Missing some string subsequences from this algorithm

Question

I am looking to come up with the following pattern of string subsequences:

# input - ALEX
# 1 - A, L, E, X
# 2 - AL, AE, AX, LE, LX, EX
# 3 - ALE, ALX, AEX, LEX
# 4 - ALEX

I wrote this, it comes up with most, but I am missing "LX","AE","ALX" and "AEX". Can I get those without adding a third loop?

public class StringPermutations {
    public void stringPermutations(String input) {
        for(int i = 0; i<input.length();i++){
            for(int j = i; j<input.length(); j++){
                String s = input.substring(i,j+1);
                System.out.println(s);
            }
        }
    }

    public static void main(String[] args) {
        new StringPermutations().stringPermutations("ALEX");
    }
}

...sorry for the confusion, the pattern I am seeking is commented at the top. — Alexander Mills
– Alexander Mills, Commented Dec 4, 2014 at 21:37

aioobe · Accepted Answer · 2018-09-04 12:18:15Z

What you're describing is more a problem of finding all subsequences. You can't really do that by two nested for loops. It's easier to solve recursively as follows:

// Find all subsets
public List<String> helper(String input) {
    if (input.isEmpty())
        return Collections.singletonList("");

    List<String> result = new ArrayList<>();
    List<String> subResult = helper(input.substring(1));
    result.addAll(subResult);
    for (String s : subResult)
        result.add(input.charAt(0) + s);
    return result;
}

// Print all subsets using the helper method
public void stringPermutations(String input) {
    for (String s : helper(input))
        System.out.println(s);
}

This solution also includes the empty string. (Not sure why you suppressed the 0 case when describing the pattern at the top of your question ;)

With some lambda foo, you can print out the result as follows:

helper(input).stream()
             .collect(groupingBy(String::length, toList()))
             .entrySet()
             .stream()
             .forEach(e -> System.out.println(e.getKey() + ": " + e.getValue()));

Output:

0: []
1: [X, E, L, A]
2: [EX, LX, LE, AX, AE, AL]
3: [LEX, AEX, ALX, ALE]
4: [ALEX]

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