Top K frequent elements

Problem Statement

Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.

1. Hashing + Sorting

Construct frequency map of numbers in input.

Convert the key-value pairings into a array of 2 elements {count, number}.

Sort that array descending by comparing "count" in each pair.

Return "number" part of the first "k" pairs as answer.

2. Hashing + Heap

As “Top K” frequent elements asked, we should think of heap

The heap should be such that the highest frequency elements placed first and then the lower frequncey ones. In other words, a max-heap with the comparison basis as frequency of the number in input array

Construct frequency map of numbers in input

Create max-heap
Traverse frequency map:
    Add elements to max-heap such that they are compared by their frequencies

From the constructed heap, till "k" iterations:
    Keep popping the top-most element of heap and add to answer

Return answer

Assuming there are ’\(m\)’ unique elements, which is at worst ’\(n\)’.

Also, ’\(k\)’ will be ’\(m\)’ at most

Time:

• \( O(n) \) to traverse array elements and construct frequency map
• \( m \) insertions into max-heap is \( O(m \cdot logm) \) , which will be \( O(n \cdot logn) \) at worst
• \( k \) elements popped from max-heap takes \( O(k \cdot logm) \) , which will be \( O(n \cdot logn) \) at worst

Space:   \( n \) elements at worst in frequency map and heap and answer i.e. \( O(2n) \)

vector<int> topKFrequent (vector<int> &nums, int k) {
    // {number : count} map
    unordered_map<int, int> freq;
    // Construct frequency map
    for (int number : nums) {
        freq[number]++;
    }
    // Construct Max-heap where elements compared by count
    priority_queue<pair<int, int> > maxHp;
    for (auto &[number, count] : freq) {
        maxHp.push({count, number});
    }
    vector<int> ans;
    // Retrieve K top elements from heap
    while (k--) {
        // Pick off element at top of heap and add to answer
        ans.push_back(maxHp.top().second);
        maxHp.pop();
    }
    return ans;
}

3. Hashing + Buckets

The frequency count of any number in the input can range between: \( 0 \text{ to } n \)

We can create \( n \) buckets where each bucket represents elements having that count. The index of the bucket will be the frequency count of it’s elements

Construct frequency map

Create n+1 buckets (or n, ignoring the zero-count bucket)
Traverse frequency map:
    Append number to the bucket (index) representing that count

Traverse buckets from the end till "k" elements in answer:
    Add elements of bucket to answer

Example:

nums = [1,1,2,5,1,2,4], k = 2

n = 7
freqMap = { 1:3, 2:2, 5:1, 4:1 }

buckets[n + 1] array will have elements alikes: (index i.e. Count) -> Elements having that count
Here, buckets[8] will be:
    [ 0 ] -> [ ]
    [ 1 ] -> [ 5, 4 ]
    [ 2 ] -> [ 2 ]
    [ 3 ] -> [ 1 ]
    [ 4 ] -> [ ]
    [ 5 ] -> [ ]
    [ 6 ] -> [ ]
    [ 7 ] -> [ ]

Traversing from end and taking k=2 elements, ans=[1,2]

Time:

• \( O(n) \) to traverse array elements and construct frequency map
• \( m \) insertions into respective buckets is \( O(m) \) , which will be \( O(n) \) at worst
• Traverse buckets and append elements into answer: would be \( O(n) \) amortized

Space:   \( n \) elements at worst in frequency map and \( n \) buckets and combined elements of all buckets would be \( n \) at worst

vector<int> topKFrequent (vector<int> &nums, int k) {
    // {number : count} map
    unordered_map<int, int> freq;
    for (int number : nums) {
        freq[number]++;
    }

    int n = nums.size();
    // n+1 buckets where each bucket stores numbers having
    // frequency count equal to index of bucket
    vector<vector<int>> buckets(n + 1);
    for (auto &[number, count] : freq) {
        // Add number to the bucket representing that frequency count
        buckets[count].push_back(number);
    }

    vector<int> ans;
    for (int i = n; i >= 0; i--) {
        // Skip processing bucket if it's empty
        if (buckets[i].empty()) {
            continue;
        }
        // Append elemets of the bucket into answer (till k total items)
        for (int number : buckets[i]) {
            ans.push_back(number);
            if (ans.size() == k) {
                return ans;
            }
        }
    }
    return ans;
}