CareerCup

Based on the question description we see that we need to consider only those integers whose value is = k since elements with value > k and value < k are rejected. So if some window has value k then answer is k else answer is 0. So the problem changes to check each window if it has k in it or not. To do so we can just keep a running window and store the last occurrence of k. Following is the code:

vector<int> average(vector<int> stream, int k, int w) {
	int i, p = 0;
	int n = stream.size();
	int lastIndexOfK = -1;
	vector<int> ans;
	for(i = 0; i < n; ++i) {
	   int p = i % w;
       if(stream[i] == k) {
           lastIndexOfK = p;
       }
       if(lastIndexOfK > p || (lastIndexOfK == 0 && p == 0)) {
          lastIndexOfK = -1;
       }
       if(lastIndexOfK != -1) {
           ans.push_back(k);
       }
	}
	return ans;
}