Regular expression

[ChrisW67]

Kunashir's option will match strings like "aaa", "abb" etc. which are not anagrams of the original string.

I don't know that you can do this in the general case with a single regular expression without literally listing every permutation. In cases of repeated characters you are able to collapse the list:

"abc" => abc|acb|bac|bca|cab|cba  // all the permutations
"abc" => a(bc|cb)|b(ac|ca)|c(ab|ba)  // a different arrangement of all permutations
 
"bob" => bob|bbo|obb|obb|bbo|bob // all the permutations, note the repeats
"bob" => b(ob|bo)|obb // a simpler RE

"abc" => abc|acb|bac|bca|cab|cba  // all the permutations
"abc" => a(bc|cb)|b(ac|ca)|c(ab|ba)  // a different arrangement of all permutations

"bob" => bob|bbo|obb|obb|bbo|bob // all the permutations, note the repeats
"bob" => b(ob|bo)|obb // a simpler RE

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The number of permutations rises rapidly with string length also. You may quickly find the limits of the regular expression engine.

[stampede]

To check if string is anagram of given input you can always use brute-force algorithm that removes each character of input from string. In the end, if you are left with empty string, then it was a permutation of input characters. In pseudo-code:

bool isAnagram( baseWord, toTest ){
    if( baseWord.length() != toTest.length() ) return false; // check obvious condition - cannot be a permutation if numbers of elements are not equal
    foreach( char c, baseWord ){
        toTest.removeExactlyOne(c);
    }
    return toTest.isEmpty();
}

bool isAnagram( baseWord, toTest ){
    if( baseWord.length() != toTest.length() ) return false; // check obvious condition - cannot be a permutation if numbers of elements are not equal
    foreach( char c, baseWord ){
        toTest.removeExactlyOne(c);
    }
    return toTest.isEmpty();
}

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Not very good complexity ( O(N^2), N - number of chars in string ), but its better than nothing.

[kunashir]

ChirsW67, let me disagree with you.
[abc] - It is any symbol from range ('a','b','c'), and repeated three times.
I check it in "RegExp Planner".

[stampede]

@kunashir:
I think ChrisW67 is right:

#include <QDebug>
#include <QRegExp>
 
int main(int argc, char **argv) {
    QRegExp ex("[abc]{3,}");
    qDebug() << ex.exactMatch("aaa");
    return 0;
}

#include <QDebug>
#include <QRegExp>

int main(int argc, char **argv) {
    QRegExp ex("[abc]{3,}");
    qDebug() << ex.exactMatch("aaa");
    return 0;
}

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prints 'true'.

Another algorithm I can think of, is to use O( n log(n) ) sorting method, eg. merge sort, something like:

bool isAnagram(baseWord, toTest){
    if( baseWord.length() == toTest.length() ){
        return sort(baseWord) == sort(toTest);
    }
    return false;
}

bool isAnagram(baseWord, toTest){
    if( baseWord.length() == toTest.length() ){
        return sort(baseWord) == sort(toTest);
    }
    return false;
}

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Complexity of this algorithm is the same as sorting method (string equality test should be linear), so its better than my previous suggestion.

[ChrisW67]

@kunashir: Your regular expression is perfectly fine but it doesn't quite answer the question posed. An anagram uses all the letters of the original word in another order.


There might be something of use here. I haven't tested that code though.