Permutations of a String

Question: Print all the permutations of a string. Write code in C.

Answer: Here is a recursive solution to print all the permutations of a string. However, this solution does not take care of duplicates. It is assumed that there are no duplicates in the string. I left out the implementation of the swap method since that implementation is not important here.

The idea is to keep the first character constant and generate permutations with the rest. Then keep the first two characters constant and generate permutations with the rest until you are out of characters :)

void permutate( char[] str, int index )
{
    int i = 0;
    if( index == strlen(str) )
    { // We have a permutation so print it
        printf(str);
        return;
    }
    for( i = index; i < strlen(str); i++ )
    {
        swap( str[index], str[i] ); // It doesn't matter how you swap.
        permutate( str, index + 1 );
        swap( str[index], str[i] );
    }
}
permutate( "abcdefgh", 0 );

Now what do we do if there are duplicates in the string? The trick is to sort the characters in the alphabetical order first. We can then ignore the duplicates easily when generate the permutation.

void permutate( char[] str, int index )
{
    int i = 0;
    static lastChar = 0;
    if( index == strlen(str) )
    { // We have a permutation so print it
        printf(str);
        return;
    }
    for( i = index; i < strlen(str); i++ )
    {
        if( lastChar == str[i] ) {
            continue;
        } else {
            lastChar = str[i];
        }
        swap( str[index], str[i] ); // It doesn't matter how you swap.
        permutate( str, index + 1 );
        swap( str[index], str[i] );
    }
}
permutate( sort("Hello World"), 0 );

Simple right? Any comments or suggestions are always welcome.

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5 Responses

  1. Jerry says:

    sorry, I meant just “char lastChar = 0;”

  2. Jerry says:

    I don’t think its correct to put “static lastChar = 0;” to detect duplicates.

    I guess just “static char lastChar = 0;” would do the trick.

    Any comments ??.

  3. Bansal says:

    In case of duplicates the function doesn’t produce some results.
    e.g for string “aaaa” the string “aaa” itself is not generated.

    Also the assumption that all duplicate characters are clubbed together (because of initial sorting) doesn’t hold while permuting.

    A simple solution would be to keep track of all characters used at a stack call while permuting.

    void permutate( char[] str, int index )
    {
    if( index == strlen(str) )
    { // We have a permutation so print it
    printf(str);
    return;
    }

    char used[255];
    bzero(used, sizeof(used));

    int i;
    for( i = index; i < strlen(str); i++ )
    {
    if( (int)used[arr[i]] != 0)
    {
    continue;
    }

    used[arr[i]] = 1;

    swap( str[index], str[i] ); // It doesn't matter how you swap.
    permutate( str, index + 1 );
    swap( str[index], str[i] );
    }
    }

  4. LeFunk says:

    Nah, the latter function creates duplicates, if some characters repeat in the word – try “poop” for example.

  5. Blech. Far too complex for a simple problem. Here’s a simpler and easier to understand solution in ActionScript 3 (nothing fancy) that yields no duplicates.


    function p(s:String, pre:String):void {
    if (s.length < 1)
    trace(pre);
    else
    for (var i:int = 0; i < s.length; i++)
    p(s.substr(0,i) + s.substr(i+1), pre + s.charAt(i));
    }

    p("abcd", "");

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