Sliding Window Median in C++

C++Server Side ProgrammingProgramming

Suppose we have a list of numbers, and we have one window size k, we have to find the list of medians using sliding window manner. So, if the distribution is like below −

Window PositionMedian
13-1-353681
13-1-35368-1
13-1-35368-1
13-1-353683
13-1-353685
13-1-353686

Here we have considered the k is 3, and the result will be [1,-1,-1,3,5,6]

To solve this, we will follow these steps −

  • Define one set arr
  • Define a function insert(), this will take x,
  • insert x into arr
  • Define a function delete_(), this will take x,
  • delete x from arr, if this exists
  • Define a function getMedian()
  • n := size of arr
  • a := jump to n/2 – 1 step forward of first element of arr, and get the value
  • b := jump to n/2 step forward of first element of arr, and get the value
  • if size of arr, then −
    • return b
  • return (a + b) * 0.5
  • From the main method do the following
  • Define an array ans
  • clear the array arr
  • for initialize i := 0, when i < k, update (increase i by 1), do −
    • call the function insert(nums[i])
  • for initialize i := k, j := 0, when i < size of nums, update (increase i by 1), (increase j by 1), do −
    • insert returned value of getMedian() at the end of ans
    • call the function delete_(nums[j])
    • call the function insert(nums[i])
  • insert returned value of getMedian() at the end of ans
  • return ans

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
   cout << "[";
   for(int i = 0; i<v.size(); i++){
      cout << v[i] << ", ";
   }
   cout << "]"<<endl;
}
class Solution {
public:
   multiset <double> arr;
   void insert(double x){
      arr.insert(x);
   }
   void delete_(double x){
      arr.erase(arr.find(x));
   }
   double getMedian(){
      int n = arr.size();
      double a = *next(arr.begin(), n / 2 - 1);
      double b = *next(arr.begin(), n / 2);
      if(arr.size() & 1)return b;
      return (a + b) * 0.5;
   }
   vector<double> medianSlidingWindow(vector<int>& nums, int k) {
      vector <double> ans;
      arr.clear();
      for(int i = 0; i < k; i++){
         insert(nums[i]);
      }
      for(int i = k, j = 0; i < nums.size(); i++, j++){
         ans.push_back(getMedian());
         delete_(nums[j]);
         insert(nums[i]);
      }
      ans.push_back(getMedian());
      return ans;
   }
};
main(){
   Solution ob;
   vector<int> v = {1,3,-1,-3,5,3,6,8};
   print_vector(ob.medianSlidingWindow(v, 3));
}

Input

{1,3,-1,-3,5,3,6,8}

Output

[1, -1, -1, 3, 5, 6, ]
raja
Published on 01-Jun-2020 11:17:46
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