A string as input The objective is to find 3 character patterns in str start with 't' and end with the char 'p'. but middle character is removed

Question

Sample Input #1

"tiptop"

Sample Output #1

tptp

Sample Input #2

"TipTop"

Sample Output #2

TipTop

Sample Input #3

"tpttipptptaptptopp"

Sample Output #3

tpttpptptptptpp

code

public String changeLetters(String str){
        String str1="";
        for(int i=0;i<str.length();i++){
            char ch1=str.charAt(i);
            char ch2=str.charAt(i+2);
            char ch3=str.charAt(i+1);
            if(str.length()==i+1)
            break;
            if((ch1==116)&&(ch2==112)){
                str1=str1+ch1+ch2;
                i=i+2;
            }else
              str1=str1+ch1;
        }
        return str1;

When I am trying to run my code only first test case "tiptop" runs while 2nd test case fails, can you suggest where I am doing mistake.

Testcase Pass/Fail Parameters Actual Output Expected Output

1 Pass 'tiptop' tptp tptp

2 Fail 'TipTop' null TipTop

Sumit Singh · Accepted Answer · 2015-08-18T09:13:36.263

0

Use following code:

public static String changeLetters(String str) {
    String str1 = "";
    for (int i = 0; i < str.length(); i++) {
        if (i + 1 >= str.length()) {
            str1 += str.charAt(i);
            return str1;
        }
        char ch3 = '\u0000';
        char ch1 = str.charAt(i);
        char ch2 = str.charAt(i + 1);
        if (i + 2 < str.length()) {
            ch3 = str.charAt(i + 2);
        }
        if ((ch1 == 116) && (ch2 == 112)) {
            i++;
            str1 = str1 + ch1 + ch2;
        } else if ((ch1 == 116) && (ch2 != 112) && (ch3 == 112)) {
            i = i + 2;
            ch2 = ch3;
            str1 = str1 + ch1 + ch2;
        } else {
            str1 = str1 + ch1;
        }
    }
    return str1;
}

See the output: Ouput link

tpttpptptptptpp
TipTop

edited Aug 18 '15 at 09:13

answered Aug 18 '15 at 08:22

Sumit Singh

15,743
6
59
89

my question is that if string is in lower case then remove middle char and if it is upper case or any other character then it will print as same. – Pallavi Singh Aug 18 '15 at 08:26
@PallaviSingh where is your " 2nd test case fails"? – Sumit Singh Aug 18 '15 at 08:32
2nd test case is that if input is TipTop, Then expected output remain TipTop. but my test case gives null. – Pallavi Singh Aug 18 '15 at 08:40

score 0 · Answer 2 · answered Aug 18 '15 at 08:35

0

You need a simple regex :

    String[] s = {"tiptop", "TipTop", "tpttipptptaptptopp"} ;
    String regex = "(?<=t)[^pt](?=p)"; // any single character which is not 'p' or 't' and preceeded by t and followed by p
    for (String str : s) {
    System.out.println(str.replaceAll(regex, ""));
    }

O/P:

tptp
TipTop
tpttpptptptptpp

answered Aug 18 '15 at 08:35

TheLostMind

35,966
12
68
104

can we do it logically without using any string functions.? – Pallavi Singh Aug 18 '15 at 08:38
@PallaviSingh - Yep.. we could but this looks more elegant.. Just one line of code :) – TheLostMind Aug 18 '15 at 08:39
You provide @PallaviSingh a code sample that is still a fail for upper cases ;) – Erwan C. Aug 18 '15 at 08:42
@ErwanC. - Did you even consider comparing my input / output to what she expects? – TheLostMind Aug 18 '15 at 08:43
@TheLost Mind - yes but it is mentioned use only logic no strings functions and even without using array. – Pallavi Singh Aug 18 '15 at 08:44
@PallaviSingh - Change your loop condition is wrong.. `i` can go above `str.length()` (you use i+1, i+2), change it – TheLostMind Aug 18 '15 at 08:47
@ TheLostMind yes your code returns "TipTop" for input "TipTop" when it is supposed to return "TpTp". – Erwan C. Aug 18 '15 at 08:49
@ErwanC. - The expected out is `TipTop` my friend – TheLostMind Aug 18 '15 at 08:50
@TheLostMind -thats why i put one break statement in case if i extend to i+1 ,i+2. – Pallavi Singh Aug 18 '15 at 08:50
@TheLost Mind nevermind op changed his expected result and his actual ouptut.... It seems Sumit Singh and i misunderstood the original question – Erwan C. Aug 18 '15 at 08:51
@PallaviSingh - But `char ch2=str.charAt(i+2);` will be executed.. and your code will break – TheLostMind Aug 18 '15 at 08:52
@PallaviSingh - Yes, remove the `break` and `if` condition and change `i – TheLostMind Aug 18 '15 at 09:13
@TheLostMind - loop will not print last two character if it goes to else part, your code gives output TipT – Pallavi Singh Aug 18 '15 at 09:18
@PallaviSingh - OK.. move the `if` and `break` to beginning of the loop and keep everything as it was earlier (use `str.length()` not `str.length-2` in loop – TheLostMind Aug 18 '15 at 09:20

score 0 · Answer 3 · answered May 20 '16 at 13:09

public class LetterPattern {

 static String testcase1 = "tip.tip..tip";

 public static void main(String args[]){
 LetterPattern testInstance= new LetterPattern();
 String result = testInstance.changeLetters(testcase1);
 System.out.println(result);
 }

 public String changeLetters(String str)
 {
 //write your code here
    String str1 = "";
    int j = 0;
    int count = 2;
    for (int i=0; i<str.length(); i++)
    {

        j=i;
        // this will execute when 'tp'occurs that is when t and p are in continuty
         if (str.charAt(i) == 't' && str.charAt(j+1) == 'p') 
        {
            str1 = str1 + "tp";
            i= j+1;
        }

        // this will execute when t and p have one more letter in between, like tip, top. 
        else if (str.charAt(i) == 't' && str.charAt(j+2) == 'p') 
        {
            str1 = str1 + "tp";
            i= j+2;
        }

        // this will execute when there is no word like tp, top, tip which exactly mean all the words except tp, top, tip etc.
        else if (!((str.charAt(i) == 't' && str.charAt(j+1) == 'p'))
        {
            str1 = str1+str.charAt(i);
        }

    }
    return str1;
 } 
}

This is another easy way of solving the above problem. I hope this will help you. — Sahil Bali, May 20 '16 at 13:12

A string as input The objective is to find 3 character patterns in str start with 't' and end with the char 'p'. but middle character is removed

1 Pass 'tiptop' tptp tptp

2 Fail 'TipTop' null TipTop

3 Answers3