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Solution: Top K Frequent Elements

Statement

Naive approach

The naive approach to finding the k most frequent elements begins by counting how often each element appears in the input array. To do this, we iterate through the array one element at a time and keep track of counts in a hash map. If an element is seen for the first time, we add it to the map with a count of 1. If it already exists in the map, we increment its count by 1. By the end of this pass, the hash map contains every distinct element from the array as keys and their corresponding frequencies as values.

After building the frequency map, the algorithm repeatedly searches for the element with the highest count. In each round, it scans through the map to find the maximum frequency, adds that element to the result list, and removes it from the map. This process is repeated until k elements have been chosen.

In terms of efficiency, constructing the frequency map requires scanning all n elements once, while finding the top k elements requires k scans over the ...

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Problem
Submissions
Solution

Solution: Top K Frequent Elements

Statement

Naive approach

The naive approach to finding the k most frequent elements begins by counting how often each element appears in the input array. To do this, we iterate through the array one element at a time and keep track of counts in a hash map. If an element is seen for the first time, we add it to the map with a count of 1. If it already exists in the map, we increment its count by 1. By the end of this pass, the hash map contains every distinct element from the array as keys and their corresponding frequencies as values.

After building the frequency map, the algorithm repeatedly searches for the element with the highest count. In each round, it scans through the map to find the maximum frequency, adds that element to the result list, and removes it from the map. This process is repeated until k elements have been chosen.

In terms of efficiency, constructing the frequency map requires scanning all n elements once, while finding the top k elements requires k scans over the ...