|
| 1 | +/* |
| 2 | +In this question we divide the already entered elements into 2 halves |
| 3 | +the elements in the half with smaller elemenrs go into a max heap |
| 4 | +the elememts in the half with bigger elemenrs go in min heap |
| 5 | +If there are even no of elements we use biggest element of smaller half and |
| 6 | +smaller element of bigger half to compute the median |
| 7 | +In case of odd no of elements we keep 1 extra element in the max heap or min heap |
| 8 | +and use the heap with more elements to calculate median |
| 9 | +*/ |
| 10 | +#include<bits/stdc++.h> |
| 11 | +using namespace std; |
1 | 12 |
|
| 13 | +priority_queue<int, vector<int>, greater <int> > min_heap; //making a min heap |
| 14 | +priority_queue<int> max_heap; //making a max heap |
| 15 | + |
| 16 | +int absh(int x){ //function to return absolute value of an int |
| 17 | + if(x>0) return x; |
| 18 | + else return (-1*x); |
| 19 | +} |
| 20 | + |
| 21 | +void add(int a) //function to add a number |
| 22 | +{ |
| 23 | + if( max_heap.size() && a >= max_heap.top()) //if the number is bigger than the current median |
| 24 | + min_heap.push(a); //add in the min heap |
| 25 | + else |
| 26 | + max_heap.push(a); // otherwise max heap |
| 27 | + |
| 28 | + if(absh(max_heap.size() - min_heap.size()) > 1) //if the difference between the number of elements |
| 29 | + { //larger than 1 adjust elements within the heaps to make it 0 |
| 30 | + if(max_heap.size() > min_heap.size()) //if max heap has more elements transfer one to min heap |
| 31 | + { |
| 32 | + int temp = max_heap.top(); |
| 33 | + max_heap.pop(); |
| 34 | + min_heap.push(temp); |
| 35 | + } |
| 36 | + else //if min heap has more elements transfer it to max heap |
| 37 | + { |
| 38 | + int temp = min_heap.top(); |
| 39 | + min_heap.pop(); |
| 40 | + max_heap.push(temp); |
| 41 | + } |
| 42 | + } |
| 43 | +} |
| 44 | + |
| 45 | +double get_median() //function to return median with current elements |
| 46 | +{ |
| 47 | + int total = min_heap.size() + max_heap.size(); |
| 48 | + double ret; |
| 49 | + if(total%2 == 1) //check if the no of elements is odd, if it is odd we'll have a unique median |
| 50 | + { |
| 51 | + if(max_heap.size() > min_heap.size()) //checking which of the 2 heaps has the median |
| 52 | + ret = max_heap.top(); |
| 53 | + else |
| 54 | + ret = min_heap.top(); |
| 55 | + } |
| 56 | + else //if it is even we have to take the mean of 2 central elements |
| 57 | + { |
| 58 | + ret = 0; |
| 59 | + if(max_heap.empty() == false) |
| 60 | + ret += max_heap.top(); //adding 1st central element |
| 61 | + if(min_heap.empty() == false) |
| 62 | + ret += min_heap.top(); //adding 2nd central element |
| 63 | + ret/=2; // taking mean of both |
| 64 | + } |
| 65 | + return ret; |
| 66 | +} |
| 67 | + |
| 68 | +int main() |
| 69 | +{ |
| 70 | + cout << setprecision(1) << fixed; |
| 71 | + int n, a; |
| 72 | + cin >> n; |
| 73 | + for(int i = 1; i<=n; i++) |
| 74 | + { |
| 75 | + cin >> a; |
| 76 | + add(a); |
| 77 | + cout << get_median() << endl; |
| 78 | + } |
| 79 | +} |
0 commit comments