using a hash map for frequency updates and a priority queue for retrieving the most frequent ID results in a time complexity of O(n log n). Space complexity is O(n) due to the storage requirements for the hash map or ordered set.

What Interviewers Usually Probe

Tests if the candidate can efficiently track changes in a dynamic collection.
Evaluates the candidate's ability to optimize frequent ID retrieval with advanced data structures.
Assesses knowledge of balancing time and space complexity in array-based problems.

Common Pitfalls or Variants

Common pitfalls

Not handling the case where the collection becomes empty and the result should be 0.
Using inefficient data structures that cause unnecessary time complexity for frequent ID retrieval.
Overcomplicating the solution when a simpler hash map or ordered set approach would suffice.

Follow-up variants

Tracking the most frequent ID in a dynamic stream of IDs.
Handling both positive and negative frequency changes in other collection types.
Optimizing for time complexity while tracking multiple frequent items at once.

FAQ

What is the main challenge in solving the Most Frequent IDs problem?

The main challenge is efficiently tracking the most frequent ID after each update, especially when the collection changes frequently. Using hash maps or ordered sets can help.

How does the frequency change affect the solution?

The frequency change, whether positive or negative, determines how the IDs are added or removed from the collection. This directly impacts the most frequent ID count.

What data structures are best for solving Most Frequent IDs?

Hash maps are great for storing frequency counts, while ordered sets or priority queues are useful for efficiently retrieving the most frequent ID.

How can I optimize my solution for time complexity?

To optimize time complexity, maintain frequency counts in an ordered data structure like a priority queue, which allows efficient retrieval of the most frequent ID.

What happens if the collection becomes empty?

If the collection becomes empty after a step, the result for that step should be 0. This case needs to be handled explicitly during frequency updates.

terminal

Solution

Solution 1: Hash Table + Priority Queue (Max Heap)

We use a hash table $cnt$ to record the occurrence times of each ID, a hash table $lazy$ to record the number of times each occurrence needs to be deleted, and a priority queue $pq$ to maintain the maximum occurrence times.

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class Solution:
    def mostFrequentIDs(self, nums: List[int], freq: List[int]) -> List[int]:
        cnt = Counter()
        lazy = Counter()
        ans = []
        pq = []
        for x, f in zip(nums, freq):
            lazy[cnt[x]] += 1
            cnt[x] += f
            heappush(pq, -cnt[x])
            while pq and lazy[-pq[0]] > 0:
                lazy[-pq[0]] -= 1
                heappop(pq)
            ans.append(0 if not pq else -pq[0])
        return ans

Continue Practicing

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