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                            Solution in C :

In  C :




#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int GetMedian(int f[], int d);

int main() {
    int n, d;
    scanf("%d %d", &n, &d);
    int s[n];
    int i;
    for(i = 0; i < n; i++) {
        scanf("%d", &s[i]);
    }
    
    int frequency[201];
    for(i = 0; i < 201; i++) {
        frequency[i] = 0;
    }
    
    for(i = 0; i < d; i++) {
        frequency[s[i]]++;
    }
    
    int m[n];
    m[d] = GetMedian(frequency, d);
    
    for(i = d + 1; i < n; i++) {
        frequency[s[i-1]]++;
        frequency[s[i-d-1]]--;
        m[i] = GetMedian(frequency, d);
    }
    int not = 0;
    for(i = d; i < n; i++) {
        if(s[i] >= m[i]){
            not++;
        }
    }
    printf("%d", not);
    return 0;
}

int GetMedian(int f[], int d) {
    int Median;
    if(d % 2 == 0) {
        int mid = d/2;
        int cF = 0;
        for(int i = 0; i < 201; i ++) {
            cF += f[i];
            if(cF > mid) {
                Median = 2*i;
                break;
            }
            else if(cF == mid) {
                for(int j = i + 1; j < 201; j++) {
                    if(f[j] != 0) {
                        //printf("\n%d %d\n", i, j);
                        Median = i + j;
                        break;
                    }
                }
                break;
            }
        }
    }
    else {
        int mid = d/2 + 1;
        int cF = 0;
        for(int i = 0; i < 201; i++) {
            cF += f[i];
            if(cF >= mid) {
                Median = 2*i;
                break;
            }
        }
    }
    //printf("%d", Median);
    return Median;
}
                        

                        Solution in C++ :

In C ++ ;





#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <algorithm>

using namespace std;

#define MAXE 210

int A[200010];
int F[MAXE];

int median2(int D) {
  int p = 0;
  for (int i = 0; i < MAXE; i++) {
    p += F[i];
    if (p * 2 > D) {
      return 2 * i;
    } else if (p * 2 == D) {
      for (int j = i + 1; ; j++) {
        if (F[j]) {
          return i + j;
        }
      }
    }
  }
  return -1;
}

int main() {
  int N, D;
  cin >> N >> D;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
  }
  int result = 0;
  for (int i = 0; i < N; i++) {
    if (i >= D) { 
      if (A[i] >= median2(D)) {
        ++result;
      }
      F[A[i - D]]--;
    }
    F[A[i]]++;
  }
  cout << result << endl;
  return 0;
}
                    

                        Solution in Java :

In Java :




import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

   public static double median(int[]b){

		int mid=(int)b.length/2;
		if(b.length%2==1){
			return b[mid];
		}
		else{
			return (b[mid-1]+b[mid])/2.0;
		}
	}
	
	public static int BS(int[] arr,int in, int start,int end){
		
		 if(start>end){
				return -1;
				}
		
			int mid=start+(end-start)/2;
			if(arr[mid]==in){
		return mid;
			}
			
			else if(arr[mid]<in){
				return BS(arr,in,mid+1,end);
			}
			else {
				return BS(arr,in,start,mid-1);
			}	
		}
	public static void replace(int[]b,int x, int y){
		//for(int i=0;i<b.length;i++)
			//System.out.print(b[i]);
		int ind=BS(b,x,0,b.length-1);
		//System.out.println(x);
		b[ind]=y;
		if((ind==0&&b[ind]<=b[ind+1])||(ind==b.length-1&&b[ind]>=b[ind-1])){
			return;
		}
		else if(b[ind]>b[ind+1]){
			int temp=0;
			while(b[ind]>b[ind+1]){

				temp=b[ind];
				b[ind]=b[ind+1];
				b[ind+1]=temp;
				if(ind==b.length-2){
					
					break;
				}
				ind++;
			}
		}
		else if(b[ind]<b[ind-1]){
			int temp=0;
			while(b[ind]<b[ind-1]){
				if(ind==1){
					break;
				}
				temp=b[ind];
				b[ind]=b[ind-1];
				b[ind-1]=temp;
				if(ind==1){
					break;
				}
				ind--;
			}
		}
	}

    public static void main(String[] args) throws IOException {
    	 Scanner scan=new Scanner(System.in);
    	 int n=scan.nextInt();
    	 int d=scan.nextInt();
    	int a[]=new int[n];
    	
for(int i=0;i<n;i++){
	a[i]=scan.nextInt();
}
int b[]=new int[d];
int not=0;

for(int i=0;i<d;i++){
	b[i]=a[i];
} 
Arrays.sort(b);
for(int i=d;i<n;i++){
if(a[i]>=2*median(b)){

	not=not+1;
}
replace(b,a[i-d],a[i]);
}

System.out.println(not);

    }
}
                    

                        Solution in Python : 
                            
In Python3 :





n,d = map(int,input().strip().split())
exp = list(map(int, input().strip().split()))
count = {}
i = 0
while i<d:
    x = exp[i]
    if x not in count:
        count[x] = 0
    count[x] += 1
    i+=1

def median():
    global count
    cnt,midval = 0,0
    for i in range(201):
        if i in count:
            cnt += count[i]
            if cnt >= d//2:
                midval = i
                break
    secondval = midval
    
    if cnt == d//2:
        x = midval + 1
        while x not in count or count[x] == 0:
            x+=1
        secondval = x
    if d%2==1:
        return secondval
    return (midval + secondval)/2
        
ans = 0
i-=1
while i<n-1:
    md = median()
    if exp[i+1] >= 2*md:
        ans += 1
    count[exp[i+1-d]]-=1
    x = exp[i+1]
    if x not in count:
        count[x] = 0
    count[x]+=1
    i+=1
    
print (ans)