int Solution::anytwo(string A)
{
//count occurence of each char
//if occurence is goes more than 2 than return 1;
int m[256]={0};
for(auto x:A)
if(++m[x]>2)
return 1;
//create extra string with occurence 2 is same as greater than 1;
string tmp="";
for(auto x:A)
if(m[x]>1)
tmp+=x;
//other extra string which is reverse of previos
string rtmp=tmp;
reverse(rtmp.begin(),rtmp.end());
//if string is palindrom than return false 0;
if(tmp==rtmp)
return 0;
//most of the time u will get 1 ;)
return 1;
}