Replace only characters that haven't been replaced

Question

I have a question in regards to replacing certain characters. I have a string of x amount of characters, within which they contain all characters within the normal English alphabet.

String x = "exampleString"

My question is, because what i want to do is replace the characters that are within this string, I'm not able to user replace, as this replaces previously replaces strings, such as.

if(x.contains(e)){
x = x.replace("e","a")
}

That will replace every E character within the string above.

I was also trying to use:

x=x.replace("e","a").replace("a","b")

but that would also replace every a character, even the previously replaced. I though this would work, because strings are immutable in Java, however it doesn't.

I also considered counter to see when the string has been replaced and omit the replaced strings, however I'm not able to implement this.

Can anyone suggest solution?

Regards ProgrammingNewbie.

Unless your mappings are circular, you can reverse it: `x=x.replace("a","b").replace("e","a")` — shmosel, Nov 21 '18 at 00:31

Pritam Banerjee · Answer 1 · 2018-11-21T00:29:11.040

1

You can try this:

 // Your String
String str = "exampleString";

// Convert it to a Character array.
char[] strCharArr = str.toCharArray();

// Have a corresponding boolean array
boolean[] arrStr = new boolean[str.length()];

// Character to be replaced
char a = 'r';

// Character to be replaced with
char b = 's';        

// Loop through the array
for(int i = 0; i < strCharArr.length; i++) {

    // Check if the characted matches and have not been replaced before.
    if(strCharArr[i] == a && !arrStr[i]) {
        strCharArr[i] = b;
        arrStr[i] = true; 
    }
}
System.out.println(String.valueOf(strCharArr));

// Now if you need to replace again. 
a = 's';
b = 'a';        
for(int i = 0; i < strCharArr.length; i++) {
    if(strCharArr[i] == a && !arrStr[i]) {
       strCharArr[i] = b;
       arrStr[i] = true; 
    }
}

System.out.println(String.valueOf(strCharArr));

After this you can convert back the character array to the corresponding string. You can write a function and replace it as many times you want.

edited Nov 21 '18 at 00:29

answered Nov 21 '18 at 00:11

Pritam Banerjee

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what is the use of `boolean[] arrStr` there is only one loop and the value is not set until after it is set. – Scary Wombat Nov 21 '18 at 00:13
@ScaryWombat Updated the explanation. – Pritam Banerjee Nov 21 '18 at 00:13
Updated my comment, still unclear – Scary Wombat Nov 21 '18 at 00:14
@ScaryWombat It will all be false. And once it is replaced the corresponding boolean will be set to true. – Pritam Banerjee Nov 21 '18 at 00:18
It does not make sense. In 100% of cases `arrStr[i]` will be `false` as expected – Scary Wombat Nov 21 '18 at 00:19
@ScaryWombat Yes, that's what I want. False means it is not replaced, so it can be replaced. Once replaced the boolean will be set to true and will not be replaced again. That is what OP needs. – Pritam Banerjee Nov 21 '18 at 00:25
But as you are only looping once, it is not used. Please modify your answer to show real usage. – Scary Wombat Nov 21 '18 at 00:29
@ScaryWombat Spoon feeding is not really the right way to teach coding to someone. That's why I added the comments and also added what needs to be done next. – Pritam Banerjee Nov 21 '18 at 00:30
1

Good edit - now it is easy to understand for generations of programmers to come. – Scary Wombat Nov 21 '18 at 00:33

score 0 · Accepted Answer · answered Nov 21 '18 at 00:33

The problem is that you are performing two replacements over the entire string, subjecting a position to multiple replacements. You need a solution that loops over the string only once, so that an "e" that is replaced with "a" won't then get replaced with a "b".

String x = "exampleString";
StringBuilder buf = new StringBuilder(x);
for (int i = 0; i < buf.length(); i++)
{
    switch (buf.charAt(i))
    {
    case 'e':
        buf.setCharAt(i, 'a');
        break;
    case 'a':
        buf.setCharAt(i, 'b');
        break;
    }
}
System.out.println(buf.toString());

This uses a StringBuilder for mutability and it bypasses the replace method. There is no need to keep track of what has been replaced. This will only perform a maximum of one replacement per character. Output when run:

axbmplaString

Replace only characters that haven't been replaced

2 Answers2