Find kth largest element in an array, two different priority_queue solutions time complexity

Question

I'm interested in two solutions using priority_queue specifically. Although they both use priority_queue, I think they have different time complexity.

Solution 1:

int findKthLargest(vector<int>& nums, int k) {
        priority_queue<int> pq(nums.begin(), nums.end()); //O(N)
        for (int i = 0; i < k - 1; i++) //O(k*log(k))
            pq.pop(); 
        return pq.top();
}

Time Complexity: O(N) + O(k*log(k))

EDIT: sorry, it should be O(N) + O(k*log(N)) thanks for pointing out!

Solution 2:

int findKthLargest(vector<int>& nums, int k) {
    priority_queue<int, vector<int>, greater<int>> p;
    int i = 0;
    while(p.size()<k) {

        p.push(nums[i++]);
    }

    for(; i<nums.size(); i++) {

        if(p.top()<nums[i]){
            p.pop();
            p.push(nums[i]);
        }

    }

    return p.top();
}

Time Complexity: O(N*log(k))

So in most cases the 1st solution is much better than the 2nd?

Answer 1

In the first case the complexity is O(n)+klog(n) not O(n)+klog(k) as there are n elements in the heap. In the worst case, k can be as large as n, so for unbounded data O(nlog(n)) is the correct worst case complexity.

In the second case, there is never more than k items in the priority queue, so the complexity is O(nlog(k)) and again for unbounded data k can be as large as n, so it is O(nlog(n)).

For smaller k, the second code will run faster, but as k becomes larger the first code becomes faster. I made some experiments, here are the results:

k=1000
Code 1 time:0.123662
998906057
Code 2 time:0.03287
998906057
========
k=11000
Code 1 time:0.137448
988159929
Code 2 time:0.0872
988159929
========
k=21000
Code 1 time:0.152471
977547704
Code 2 time:0.131074
977547704
========
k=31000
Code 1 time:0.168929
966815132
Code 2 time:0.168899
966815132
========
k=41000
Code 1 time:0.185737
956136410
Code 2 time:0.205008
956136410
========
k=51000
Code 1 time:0.202973
945313516
Code 2 time:0.236578
945313516
========
k=61000
Code 1 time:0.216686
934315450
Code 2 time:0.27039
934315450
========
k=71000
Code 1 time:0.231253
923596252
Code 2 time:0.293189
923596252
========
k=81000
Code 1 time:0.246896
912964978
Code 2 time:0.321346
912964978
========
k=91000
Code 1 time:0.263312
902191629
Code 2 time:0.343613
902191629
========

I modified the second code a little bit to make to similar to code1:

int findKthLargest2(vector<int>& nums, int k) {
    double st=clock();
    priority_queue<int, vector<int>, greater<int>> p(nums.begin(), nums.begin()+k);

    int i=k;
    for(; i<nums.size(); i++) {
        if(p.top()<nums[i]){
            p.pop();
            p.push(nums[i]);
        }

    }
    cerr<<"Code 2 time:"<<(clock()-st)/CLOCKS_PER_SEC<<endl;
    return p.top();
}
int findKthLargest1(vector<int>& nums, int k) {
        double st=clock();
        priority_queue<int> pq(nums.begin(), nums.end()); //O(N)
        for (int i = 0; i < k - 1; i++) //O(k*log(k))
            pq.pop(); 

        cerr<<"Code 1 time:"<<(clock()-st)/CLOCKS_PER_SEC<<endl;
        return pq.top();
}

int main() {   

 READ("in");
 vector<int>v;
 int n;
 cin>>n;
 repl(i,n)
 {
     int x;
     scanf("%d",&x);
     v.pb(x);
 }

 for(int k=1000;k<=100000;k+=10000)
 {
     cout<<"k="<<k<<endl;
    cout<<findKthLargest1(v,k)<<endl;
    cout<<findKthLargest2(v,k)<<endl;
    puts("========");
  }
}

I used 1000000 random integers between 0 to 10^9 as dataset, generated by C++ rand() function.

Answer 2

well, no the first is O(N)+O(k*log(N)) because the pop is O(log(N))

int findKthLargest(vector<int>& nums, int k) {
        priority_queue<int> pq(nums.begin(), nums.end()); //O(N)
        for (int i = 0; i < k - 1; i++) //O(k*log(N))
            pq.pop(); // this line is O(log(N))
        return pq.top();
}

it's still better than the second in most cases.