Question

#include <iostream>
#include <string>
#include <algorithm>
#include <cstdlib>
#include <cstdio>

using namespace std;

static bool isanagram(string a, string b);

int main(void)
{
    int i,n,j,s;
    cin >> n;
    string a, b;
    cin >> a >> b;
    if(!isanagram(a,b)) cout << "False" << endl;
    else cout << "True" << endl;
    return 0;


}

static bool isanagram(string a, string b)
{
    int i, j, size, s=0;
    size = a.size();
    bool k;
    for(i=0;i<size;i++)
    {
        k=false;
        for(j=0;j<size;j++)
        {
            if(a[i] == b[j]) { k = true; break; }
        }
        if(k==true) s+=1;
    }
    cout << a[2] << b[2] << endl;
    if(s == size) return true;
    else return false;

}

I don’t know where exactly is the problem so i just pasted the whole code.

It should be a simple program capable for finding if two strings are anagrams, but it’s not working and i don’t know why. I used pointers in the program so thought the might be the problem and removed them, i removed other things additionally but still it’s not working. If you can give it a look-see and tell me some idea where i might’ve gone wrong with my code ?

Thank you in advance.

Answer 1

First things first: don’t declare the method static. It’s a confusing keyword at the best of times given all the roles it can fulfill… so reserve for times when you really have to (method or attribute of a class that is not tied to any instance for example).

Regarding the algorithm: you’re nearly there, but presence only is not sufficient, you need to take the number of characters in account too.

Let’s do it simply:

bool anagram(std::string const& lhs, std::string const& rhs)
{
  if (lhs.size() != rhs.size()) return false; // does not cost much...

  std::vector<int> count(256, 0); // count of characters
  for (size_t i = 0, max = lhs.size(); i != max; ++i)
  {
    ++count[lhs[i]];
    --count[rhs[i]];
  }

  for (size_t i = 0, max = count.size(); i != max; ++i)
    if (count[i] != 0) return false;

  return true;
} // anagram

Let’s see it at work: anagram("abc","cab")

1. Initialization: count = [0, 0, ...., 0]
2. First loop i == 0 > ['a': 1, 'c': -1]
3. First loop i == 1 > ['a': 0, 'b': 1, 'c': -1]
4. First loop i == 2 > ['a': 0, 'b': 0, 'c': 0 ]

And the second loop will pass without any problem.

Variants include maintaining 2 counts arrays (one for each strings) and then comparing them. It’s slightly less efficient… does not really matter though.

int main(int argc, char* argv[])
{
  if (argc != 3) std::cout << "Usage: Program Word1 Word2" << std::endl;
  else std::cout << argv[1] << " and " << argv[2] << " are "
                 << (anagram(argv[1], argv[2]) ? "" : "not ")
                 << "anagrams" << std::endl;
}