Improving performance hacker rank Median
There are a lot of stylistic changes I'd normally suggest, but I'm going to forgo that to mention three very simple optimizations:


remove new Char[] { ' ', '\t', '\n' } from within the loop. No need to recreate this object every time as it's an invariant.
you know the maximum size of the a list, so create it with that capacity in mind to prevent resizing as it's being accessed: List&lt;double&gt; a= new List&lt;double&gt;(N);
Repeated code in both the if and else branches pulled out (no performance optimization, but basic tenet of DRY (Don't Repeat Yourself)): int index= a.BinarySearch(x[i]);


Resulting code looks like this:

using System;
using System.Collections.Generic;
using System.IO;
using System.Text;
class Solution {

   static void Main(String[] args) {
      int N;
      StringBuilder st= new StringBuilder();

      N = Convert.ToInt32(Console.ReadLine());
      int[] x = new int[N];
       List&lt;double&gt; a= new List&lt;double&gt;(N);
      string[] s = new string[N];

      Char[] splitChars = new Char[] { ' ', '\t', '\n' };

      for(int i=0; i&lt;N ;i++){

         string tmp = Console.ReadLine(); 
         string[] split = tmp.Split(splitChars);

         s[i] = split[0].Trim();
         x[i] = Convert.ToInt32(split[1].Trim());
          bool r=true;

          int index= a.BinarySearch(x[i]);
          if(s[i]=="r"){
              if(index&gt;=0){
                  a.RemoveAt(a.LastIndexOf(x[i]));    
              }
              else{
              r=false;
              }

          }else{
                if (index &lt; 0) index = ~index;
                  a.Insert(index,x[i]);
          }
          if(!r || a.Count==0){
              st.AppendLine("Wrong!");

          }
          else{
          st.AppendLine(calcularModa(a).ToString());
          }
      }      
       Console.WriteLine(st.ToString());

   }
   static double calcularModa(List&lt;double&gt; a){

   int i= a.Count/2;
       if(a.Count % 2 !=0){
           return a[i];
       }else{
           return ((a[i - 1] + a[i]))/2;
       }

    }
}