CODEWITHSUNDEEP
javastring
Rodrick Gorden
Rodrick Gorden

Reputation: 25

count number of times aaa appears in a string with and with overlaping

Count the number of times aaa appears in a string, 1. without repetition and 2. when repetition is allowed. I tried the following code which works well for 1st case but does'nt work well for the second case.

public static void main(String[] args) {
    String str="cbaadaaabefaaaag";
    int count=0,count1=0;
    for(int i=0;i<str.length();i++)
    {
        if(str.charAt(i)=='a' && str.charAt(i+1)=='a' && str.charAt(i+2)=='a')
        {
            count++;
        }
    }
    System.out.println(count);
    for(int i=0;i<str.length();i++)
    {
        if(i==0)
        {
            if(str.charAt(i)=='a' && str.charAt(i+1)=='a' && str.charAt(i+2)=='a')
            {
                count1++;
            }
        }
        else
        {
            if(str.charAt(i-1)=='a' && str.charAt(i)=='a' && str.charAt(i+1)=='a')
            {
                count1++;
            }
        }
    }
    System.out.println(count1);

}

The input I tried is cbaadaaabefaaaag which should give 3 for 1st case and 2 for 2nd case, also when the input is aaaa it should be 1 for 1st case and 2 for overlaping case.

Upvotes: 0

Views: 383

Answers (2)

Pratham Mittal
Pratham Mittal

Reputation: 17

public static void main(String[] args) {

    Scanner scn = new Scanner(System.in);

    String str = scn.nextLine();

    int count = 0, count1 = 0;

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

        if (str.charAt(i) == 'a' && str.charAt(i + 1) == 'a' && str.charAt(i + 2) == 'a') {

            count++;

        }

    }

    System.out.println(count);

    for (int i = 0; i <= str.length() -1; i++) {
        if (str.charAt(i) == 'a' && str.charAt(i + 1) == 'a' && str.charAt(i + 2) == 'a') {
            i += 2;
            count1++;

        }

    }
    System.out.println(count1);
}

Upvotes: 0

thatguy
thatguy

Reputation: 22099

This code should help you:

public static void main(String args[]) throws Exception {

    String str = "cbaadaaabefaaaaaaag";
    System.out.println(count(str, true));
    System.out.println(count(str, false));

}

public static int count(String str, boolean overlap) {

    int count = 0;

    for (int i = 0; i < str.length() - 2; i++) {
        if (str.charAt(i) == 'a' && str.charAt(i + 1) == 'a'
                && str.charAt(i + 2) == 'a') {
            if (!overlap) {
                i += 2;
            }
            count++;
        }
    }

    return count;

}

The condition i < str.length() - 2 ensures that you do not get a StringIndexOutOfBoundsException, because if there are only two characters left in the string, there cannot be anther substring aaa.

The overlap condition adds two to the current index, where a substring aaa was found, so your index points to the next character after the last found sequence aaa, which prevents overlapping.

Upvotes: 3

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