sum of top k items in a steam

[ishu9bansal]

It will be implemented by a priority queue of size k.

Example use case https://leetcode.com/contest/biweekly-contest-93/problems/maximum-star-sum-of-a-graph/

[ishu9bansal]

aggregation of top k elements in a stream or a floating window, with different flexibility. It is possible in O(n) where n is the size of total iteration. With query and updates in O(1) average during the iteration. This impl achieves the query and updates in O(log n) https://leetcode.com/submissions/detail/1158279622/

class topK{
private:
    multiset<int> l,r;
    int k;
    long long sum;
public:
    topK(int K=0) : k(K), sum(0) {
        if(k<0) k = 0;
    }
    long long get() {
        return sum;
    }
    void add(int x) {
        l.insert(-x);   // trick for sorting in descending order
        sum += x;   // update aggregator
        int y;
        while(l.size()>k){  // maintain container size
            y = *l.begin();
            sum += y;   // adjust aggregator before removal
            l.erase(l.begin()); // remove extra and move it to other container
            r.insert(-y);    // reversing again to sort in ascending order
        }
    }
    void remove(int x) {
        auto it = l.find(-x);   // find in primary container
        if(it!=l.end()){    // if found
            int y = *it;
            sum += y;   // adjust aggregator before removal
            l.erase(it);    // remove the required and expect from other container
            while(l.size()<k && r.size()>0) {   // container size check
                y = *r.begin(); // get from secondary container
                l.insert(-y);   // insert in primary container
                sum += y;       // update aggregator
                r.erase(r.begin()); // erase from secondary container
            }
        } else {    // if not found in primary search and remove from secondary
            auto jt = r.find(x);
            if(jt!=r.end()) r.erase(jt);
        }
    }
    int expectation() {
        // this number tells that how many elements short we are to complete the k elements
        // if number is negative, it means that we overshoot and want to drop a few elements
        return k-l.size()-r.size();
    }
};

[ishu9bansal]

https://leetcode.com/problems/divide-an-array-into-subarrays-with-minimum-cost-ii/submissions/1158327361/

abstraction of the aggregation methods

template <class T>
class aggregator{
protected:
    T aggregate;
public:
    virtual void add(int, int) = 0;
    virtual void remove(int, int) = 0;
    T get() {
        return aggregate;
    }
};

class Sum: public aggregator<long long> {
public:
    Sum() {
        this->aggregate = 0;
    }
    void add(int x, int bound) {
        aggregate += x;
    }
    void remove(int x, int bound) {
        aggregate -= x;
    }
};

class Count: public aggregator<int> {
public:
    Count() {
        this->aggregate = 0;
    }
    void add(int x, int bound) {
        aggregate++;
    }
    void remove(int x, int bound) {
        aggregate--;
    }
};

class Max: public aggregator<int> {
public:
    Max() {
        this->aggregate = INT_MIN;
    }
    void add(int x, int bound) {
        aggregate = bound;
    }
    void remove(int x, int bound) {
        aggregate = bound;
    }
};
class topK{
private:
    multiset<int> l,r;
    int k;
    Sum sum;
    Count count;
    Max max;
    void aggregateAdd(int x) {
        int bound = 0;
        if(l.size())  bound = -*l.begin();
        sum.add(x,bound);
        count.add(x,bound);
        if(l.empty())   bound = INT_MIN;
        max.add(x,bound);
    }
    void aggregateRemove(int x) {
        int bound = 0;
        if(l.size())  bound = -*l.begin();
        sum.remove(x,bound);
        count.remove(x,bound);
        if(l.empty())   bound = INT_MIN;
        max.remove(x,bound);
    }
public:
    topK(int K=0) : k(K) {
        if(k<0) k = 0;
    }
    long long getSum() {
        return sum.get();
    }
    int getCount() {
        return count.get();
    }
    int getMax() {
        return max.get();
    }
    void add(int x) {
        l.insert(-x);   // trick for sorting in descending order
        aggregateAdd(x);   // update aggregator, positioning is important!
        int y;
        while(l.size()>k){  // maintain container size
            y = -*l.begin();
            l.erase(l.begin()); // remove extra and move it to other container
            aggregateRemove(y);   // adjust aggregator after removal, positioning is important!
            r.insert(y);    // reversing again to sort in ascending order
        }
    }
    void remove(int x) {
        auto it = l.find(-x);   // find in primary container
        if(it!=l.end()){    // if found
            int y = -*it;
            l.erase(it);    // remove the required and expect from other container
            aggregateRemove(y);   // adjust aggregator after removal, positioning is important!
            while(l.size()<k && r.size()>0) {   // container size check
                y = *r.begin(); // get from secondary container
                l.insert(-y);   // insert in primary container
                aggregateAdd(y);       // update aggregator, positioning is important!
                r.erase(r.begin()); // erase from secondary container
            }
        } else {    // if not found in primary search and remove from secondary
            auto jt = r.find(x);
            if(jt!=r.end()) r.erase(jt);
        }
    }
};

[ishu9bansal]

Possible improvements

• Interpretation of k=0?
• Stream of numbers with top k limit, add option to make right - r set optional so simulate a stream (we don't need remove method in that case)
• Add possibility of adding more aggregations, custom aggregations.
• Improve the aggregation plug and play.
• Major improvement, bring the O(n) version. Convert this to queue based implementation
• Flexibility of sort direction
• Flexibility of other data classes, why only int?