StringCompression.java

/* (C) 2024 YourCompanyName */
package string;

/**
 * Created by gouthamvidyapradhan on 12/04/2018. Given an array of characters, compress it in-place.
 *
 * <p>The length after compression must always be smaller than or equal to the original array.
 *
 * <p>Every element of the array should be a character (not int) of length 1.
 *
 * <p>After you are done modifying the input array in-place, return the new length of the array.
 *
 * <p>Follow up: Could you solve it using only O(1) extra space?
 *
 * <p>Example 1: Input: ["a","a","b","b","c","c","c"]
 *
 * <p>Output: Return 6, and the first 6 characters of the input array should be:
 * ["a","2","b","2","c","3"]
 *
 * <p>Explanation: "aa" is replaced by "a2". "bb" is replaced by "b2". "ccc" is replaced by "c3".
 * Example 2: Input: ["a"]
 *
 * <p>Output: Return 1, and the first 1 characters of the input array should be: ["a"]
 *
 * <p>Explanation: Nothing is replaced. Example 3: Input:
 * ["a","b","b","b","b","b","b","b","b","b","b","b","b"]
 *
 * <p>Output: Return 4, and the first 4 characters of the input array should be: ["a","b","1","2"].
 *
 * <p>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.
 *
 * <p>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;
  }
}