CODEWITHSUNDEEP
javastringalgorithm
Klaus
Klaus

Reputation: 1731

Generating Strings By Rotating Elements

I am trying to develop an algorithm which takes in an input string S and generate Strings by eliminating two characters at a time. Here's my code so far.

    String[] strings = {"a", "b", "c", "d", "e", "f"};
    int charsToBeRemoved = strings.length - 2;

    for( int i = 0; i < strings.length; i++ )
    {
        int counter = 1;
        StringJoiner p = new StringJoiner( "|" );
        p.add( strings[i] );
        for( int j = i + 1; j < strings.length; j++ )
        {
            p.add( strings[j] );
            counter += 1;
            if( counter == charsToBeRemoved )
            {
                counter = 1;
                j = ( charsToBeRemoved - 2 ) > 0 ? j - ( charsToBeRemoved - 2 ) : j;
                regexList.add( p.toString() );
                p = new StringJoiner( "|" );
                p.add( strings[i] );
            }
        }
    }

This provides the output : [a|b|c|d, a|c|d|e, a|d|e|f, b|c|d|e, b|d|e|f, c|d|e|f] . However, what I am trying to get is a rotating array based solution. Something like [a|b|c|d, a|c|d|e, a|d|e|f, b|c|d|e, b|d|e|f, b|e|f|a, c|d|e|f, c|e|f|a, c|f|a|b], d|e|f|a, d|f|a|b, d|a|b|c. In my current code, these outputs b|e|f|a, c|e|f|a, c|f|a|b, d|e|f|a, d|f|a|b, d|a|b|care not formed because I need a way to access the previous elements of the array.

Can anyone suggest a way to do so, or how I should change my approach?

Upvotes: 0

Views: 53

Answers (1)

RaffleBuffle
RaffleBuffle

Reputation: 5455

To "wrap around" to the beginning of the array you'd use the % mod operator.

String[] strings = { "a", "b", "c", "d", "e", "f" };
int reduction = 2;

for(int i=0; i<strings.length; i++)
{
    for(int j=0; j<=reduction; j++)
    {
        StringJoiner p = new StringJoiner("|");
        p.add(strings[i]);
        for(int k=1; k<strings.length-reduction; k++)
            p.add(strings[(i + j + k) % strings.length]);
        System.out.println(p);
    }
}

Output:

a|b|c|d
a|c|d|e
a|d|e|f
b|c|d|e
b|d|e|f
b|e|f|a
c|d|e|f
c|e|f|a
c|f|a|b
d|e|f|a
d|f|a|b
d|a|b|c
e|f|a|b
e|a|b|c
e|b|c|d
f|a|b|c
f|b|c|d
f|c|d|e

Upvotes: 1

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