6-lines C++ Solution using map, 0ms

  • 0

    Even presence of a character is cancelled out, while odd presence of a character remains. If the number of characters with odd presence in the string is more than 1, then Palindrome is not possible. Otherwise, Palindrome is possible.

    bool canPermutePalindrome(string s) {
            map<char, int> result;
            for(char& c:s)
                else result.erase(c);   
            return result.size()<=1;

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