median

I am trying to find median of constant size array. But array is always uptading. I mean new numbers are replaced with old numbers. I call this process running median, or we can say on the fly median.. Here is my code and inside the code, when rand() function generates 78, the code cannot find the correct median. (Before 78; 41, 67, 34, 0, 69, 24 was generated) #include <iostream> #include <stdlib.h> #include <algorithm> #define MAX_SIZE 5 using namespace std; bool isOdd( int integer ) { if ( integer % 2 == 0 ) return false; else return true; } int main() { int median; int *minArray ; int

how do you setup minimum number of comparisons in general? James McNellis To cite Donald Knuth (by way of Wikipedia, since I don't have my copy of TAOCP at the moment), the lower bound for the number of comparisons is six: http://en.wikipedia.org/wiki/Selection_algorithm (scroll down to the section entitled "Lower Bounds"). Your goal is, effectively, to find the k lowest values where k is half the size of the list, rounded up, (so, k = 3; n = 5) and then take the max of those. Plugging that into the formula there on the page, you get: (5 - 3) + 1 + 1 + 2 = 6 In this case, the actual minimum

A way of finding the median of a given set of n numbers is to distribute them among 2 heaps. 1 is a max-heap containing the lower n/2 (ceil(n/2)) numbers and a min-heap containing the rest. If maintained in this way the median is the max of the first heap (along with the min of the second heap if n is even). Here's my c++ code that does this: priority_queue<int, vector<int> > left; priority_queue<int,vector<int>, greater<int> > right; cin>>n; //n= number of items for (int i=0;i<n;i++) { cin>>a; if (left.empty()) left.push(a); else if (left.size()<=right.size()) { if (a<=right.top()) left.push