Implement a method to perform basic string compression using the counts of repeated characters. For example, the string aabcccccaaa would become a2b1c5a3. If the "compressed" string would not become smaller than the original string, your method should return the original string. The method signature is: “public static String compress(String inpu…

CompressString.java

/*
* Implement a method to perform basic string compression using
* the counts of repeated characters. For example, the string aabcccccaaa
* would become a2b1c5a3. If the "compressed" string would not become
* smaller than the original string, your method should return the original string.
*/
public static String compresString(String text) {
	int length = text.length();
	/*
	*  Compression makes sense at length of larger than or equal to 3.
	*  For instance: "aaa" -> "a3"
	*/
	if(length > 2) {
		String compressedText = "";	
		char lastChar = text.charAt(0);	
		int charCount = 1;
		for(int i = 1; i < length; i++) {
			if(text.charAt(i) == lastChar) charCount++;	
			else {
				compressedText += Character.toString(lastChar) + Integer.toString(charCount);
				lastChar = text.charAt(i);	
				charCount = 1;
			}
		}	
		compressedText += Character.toString(lastChar) + Integer.toString(charCount);
		if(compressedText.length() < length)
			return compressedText;		
	}		return text;
}

It would be better to implement StringBuilder/StringBuffer for time/space complexity.

public static String compress(String str) {

	StringBuilder output = new StringBuilder();

	for(int i = 0; i<str.length(); i++){
		int val = 1;
		while(i != str.length()-1 && str.charAt(i) == str.charAt(i+1)){
			i++;
			val++;
		}
		output.append(str.charAt(i));
		
		String st = String.valueOf(val);
		for(int k = 0;k<st.length();k++){
			output.append(st.charAt(k));

		}
	}
	return output.toString();
}

public static String compress1(String str) { // using MAP Solution MIMP

	Map<Character, Integer> map = new LinkedHashMap<>();
	for(int i = 0; i<str.length(); i++){

		if(map.get(str.charAt(i)) != null) {
			map.put(str.charAt(i), map.get(str.charAt(i))+1);
		}
		else {
			map.put(str.charAt(i), +1);
		}
	}
	StringBuilder output = new StringBuilder();
	for(Entry<Character, Integer> entry : map.entrySet()) {
		output.append(entry.getKey());
		output.append(entry.getValue());
	}

	return output.toString();
}

Thanks @edihasaj and @vinaygupta2255 - Yes, using StringBuilder makes sense!