Basic string compression counting repeated characters

My task was to implement a method which performs basic string compression by counting sequences of repeating characters. Given "aaabbbccc" it should return "a3b3c3". I have included some sample tests I made up. Please let me know if I have missed any cases. I am looking for the fastest implementation possible with the most concise code. I'm looking to cut down on if else statements or give them simpler logic if possible.

```
public class StringCompression {
public String compress(String str){
int count = 1;
StringBuilder builder = new StringBuilder();

for(int i = 1; i<str.length(); i++){

if(str.charAt(i) == str.charAt(i-1) && i < str.length()-1){
count++;
}
// case when the last letter is in the sequence preceding it. Add that sequence to
// the compressed string
else if(i == str.length()-1 && str.charAt(i) == str.charAt(i-1)){
count++;
builder.append(str.charAt(i));
builder.append(count);
}

// case where the last letter is NOT in the sequence preceding it. Add it to string.
else if(i == str.length()-1 && str.charAt(i) != str.charAt(i-1)){
builder.append(str.charAt(i-1));
builder.append(count);
count = 1;
builder.append(str.charAt(i));
builder.append(count);
}
else{
// appending the character and THEN appending the count works.
builder.append(str.charAt(i-1));
builder.append(count);
count = 1;
}

}

str = builder.toString();
System.out.println(str);

return str;
}

public static void main(String[] args){
StringCompression test = new StringCompression();

test.compress("aabcccccaaa");
test.compress("aaaaa");
test.compress("

As @vnp said, don't print the result for "testing". Convert each statement in the main method to proper unit tests, for example:

@Test
public void test_aabcccccaaa() {
    assertEquals("a2b1c5a3", compress("aabcccccaaa"));
}

@Test
public void test_a5() {
    assertEquals("a5", compress("aaaaa"));
}

Once this is done, you can go ahead and safely refactor the rest of the code,
having an easy way to repeat the tests.

Being aware of, and working knowledge of unit testing should definitely score you extra points in a job interview, or might be even required.

Bug

For single letter inputs, the method seems to return an empty string. That looks incorrect. Judging by that for "abc" it returns "a1b1c1", it would seem that for "a" it should return "a1" instead of an empty string

Simplify

The algorithm can be simplified to these steps:

• Loop over the characters, from the 2nd till the end

• If the current character is the same as the previous, increment the count

• If different, append the count and append the previous character

• After the end of the loop, append the count

The implementation can be something like this:

public String compress(String str) {
    if (str.isEmpty()) {
        return "";
    }

    char[] chars = str.toCharArray();
    StringBuilder builder = new StringBuilder();

    int count = 1;
    char prev = chars[0];
    for (int i = 1; i < chars.length; i++) {
        char current = chars[i];
        if (current == prev) {
            count++;
        } else {
            builder.append(prev).append(count);
            count = 1;
        }
        prev = current;
    }
    return builder.append(prev).append(count).toString();
}

StackExchange Code Review Q#65335, answer score: 9