CareerCup

Use Hashtable

I think we can sort the array and remove the duplicates like...

sort(a);
l=a.length();
j=0;
for(i=0;i<l;i++)
{
if(a[j]!=a[i])



}

I think we can sort the array and remove the duplicates like...

sort(a);
l=a.length();
j=0;
for(i=0;i<l;i++)
{
if(a[j]!=a[i])
{
j++;
a[j]=a[i];
}
}
so by this way we need o(nlogn) to sort the array and again a o(n2) to remove duplicates...

the other possibilities can be like...
i=0;
while(i<a.length)
{
if( !contain(a[i]) )
{
m[r]=a[i];r++;
}
i++;
}


int contain(int n)
{
int j;
for(j=0;j<r;j++)
{
if(m[j]==n)
return 1;
}
return 0;
}

sorting takes o(nlogn) time you can use hashtable

1. start from array check if number is already present in hashtable or not if no put inplace in array
2.if yes check next element..
so o(n) time and o(1) space

In python...

def duplicate(lst):
    lst1 = []
    l = len(lst)
    i = 0
    while i < l:
        if lst[i] not in lst1:
            lst1.append(lst[i])
        i += 1
    return lst1

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Collections;

namespace RemoveDupsinArray
{
    class Program
    {
        
        static void Main(string[] args)
        {
            int[] array = { -4, 0, 1, 6, 0, 1, 6, -2 };
            RemoveDups(array);
            Console.Read();
        }
        public static void RemoveDups(int[] array)
        {
            Hashtable ht = new Hashtable();
            string output = "";
            for (int i = 0; i < array.Length; i++)
            {
                if (!ht.ContainsKey(array[i]))
                {
                    ht.Add(array[i],0);
                    output += " " + array[i]; 
                }
                
            }
            Console.WriteLine("{0} ", output);
            
        }
    }
}
output: -4 0 1 6 -2