Introduction

In this blog post, we will explore how to solve the LeetCode problem "3092" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 ith 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.

Explanation

Here's the breakdown of the solution:

• Maintain a frequency map: Use a dictionary to store the current frequency of each ID.
• Iterate and update: For each step, update the frequency of the corresponding ID.

• Track the maximum: After each update, find the maximum frequency among all IDs and store it in the result array.

• Runtime Complexity: O(n*m), where n is the length of nums and freq arrays and m is the number of unique IDs encountered. In the worst case, m can be equal to n. Storage Complexity: O(m), where m is the number of unique IDs encountered. In the worst case, m can be equal to n.

Code

    def most_frequent_id(nums, freq):
    """
    Calculates the count of the most frequent ID in a collection after each step.

    Args:
        nums: An array of IDs.
        freq: An array of frequencies, indicating how many times each ID
            should be added or removed at each step.

    Returns:
        An array where each element represents the count of the most frequent
        ID after each step.
    """
    n = len(nums)
    ans = []
    counts = {}
    for i in range(n):
        id_val = nums[i]
        freq_val = freq[i]

        counts[id_val] = counts.get(id_val, 0) + freq_val

        max_freq = 0
        for count in counts.values():
            max_freq = max(max_freq, count)

        ans.append(max_freq)

    return ans