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Friday, October 12, 2012

String permutation with backtracking

This is probably one of classic interview questions. There are many solutions for this problem out there but today, a friend of mine mentioned this problem and I know that I solved this long time ago but my memory was fading so I decided to try this out.

As I thought about the problem, I decided to tackle this with backtracking. Basically the idea is that I take one character out from the initial word and mark that character to note that the character has been taken out for permutation. All along, I am passing just one character array and each time I reach the end, I simply print out the resulted string.

So let me show my code.
void perm_internal(string word, char *output, int n, int k)
{
    if (n == k) {
        cout << "[" << output << "]" << endl;
        return;
    }

    for (int i = 0; i < n; i++) {
        if (word[i] == 0) {
            continue;
        }
        char tmp = word[i];
        output[k] = tmp;
        word[i] = 0;
        perm_internal(word, output, n, k+1);
        word[i] = tmp;
    }
}

void perm(string word)
{
    int n = word.length();
    char *output;

    if (n == 0)
        return;
    output = (char *)malloc(n+1);
    memset(output, 0x0, n+1); 
    cout << "input word: " << word << "(" << n << ")" << endl;
    perm_internal(word, output, n, 0);
}


I want to try out different approaches to solve this problem but it's getting late so I will do that next time.

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