int Solution::anytwo(string A) {
map<string,pair<int,int>> m;
// Check only for subseq with len =2
for(int i=1;i<A.size();++i)
{
for(int j=0;j<i;++j)
{
string s="";
s=s+A[j]+A[i];
//if subseq s is present in m
if(m.find(s)!=m.end())
{
//check if it have same indexed character in the sequence
//or check if two subsequences overlap
// [1st subseq= present in map] [2nd subseq= s]
if(m[s].first!=j && m[s].second!=i )
return 1;
}
//if subseq not present in map
//insert into map with values of index of first and second char
// Note : subseq of only two char so indexes stored in a pair
else
{
m[s].first=j;
m[s].second=i;
}
}
}
//if no subseq of len two satisfy the question return 0
return 0;
}