从具有特定元素的大集合中查找所有顺序子集

Question

The problem is defined as follows: Input Element, assume m

Large set, assume {a, b, c, d, ..z}

I want to find all the subsets of length ranging from 2-5 elements containing the input word, m. Condition: The sequence of elements should remain same.

Output:

• {l, m}, {m,n},

• {k, l, m}, {l, m, n}, {m, n, o},

• {j, k, l, m}, {k, l, m, n}, {l, m, n, o}, {m, n, o, p}... and so on

I was able to get the subsets starting at the input word by the following code:

    ArrayList<String> phrases = new ArrayList<>();

    for (int j=1; j<=k-i; j++)  {
        String newSet = set[i] +" ";
        for (int x=1; x<=j; x++)    {
            newSet=newSet+set[i+x]+" ";
        }
        phrases.add(newSet.trim());
    }
    return phrases;
}

Answer 1

You write: "The sequence of elements should remain". So, I assume that you don't mean sets because sets don't have an order, but rather something like lists and sequences.

My proposal is the following: Find the sequences with length 2 first, then the ones with length 3 and so on. This can be done for, let's say length 4 the following way. Find the index of input m in the big "set". Then start with the sequence that ends with m as first element of the result. This is jklm . Move the "window" one step to the right until the found sequence begins with m . So, you get klmn , lmno and mnop .

This can be done by maintaining the begin of the current found sequence as index. It has to initialized with the index of m minus 4 as the current length plus 1. Then you have to iterate 4 times.

Answer 2

As pointed out by the other answer and also the comments,

1. first get the index of the input word
2. take sublists of your set both to the left and right from the input word incrementing one step until you reach their respective ends(the first and the last elements of the list) or until you reach your maximum length which is 5.

3. you should expect a list of lists containing strings.

    private static List<List<String>> getSubSet(List<String> set, String word){ List<List<String>> phrases = new ArrayList<>(); int indexOfWord = set.indexOf(word); int len = 1; while(indexOfWord-len>=0 || indexOfWord+len<=set.size()){ if (indexOfWord-len>=0) phrases.add(set.subList(indexOfWord-len, indexOfWord+1)); if (indexOfWord+len<set.size()) phrases.add(set.subList(indexOfWord, indexOfWord+len+1)); len++; if(len>4) break; } return phrases; } 

   To see a sample test case, I have taken strings of single elements just like you:

   public static void main(String[] args) { List list= Arrays.asList("a","b","c","d","e","f","g","h","i","j"); String inWord = "c"; List> phrases = getSubSet(list,inWord); for(List p: phrases){ System.out.println(p); } }

Output is:

[b, c]
[c, d]
[a, b, c]
[c, d, e]
[c, d, e, f]
[c, d, e, f, g]