Solution: Group Anagrams

#include <iostream>
#include <vector>
using namespace std;
void printStrArr(string arr[], int arrSize) {
  for (int i = 0; i < arrSize; i++)
    cout << arr[i] << endl;
}
vector <string> groupAnagrams(string arr[], int arrSize) {
  string strArr[arrSize];
  vector <string> output;
  int count = 0, k = 0;
  for (int i = 0; i < arrSize; i++) {
    for (int j = 0; j < arrSize; j++) {
      if(i == j)
        break;
      if(arr[i].length() == arr[j].length()) {
        // If a character of str1 not in c2
        for(int c1 = 0; c1 < arr[i].length(); c1++ ) {
          for(int c2 = 0; c2 < arr[j].length(); c2++) {
            if(arr[i][c1] == arr[j][c2])
              count += 1;
          }
        }
        if(count == arr[j].length()) {
          output.push_back(arr[j]);
          output.push_back(arr[i]);
        }
        count = 0;
      }
    }
  }
  return output;
}
int main() {
  string arr[] = {
    "tom marvolo riddle ",
    "abc",
    "def",
    "cab",
    "fed",
    "clint eastwood ",
    "i am lord voldemort",
    "elvis",
    "old west action",
    "lives"
  };
  int arrSize = 10;
  vector<string> vec;
  vec = groupAnagrams(arr, arrSize);
  for (std::vector<string>::const_iterator i = vec.begin(); i != vec.end(); ++i)
    std::cout << *i << ' ';
}