package string;

/**
 * Created by gouthamvidyapradhan on 12/04/2018.
 * Given an array of characters, compress it in-place.

 The length after compression must always be smaller than or equal to the original array.

 Every element of the array should be a character (not int) of length 1.

 After you are done modifying the input array in-place, return the new length of the array.


 Follow up:
 Could you solve it using only O(1) extra space?


 Example 1:
 Input:
 ["a","a","b","b","c","c","c"]

 Output:
 Return 6, and the first 6 characters of the input array should be: ["a","2","b","2","c","3"]

 Explanation:
 "aa" is replaced by "a2". "bb" is replaced by "b2". "ccc" is replaced by "c3".
 Example 2:
 Input:
 ["a"]

 Output:
 Return 1, and the first 1 characters of the input array should be: ["a"]

 Explanation:
 Nothing is replaced.
 Example 3:
 Input:
 ["a","b","b","b","b","b","b","b","b","b","b","b","b"]

 Output:
 Return 4, and the first 4 characters of the input array should be: ["a","b","1","2"].

 Explanation:
 Since the character "a" does not repeat, it is not compressed. "bbbbbbbbbbbb" is replaced by "b12".
 Notice each digit has it's own entry in the array.
 Note:
 All characters have an ASCII value in [35, 126].
 1 <= len(chars) <= 1000.

 Solution O(N) time complexity and O(1) space complexity. Maintain read and write pointers. Read from read pointer and
 increment count when a repetition is found, when there is no repetition write the count value using write pointer.
 */
public class StringCompression {

    /**
     * Main method
     * @param args
     * @throws Exception
     */
    public static void main(String[] args) throws Exception{
        char[] A = {'a','a','b','b','c','c','c'};
        System.out.println(new StringCompression().compress(A));
    }

    public int compress(char[] chars) {
        int count = 0;
        int i = 0;
        int p = 0;
        for(int j = 0; j < chars.length; j ++){
            if(chars[i] == chars[j]){
                count ++;
            } else{
                chars[p] = chars[i];
                p++;
                if(count > 1){
                    String countStr = String.valueOf(count);
                    for (int l = 0; l < countStr.length(); l++){
                        chars[p++] = countStr.charAt(l);
                    }
                }
                i = j;
                count = 1;
            }
        }
        chars[p] = chars[i];
        p++;
        if(count > 1){
            String countStr = String.valueOf(count);
            for (int l = 0; l < countStr.length(); l++){
                chars[p++] = countStr.charAt(l);
            }
        }
        return p;
    }
}