How do you solve Most Frequent IDs on LeetCode?

Most Frequent IDs (LeetCode #3092) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using a hash map allows for efficient frequency counting.

What is the brute force solution for Most Frequent IDs?

The brute force approach for Most Frequent IDs is Brute Force, with O(n²) time complexity and O(n) space complexity. In this approach, we will iterate through the collection after each step and count the occurrences of each ID to determine the most frequent one. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Most Frequent IDs?

Most Frequent IDs uses the following concepts: Array, Hash Table, Heap (Priority Queue), Ordered Set. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Most Frequent IDs?

Clarify the problem requirements and constraints before coding. Think about edge cases, such as empty collections or all negative frequencies. Explain your thought process while coding to demonstrate your understanding.

What are common mistakes when solving Most Frequent IDs?

Failing to update the frequency map correctly. Not handling the case when the collection becomes empty.

#3092

Most Frequent IDs

Medium

ArrayHash TableHeap (Priority Queue)Ordered SetHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(n)	O(n)

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Intuition

Time O(n log n)Space O(n)

This approach uses a hash map to efficiently track the frequencies of IDs and a max-heap to quickly retrieve the most frequent ID. This reduces the need to repeatedly scan the entire collection.

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Algorithm

5 steps

1Step 1: Initialize a frequency map to track occurrences of each ID.
2Step 2: Use a max-heap to keep track of the most frequent IDs.
3Step 3: For each step i, update the frequency map based on freq[i].
4Step 4: Update the max-heap and retrieve the maximum frequency efficiently.
5Step 5: Append the maximum frequency to ans.

solution.py15 lines

1import heapq
2from collections import defaultdict
3
4def most_frequent_ids(nums, freq):
5    ans = []
6    collection = defaultdict(int)
7    max_heap = []
8    for i in range(len(nums)):
9        collection[nums[i]] += freq[i]
10        if freq[i] > 0:
11            heapq.heappush(max_heap, (-collection[nums[i]], nums[i]))
12        while max_heap and -max_heap[0][0] != collection[max_heap[0][1]]:
13            heapq.heappop(max_heap)
14        ans.append(-max_heap[0][0] if max_heap else 0)
15    return ans

ℹ

Complexity note: The time complexity is O(n log n) due to the use of a max-heap to maintain the most frequent IDs, where n is the number of steps. Each insertion and removal operation in the heap takes log n time.

1Using a hash map allows for efficient frequency counting.
2A max-heap helps in quickly retrieving the most frequent ID.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.