Most Frequent IDs

Medium

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Arrays

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the data type of the IDs? Can they be integers, strings, or something else?
What is the expected size of the input data (number of IDs)? Is memory a constraint?
If multiple IDs have the same highest frequency, how should the result be returned? Return all, or just one?
What should be returned if the input list of IDs is empty or null?
Is there a defined range for the frequency of any specific ID? Could all IDs be unique?

Brute Force Solution

Approach

To find the most frequent IDs using a brute force method, we'll count how many times each ID appears by checking it against all the others. We will go through each ID in the list and compare it to every other ID in the list to count the matches.

Here's how the algorithm would work step-by-step:

Take the first ID from the list.
Compare that ID to every other ID in the list, including itself.
Each time you find a match, increase a counter for that specific ID.
Once you've compared the first ID to all the others, record the final count for that ID.
Repeat these steps for the second ID, then the third ID, and so on, until you have checked every ID in the list.
Once you have all the counts, find the ID (or IDs) with the highest count. Those are the most frequent IDs.

Code Implementation

def find_most_frequent_ids_brute_force(list_of_ids):
    id_counts = {}

    for current_id in list_of_ids:
        count = 0

        # Iterate through all IDs to compare
        for other_id in list_of_ids:

            # Count occurrences of the current ID
            if current_id == other_id:
                count += 1

        id_counts[current_id] = count

    max_frequency = 0
    most_frequent_ids = []

    # Find maximum frequency to optimize result selection
    for current_id, frequency in id_counts.items():
        if frequency > max_frequency:
            most_frequent_ids = [current_id]
            max_frequency = frequency

        # Handle ties for most frequent
        elif frequency == max_frequency:
            most_frequent_ids.append(current_id)

    return most_frequent_ids

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n IDs in the input list. For each ID, it compares it to every other ID in the list, which involves n comparisons. Therefore, the total number of comparisons is proportional to n * n. The dominant term in this calculation is n², so the Big O time complexity is O(n²).

Space Complexity

O(1) – The brute force method, as described, only uses a counter variable for each ID and a variable to track the maximum count encountered. These variables require constant space, irrespective of the number of IDs (N) in the input list. Therefore, the auxiliary space complexity is O(1), indicating constant space usage.

Optimal Solution

Approach

To find the most frequent IDs efficiently, we'll count how often each ID appears. Then, we'll quickly identify the ID or IDs that show up the most.

Here's how the algorithm would work step-by-step:

First, go through the list of IDs and count how many times each ID appears. Think of it like making a tally sheet for each unique ID.
Next, find the highest count from your tally sheet. This represents the maximum number of times any ID appeared.
Finally, go back through your tally sheet and identify all the IDs that have this maximum count. These are your most frequent IDs. If only one ID has this maximum count, then it is the most frequent one.

Code Implementation

def find_most_frequent_ids(list_of_ids):
    id_counts = {}
    for id_value in list_of_ids:
        if id_value in id_counts:
            id_counts[id_value] += 1
        else:
            id_counts[id_value] = 1

    # Find the maximum count to identify most frequent IDs.
    maximum_count = 0
    for id_value in id_counts:
        if id_counts[id_value] > maximum_count:
            maximum_count = id_counts[id_value]

    most_frequent = []
    # Collect all IDs that appear with the maximum frequency.
    for id_value in id_counts:

        if id_counts[id_value] == maximum_count:

            most_frequent.append(id_value)

    return most_frequent

Big(O) Analysis

Time Complexity

O(n) – The solution involves iterating through the list of IDs once to count the frequency of each ID, where n is the number of IDs. Then, finding the maximum count requires iterating through the counts, which is proportional to the number of unique IDs which is at most n. Finally, iterating through the counts again to find the IDs with the maximum count also takes at most n time. Therefore, the overall time complexity is dominated by the initial count and subsequent lookups, all of which are linear with respect to n.

Space Complexity

O(N) – The algorithm uses a tally sheet to count the occurrences of each unique ID. In the worst-case scenario, all N IDs in the input list are unique, resulting in a hash map (the tally sheet) storing N distinct key-value pairs where each key is an ID and each value is a count. Additionally, to store the most frequent IDs, another data structure is required with a maximum size of N in the worst case, where all IDs have the same frequency. Therefore, the auxiliary space used is proportional to N, leading to a space complexity of O(N).

Edge Cases

Case	How to Handle
Empty input list	Return an empty list or a predefined value (e.g., null) indicating no frequent IDs exist.
Input list with only one ID	Return the single ID if the frequency requirement is met (e.g., finding IDs that occur more than 0 times), or an empty list otherwise.
Input list where all IDs are unique	Return an empty list if the frequency threshold is not met by any ID.
Input list where all IDs are the same	Return the single ID if its frequency meets or exceeds the threshold.
Large input list exceeding available memory	Consider using streaming algorithms or external memory techniques to process the data in chunks.
Integer overflow when counting frequencies for very large datasets	Use larger data types (e.g., long) or consider alternative approaches to prevent overflow.
Multiple IDs having the exact same maximum frequency	Return all IDs with the maximum frequency, maintaining a consistent order (e.g., sorted).
Negative ID values	The chosen data structure (e.g., hash map) should correctly handle negative ID values without unexpected behavior.

Empty input list

How to Handle:

Return an empty list or a predefined value (e.g., null) indicating no frequent IDs exist.

Input list with only one ID

How to Handle:

Return the single ID if the frequency requirement is met (e.g., finding IDs that occur more than 0 times), or an empty list otherwise.

Input list where all IDs are unique

How to Handle:

Return an empty list if the frequency threshold is not met by any ID.

Input list where all IDs are the same

How to Handle:

Return the single ID if its frequency meets or exceeds the threshold.

Large input list exceeding available memory

How to Handle:

Consider using streaming algorithms or external memory techniques to process the data in chunks.

Integer overflow when counting frequencies for very large datasets

How to Handle:

Use larger data types (e.g., long) or consider alternative approaches to prevent overflow.

Multiple IDs having the exact same maximum frequency

How to Handle:

Return all IDs with the maximum frequency, maintaining a consistent order (e.g., sorted).

Negative ID values

How to Handle:

The chosen data structure (e.g., hash map) should correctly handle negative ID values without unexpected behavior.