CarrerCup 1.3 Given two strings, write a method to decide if one is a permutation of the other.

1_3.cpp

bool permute_str(const string &str1, const string &str2){
    if (str1.empty() || str2.empty()){
        return false;
    }
    if (str1.length() != str2.length()){
        return false;
    }
    int ascii[256] = {0};
    for (size_t i = 0; i < str1.length(); ++i){
        ++ascii[int(str1[i])];
        --ascii[int(str2[i])];
    }
    for (size_t i = 0; i < 256; ++i){
        if (ascii[i] != 0)
            return false;
    }
    return true;
}

Your time complexity is O(M + N), instead of O(N)

The cost of first loop is N=length of string and the cost of second loop is at most 256. So the time complexity is O(N+256) = O(N). I read your code and I think we use the same algorithm.