CODEWITHSUNDEEP
javastringchar
vicky
vicky

Reputation: 1081

To print only 3 consecutive character from given string

I want to print only 3 same consecutive characters from the given String as below if input is:

"aasssfddddvvv"

then I should get output as output:

sss
vvv 
count=2

This is my requirement, please help me in this.

my code:

import java.util.Scanner;

class Test {

    public static void main(String args[]) {
        Scanner scan = new Scanner(System.in);
        System.out.println("Enter the String: ");
        char s[] = scan.nextLine().toCharArray();

        for (int i = 0; i < s.length - 1; i++) {

            if (s[i] == s[i + 1] && s[i] == s[i + 2] && s[i] == s[i + 3]) {
                System.out.println(s[i]);

            }
        }
        scan.close();
    }
}

Upvotes: 2

Views: 2244

Answers (7)

Magdrop
Magdrop

Reputation: 578

It seems you want only a strict count of 3 consecutive character. If that's the case, even after fixing the issues in your code (without changing the overall logic), it will print the ddd twice. Here is the output of the fixed code:

sss 
ddd
ddd
vvv
count=4

What I meant by the fixed code is in handling these issues

  1. Comparison is too long, fix: remove && s[i] == s[i + 3].
  2. Condition needs to be adjusted to i < s.length - 2
  3. Print out only a single char (probably is not a real problem)
  4. Add a counter each time a match found

Here is the fixed code (remember this is still not producing the right output you want)

int occurence_count = 0;
for (int i = 0; i < s.length - 2; i++) {

    if (s[i] == s[i + 1] && s[i] == s[i + 2]) {
        occurence_count++;
        System.out.println(s[i]);
    }
}
System.out.println("count=" + occurence_count);

One logic change is to have a separate counter that counts the consecutive characters and a variable that remembers what is the last character.

int char_count = 0;
char last_char = '!';

The basic idea is to go through our string and increment the char_count by one if it is the same and reset if it is different. We also need to initialize it with a character that you're not planing to use as input, such as !.

Since we're going through the entire string, the for loop condition needs to go to the very end using i < s.length.

When we're resetting (when a different character is detected), we want to also check whether the previous one has char_count=3 or not. If it is a match, then increment the occurence_count. We also need to do this one more time after the for loop because the check might not happens for consecutive that happens at the end of the input string. Here is the code:

    int occurence_count = 0;
    int char_count = 0;
    char last_char = '!';  
    for (int i = 0; i < s.length; i++) {
        if (s[i] != last_char) {                
            if (char_count == 3) {
                occurence_count++;
                System.out.println(""+last_char+last_char+last_char);                   
            }               
            last_char = s[i];
            char_count = 0;
        }

        char_count++;

    }
    if (char_count == 3) {
        occurence_count++;
        System.out.println(""+last_char+last_char+last_char);                   
    }

    System.out.println("count=" + occurence_count);

Upvotes: 0

Guy haimovitz
Guy haimovitz

Reputation: 1625

You cant check.cells i + 2 and i + 3 because you will cross the array bounds when you will reach i = length -2, You should count the sequence and check if the length is sufficient, if so print, also, in your example you are comparing 4 cells so the length will be 4 and not 3, start the seq in 1 because when you are reading a single char its a sequence of 1:

import java.util.Scanner;

class Test {

    public static void main(String args[]) {
        Scanner scan = new Scanner(System.in);
        System.out.println("Enter the String: ");
        char s[] = scan.nextLine().toCharArray();
        int seqLen = 3;
        int cnt = 1;
        for (int i = 1; i < s.length; i++) {

            if (s[i] == s[i - 1]) {
                cnt++;
             }
            if((cnt==seqLen)&&((s[i] !=s[i - 1] || i == s.length)){
                 if(cnt == seqLen)
                 {
                     for(int j= 0; j<seqLen; j++)
                     {
                         System.out.println(s[i - 1]);
                      }
                 }
                 cnt=1;
            }          
        }
    scan.close();
    }
}

Upvotes: 0

Tim Biegeleisen
Tim Biegeleisen

Reputation: 522331

This answer is a departure from the more manual method being used in the OP. But, this is fairly straightforward to handle uses a regex pattern matcher in Java. We can match on the pattern (.)\\2*, to capture all groups of similar letters. Then, just print out the groups which occur as 3 only.

String line = "aasssfddddvvv";
String pattern = "((.)\\2*)";
Pattern r = Pattern.compile(pattern);
Matcher m = r.matcher(line);

while (m.find()) {
    if (m.group(1).length() == 3) {
        System.out.println(m.group(1));
    }
}

sss
vvv

Demo

Upvotes: 3

gifpif
gifpif

Reputation: 4917

import java.util.Scanner;
class Test {

    public static void main(String args[]) {
        Scanner scan = new Scanner(System.in);
        System.out.println("Enter the String: ");
        char s[] = scan.nextLine().toCharArray();
        int count=0;

        for (int i = 0; i < s.length - 3; i++) {

            if (s[i] == s[i + 1] && s[i] == s[i + 2] && s[i]!=s[i + 3]) {
                System.out.println(s[i]+s[i]+s[i]);
                count++;
            }
        }
        if(s[s.length - 3] == s[s.length - 2] && s[s.length - 3] == s[s.length - 1]){
            System.out.println(s[s.length]+s[s.length]+s[s.length]);
            count++;
        }
        scan.close();
        System.out.println(count);
    }
}

Upvotes: 0

Swapnil Pandey
Swapnil Pandey

Reputation: 599

Replace the main logic of your code with this. Here we are checking if there are only 3 repeating characters. by making sure that the fourth character is different.

int count=0;
for (int i = 0; i < s.length - 3; i++) {
    if (s[i] == s[i + 1] && s[i] == s[i + 2] && s[i] != s[i + 3]) {
        System.out.println(s[i]+s[i]+s[i]);
        count++;
    }
}
System.out.println("count="+count);

Upvotes: 0

Roshana Pitigala
Roshana Pitigala

Reputation: 8806

First Iterate through your char[] array. Then, in your if statement,

  • First make sure that there are more than 2 characters left in the array.
  • Then compare the s[i] with the next char(s[i+1]) and the char after next(s[i+2])

Replace this part of your code,

for (int i = 0; i < s.length; i++) {
    if (i < s.length - 2 && s[i] == s[i + 1] && s[i] == s[i + 2]) {
        System.out.println(s[i]);
    }
}

Upvotes: 0

E141
E141

Reputation: 109

First problem is that, you try to reach lenght+2th index of array in for which is not possible.

Assume String lenght is 5

0:  0 1 2
1:  1 2 3
2:  2 3 4
3:  3 4 (5)
4:  4 **(5) (6)**

As you can see, you try to reach 5 which is not part of string. So try

for (int i = 0; i < s.length - 2; i++)

instead of 

for (int i = 0; i < s.length - 1; i++)

Tip: You can get this from error.

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 13

It says you try to reach index which you can't.

Second problem you check 4 char in if condition. Use:

if (s[i] == s[i + 1] && s[i] == s[i + 2])

instead of

if (s[i] == s[i + 1] && s[i] == s[i + 2] && s[i] == s[i + 3])

Upvotes: 2

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