Given two strings like x=?hello? and y=?open?, remove any
character from string x which is also used in string y,
thus making the result x=?hll?.
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Answer / jethvamohit
using System;
using System.Collections.Generic;
using System.Text;
namespace ConsoleApplication1
{
class Program
{
static void Main(string[] args)
{
string str = "mohim";
string str1 = "mo";
remove(str, str1);
}
private static void remove(string str,string str1)
{
char[] abc;
abc = str.ToCharArray();
char[] bcd;
string str4 = "";
bcd = str1.ToCharArray();
for (int i = 0; i < bcd.Length; i++)
{
for (int j = 0; j < abc.Length; j++)
{
if (abc[j] != bcd[i])
{
str4 = str4 + abc[j];
}
}
abc = str4.ToCharArray();
str4 = "";
}
Console.Write(abc);
Console.ReadLine();
}
}
}
| Is This Answer Correct ? | 4 Yes | 1 No |
Answer / preeti
Lttle correction in above answer.
using System;
using System.Collections.Generic;
using System.Text;
namespace ConsoleApplication1
{
class Program
{
static void Main(string[] args)
{
string str = "mohim";
string str1 = "mo";
remove(str, str1);
}
private static void remove(string str,string str1)
{
char[] abc;
abc = str.ToCharArray();
char[] bcd;
string str4 = "";
bcd = str1.ToCharArray();
for (int i = 0; i < bcd.Length; i++)
{
for (int j = 0; j < abc.Length; j++)
{
if (abc[j] != bcd[i])
{
str4 = str4 + abc[j];
}
}
abc = str4.ToCharArray();
str4 = "";
}
str = "";
for (int j = 0; j < abc.Length; j++)
{
str = str + abc[j];
}
Console.Write("str : ");
Console.Write(str)
}
}
}
output will be
------------------------
str : hi
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / amadou
. static void Main(string[] args)
{
string x = "hello";
string y = "open";
string temp = "";
for (int i = 0; i < x.Length; i++)
{
if (!y.ToString().Contains(x[i]))
{
temp += x[i];
}
Console.WriteLine("temp={0}",
temp);
}
Console.Read();
| Is This Answer Correct ? | 3 Yes | 1 No |
Answer / sourabh bose
int countX=0;
int countY=0;
string tmp=x;
for(countX=1;countX<=x.lenth();countX++)
{
for(countY=1;countY<=y.length();countY++)
{
if x.substring(1,countX).Equels(y.substring(1,countY)
{
tmp=tmp.Replace(tmp.substring(1,countX),"");
}
}
}
x=tmp;
| Is This Answer Correct ? | 4 Yes | 3 No |
Answer / guest
public static void main(String[] args) {
String str_x = "hello";
String str_y = "open";
StringBuffer result = new StringBuffer();
for (int i=0; i<str_x.length();i++){
if (str_y.indexOf(str_x.charAt(i)) < 0)
result.append(str_x.charAt(i));
}
System.out.print(result);
}
| Is This Answer Correct ? | 5 Yes | 4 No |
Answer / skybeaver
static public string StupidQuestion(string x, string y)
{
StringBuilder sb = new StringBuilder();
Hashtable ht = new Hashtable();
// just one pass through y
foreach( char c in y.ToCharArray() )
if( !ht.Contains(c) )
ht.Add(c, c)
// just one pass thru x
foreach( char c in x.ToCharArray() )
if( !ht.Contains(c) )
sb.Append(c);
return sb.ToString();
}
| Is This Answer Correct ? | 2 Yes | 2 No |
Answer / vijay bhatia
string processString(string x, string y) {
string res = "";
int occuranceCount[26];
for(int i=0; i<26; i++) {
occuranceCount[i] = 0;
}
for(i=0; i<y.size(); i++) {
occuranceCount[y[i]-'a']++;
}
for(i=0; i<x.size(); i++) {
if (occuranceCount[x[i]-'a'] == 0) {
res += x[i];
}
}
return res;
}
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / newpolaris
bool IsInStr(char ch, const std::string& B)
{
return std::string::npos != B.find(ch);
}
// act fuction
std::string remove_same_char(const std::string& A, const
std::string& B)
{
typedef std::string::const_iterator cstr_const_it;
cstr_const_it iCSTR = A.begin();
// FOR OPTIMIZATION NRVO IS NEEDED
// ? IS POSSIBLE?
std::string _rt;
while ( iCSTR != A.end() )
{
if (!IsInStr(*iCSTR,B)) _rt+=*iCSTR;
iCSTR++;
}
return _rt;
}
int main()
{
std::string x = "hello";
const std::string y = "open";
x = remove_same_char(x, y);
}
| Is This Answer Correct ? | 0 Yes | 0 No |
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