Super Reduced String

Online challenge on Hacker Rank.

Problem Statement


Steve has a string, S, consisting of n lowercase English alphabetic >letters. In one operation, he can delete any pair of adjacent letters with same >value. For example, string "aabcc" would become either "aab" or "bcc" >after operation.


Steve wants to reduce as much as possible. To do this, he will repeat the >above operation as many times as it can be performed. Help Steve out by finding >and printing S's non-reducible form!


Note: If the final string is empty, print Empty String.

Input Format


A single string, S.

Constraints


\$1 ≤ n ≤ 100\$

Output Format


If the final string is empty, print Empty String; otherwise, print the final >non-reducible string.

Sample Input


aaabccddd

Sample Output


abd

Solution Code

public class Solution {
    private static String solve(String input) {
        int len = input.length();

        int i = 0;
        while (i < len - 1) {
            char current = input.charAt(i);
            char next = input.charAt(i+1);

            if (current == next) {
                input = input.substring(0, i) + input.substring(i+2);
                len = input.length();
                i = 0;
                continue;
            }
            i++;
        }
        if (input.length() == 0) {
            return "Empty String";
        }
        return input;
    }

    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        System.out.println(solve(s.next()));
    }
}

This is a brute-force solution, as I feel kind of confused when dealing with string algorithms. Hope for some good suggestions here.

Doc comments for public methods are indispensable.

I'd probably just use a foreach-loop and a DIY stack:

/** @return input stripped of all abutting identical characters */
static String solve(String input) {
    char []remains = new char[input.length()];
    int top = -1; // highest valid index

    for (char c: input.toCharArray())
        if (top < 0 || c != remains[top])
            remains[++top] = c;
        else
            --top;

    return top < 0 ? "Empty String"
        : new String(remains, 0, top+1);
}

StackExchange Code Review Q#150921, answer score: 5