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Java

Posted by Varun Dixit
Varun Dixit
151 Points

Can anyone solve this! man this is very hard to me all PRO-grammers its a challenge!!!!

Craig Dennis please help me sir... Write a program to do basic string compression. For a character which is repeated more than once, replace consecutive duplicate occurrences with the count of repetitions e.g. if a String has 'x' repeated 5 times, replace this "xxxxx" with "x5".

1 Answer

Craig Dennis
STAFF
Craig Dennis
Treehouse Teacher
Craig Dennis
Craig Dennis
Treehouse Teacher

I'll give you some pseudo code....

Create a new string var
Create a previous character var
Create counter var
Loop through each character in original string
     Is current char equal to previous?
           Increment counter
     Otherwise
            Write letter
             Is counter greater than one?
                   Write counter
                   Reset counter
Varun Dixit
Varun Dixit
151 Points

Craig Dennis sir can you please tell what is wrong in my logic that i build in more than 6hr i'm really feeling very bad about me :(

class Solution {
    public static String compress(String inputString) {
        char arr[] = inputString.toCharArray();
        int i = 0, j = 0, sign = 0;
        String returningString = "";   
        while(i<arr.length) {  
            sign = 1;
            j = i;
            while(arr[j] == arr[j+1]) {
                sign++;
                j++;
            }
            if(sign != 1) {
                returningString += returningString.concat(Integer.toString(sign));
                returningString += returningString.concat(Character.toString(arr[i]));
            }
            if(sign == 1) {
                returningString += returningString.concat(Character.toString(arr[i]));
            }
            i += sign;
        }
        returningString += returningString.concat(Character.toString(arr[i]));
        return returningString;
    }
    public static void main(String[] args) {
        Solution s = new Solution();
        System.out.println(s.compress("aaabbccds"));
    }
}```