Generating all permutations of a certain length

Question

Suppose we have an alphabet "abcdefghiklimnop". How can I recursively generate permutations with repetition of this alphabet in groups of FIVE in an efficient way?

I have been struggling with this a few days now. Any feedback would be helpful.

Essentially this is the same as: Generating all permutations of a given string

However, I just want the permutations in lengths of FIVE of the entire string. And I have not been able to figure this out.

SO for all substrings of length 5 of "abcdefghiklimnop", find the permutations of the substring. For example, if the substring was abcdef, I would want all of the permutations of that, or if the substring was defli, I would want all of the permutations of that substring. The code below gives me all permutations of a string but I would like to use to find all permutations of all substrings of size 5 of a string.

    public static void permutation(String str) { 
    permutation("", str); 
}
private static void permutation(String prefix, String str) {
    int n = str.length();
    if (n == 0) System.out.println(prefix);
    else {
        for (int i = 0; i < n; i++)
            permutation(prefix + str.charAt(i), str.substring(0, i) + str.substring(i+1, n));
    }
}

Can you not generate them all, then loop over them all and take the first 5 characters? — OneCricketeer, Feb 10 '16 at 18:46
@cricket_007 This would produce a lot of repetitions. Besides, OP asks for an efficient way of generating them all. — Sergey Kalinichenko, Feb 10 '16 at 18:47
Use the algorithm you found, but [generate all substrings of length 5](http://stackoverflow.com/questions/2560262/generate-all-unique-substrings-for-given-string), first. — OneCricketeer, Feb 10 '16 at 18:50

score 10 · Accepted Answer · answered Feb 10 '16 at 18:53

In order to pick five characters from a string recursively, follow a simple algorithm:

Your method should get a portion filled in so far, and the first position in the five-character permutation that needs a character
If the first position that needs a character is above five, you are done; print the combination that you have so far, and return
Otherwise, put each character into the current position in the permutation, and make a recursive call

This is a lot shorter in Java:

private static void permutation(char[] perm, int pos, String str) {
    if (pos == perm.length) {
        System.out.println(new String(perm));
    } else {
        for (int i = 0 ; i < str.length() ; i++) {
            perm[pos] = str.charAt(i);
            permutation(perm, pos+1, str);
        }
    }
}

The caller controls the desired length of permutation by changing the number of elements in perm:

char[] perm = new char[5];
permutation(perm, 0, "abcdefghiklimnop");

Demo.

@cdaiga Follow the demo link and scroll to the bottom to see the results of running this program. — Sergey Kalinichenko, Feb 10 '16 at 19:05

score 1 · Answer 2 · answered Feb 10 '16 at 18:59

All permutations of five characters will be contained in the set of the first five characters of every permutation. For example, if you want all two character permutations of a four character string 'abcd' you can obtain them from all permutations: 'abcd', 'abdc', 'acbd','acdb' ... 'dcba'

So instead of printing them in your method you can store them to a list after checking to see if that permutation is already stored. The list can either be passed in to the function or a static field, depending on your specification.

score 0 · Answer 3 · answered May 08 '17 at 10:59

This is can be easily done using bit manipulation.

private void getPermutation(String str, int length)
        {
            if(str==null)
                return;
            Set<String> StrList = new HashSet<String>();
            StringBuilder strB= new StringBuilder();
            for(int i = 0;i < (1 << str.length()); ++i)
            {
                strB.setLength(0); //clear the StringBuilder
                if(getNumberOfOnes(i)==length){
                  for(int j = 0;j < str.length() ;++j){
                    if((i & (1 << j))>0){  // to check whether jth bit is set (is 1 or not)
                        strB.append(str.charAt(j));
                    }
                }
                StrList.add(strB.toString());
                }
            }
            System.out.println(Arrays.toString(StrList.toArray()));
        }

    private int getNumberOfOnes (int n) // to count how many numbers of 1 in binary representation of n
    {
        int count=0;
        while( n>0 )
        {
        n = n&(n-1);
           count++;
        }
        return count;
    }

The question is asking for permutations. But this piece of code would give the combinations — midhun mathew, Jan 13 '21 at 05:45

score 0 · Answer 4 · answered Feb 01 '21 at 06:29

class StringPermutationOfKLength
{
    // The main recursive method
    // to print all possible
    // strings of length k
    static void printAllKLengthRec(char[] set,String prefix,
                                   int n, int k)
    {

        // Base case: k is 0,
        // print prefix
        if (k == 0)
        {
            System.out.println(prefix);
            return;
        }

        // One by one add all characters
        // from set and recursively
        // call for k equals to k-1
        for (int i = 0; i < n; i++)
        {

            // Next character of input added
            String newPrefix = prefix + set[i];

            // k is decreased, because
            // we have added a new character
            printAllKLengthRec(set, newPrefix,
                n, k - 1);
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        System.out.println("First Test");
        char[] set1 = {'a', 'b','c', 'd'};
        int k = 2;
        printAllKLengthRec(set1, "", set1.length, k);

        System.out.println("\nSecond Test");
        char[] set2 = {'a', 'b', 'c', 'd'};
        k = 1;
        printAllKLengthRec(set2, "", set2.length, k);
    }

Generating all permutations of a certain length

4 Answers4

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