Problem Description

The task is to manage a collection of IDs that undergoes changes over time based on two input arrays, nums and freq, which are of equal length n. Each element in nums represents an ID, and the corresponding element in freq dictates how many IDs are added to or removed from the collection at each step.

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

The goal is to return an array ans of length n, where ans[i] is 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.

Intuition

To solve this problem, we need an efficient way to keep track of the frequency of IDs and determine the most frequent one after each operation. This involves using a combination of data structures:

1. Hash Table (cnt): This will track the current frequency of each ID in the collection. For each operation, we update the frequency of the ID accordingly (increase or decrease).

2. Lazy Deletion with Hash Table (lazy): This auxiliary structure helps manage frequencies that need to be archived. Whenever an ID frequency is updated, it means an old frequency must be decreased since a different frequency becomes relevant. We use this table to delay the deletion of out-of-date frequencies in our max-heap.

3. Priority Queue (Max Heap): Maintaining a max-heap of frequencies allows real-time access to the highest current frequency. Since Python's heapq is a min-heap by default, we store negative values of frequencies to effectively simulate a max-heap.

During each step:

• Update the cnt for the current ID according to freq.
• Reflect changes in lazy to account for past frequencies that are now outdated.
• Push the new frequency onto the heap.
• Remove elements from the heap whose frequencies are no longer valid as indicated by lazy.
• Append the top of the heap (maximum frequency) to ans, unless the heap is empty (in which case append 0).

This approach ensures we efficiently track the most frequent ID throughout a series of additions and removals.

Learn more about Heap (Priority Queue) patterns.

Solution Approach

The solution approach leverages a combination of data structures: a hash table, a priority queue, and a lazy deletion mechanic. Here's a step-by-step walkthrough:

1. Data Structures:

   • Hash Table (cnt): Utilized to store the frequency count of each ID. It is updated for every operation to reflect the current number of each ID in the collection.
   • Lazy Deletion Hash Table (lazy): Used to keep track of obsolete frequencies that need removal from the priority queue. This helps in efficiently maintaining the correct state of the max-heap.
   • Priority Queue (Max Heap): Implemented with a min-heap by using negative frequencies; this allows quick access to the ID with the maximum count after any operation.

2. Algorithm:

   • Iterate Over Every Operation: For each pair of ID and frequency from nums and freq:
     • Update lazy: Increment the count of the current frequency of ID in lazy to mark this frequency for potential removal.
     • Update Frequency in cnt: Adjust the count of ID in cnt using the current frequency (freq[i]).
     • Push New Frequency to Heap: Add the updated negative frequency of ID to the priority queue (pq).
     • Lazy Deletion in Priority Queue: While the top element in the priority queue is outdated (as indicated by lazy), remove it:
       • Decrement the count in lazy for the current top element and pop it from the queue.
     • Update Answer Array: Append the current most frequent count (negative of the top of the heap) to the resulting array, or append 0 if the heap is empty.

3. Efficiency:

   • The lazy deletion helps manage the validity of the top element in the priority queue without requiring the queue to be reconstructed from scratch after each update.
   • This ensures the algorithm can efficiently handle a series of operations, maintaining an overall time complexity of approximately (O(n \log n)), where (n) is the length of the nums and freq arrays.

This solution effectively manages the dynamic nature of the collection, ensuring the current most frequent ID can be retrieved efficiently at each step.

Example Walkthrough

Let's walk through a simple example using the solution approach described above.

Given:

• nums = [1, 2, 1, 3]
• freq = [3, 2, -1, 1]

Let's break down each step according to the solution approach:

1. Initialize Data Structures:

   • cnt = {}: A hash table to store current frequencies of IDs.
   • lazy = {}: A hash table for lazy deletion.
   • pq = []: An empty priority queue (implemented with a min-heap using negative values).
   • ans = []: Result array to store the most frequent count.

2. Step-by-Step Execution:

   • Step 0 (nums[0] = 1, freq[0] = 3):

     • Update the frequency of ID 1: cnt[1] = 3.
     • Add -3 to the priority queue: pq = [-3].
     • No elements to remove in lazy deletion.
     • Max frequency is 3 (top of pq), so ans = [3].

   • Step 1 (nums[1] = 2, freq[1] = 2):

     • Update the frequency of ID 2: cnt[2] = 2.
     • Add -2 to the priority queue: pq = [-3, -2].
     • Remove outdated frequencies using lazy, if any.
     • Max frequency remains 3 (value of cnt[1]), hence ans = [3, 3].

   • Step 2 (nums[2] = 1, freq[2] = -1):

     • Update lazy: lazy[cnt[1]] = 1 (cnt[1] is now outdated, frequency decreased from 3 to 2).
     • Reduce the frequency of ID 1: cnt[1] = 2.
     • Add -2 to the priority queue: pq = [-3, -2, -2].
     • Perform lazy deletion: Remove -3 from pq because it is outdated (as indicated by lazy), decrement lazy[-3].
     • Max frequency is 2, hence ans = [3, 3, 2].

   • Step 3 (nums[3] = 3, freq[3] = 1):

     • Update the frequency of ID 3: cnt[3] = 1.
     • Add -1 to the priority queue: pq = [-2, -2, -1].
     • No further elements require removal in this step.
     • Max frequency is still 2, hence ans = [3, 3, 2, 2].

The final result array ans represents the most frequent ID count after each step.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n * log n). This complexity arises due to the use of a priority queue (heap) to track the most frequent IDs. Every insertion and removal operation from the heap takes O(log n) time, and since such operations are performed for each element in nums (where n is the length of nums), the overall complexity results in O(n * log n).

The space complexity of the code is O(n). This is because the code utilizes a Counter to store frequencies of IDs, a separate lazy counter for tracking lazy updates, and a priority queue pq. Each of these data structures can store up to n elements, leading to a total space usage of O(n).

Learn more about how to find time and space complexity quickly.