In this HackerEarth A New Order problem solution Consider a new order of english alphabets (a to z), that we are not aware of. What we have though, dictionary of words, ordered according to the new order.

Our task is to give each alphabet a rank, ordered list of words. The rank of an alphabet is the minimum integer R that can be given to it, such that:

if alphabet Ai has rank Ri and alphabet Aj has Rj, then
  • Rj > Ri if and only if Aj > Ai
  • Rj < Ri if and only if Aj < Ai

HackerEarth A New Order problem solution


HackerEarth A New Order problem solution.

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
string str[100009];
int n,graph[26][26]={};
vector < vector < int > > ans;
void foo(int l, int r, int ind)
{
int start[26],end[26];
memset(start,-1,sizeof(start));
memset(end,-1,sizeof(end));
for(int i=0; i<26; i++)
for(int j=l; j<=r; j++)
if(str[j][ind]==(char)('a'+i))
{
if(start[i]==-1)start[i]=j;
end[i]=j;
}
for(int i=0; i<26; i++)
for(int j=0; j<26; j++)
{
if(start[i]==-1 || start[j]==-1 || i==j)continue;
if(start[j]>end[i])graph[i][j]=1;
else if(start[i]>end[j])graph[j][i]=1;
}
for(int i=0; i<26; i++)
if(start[i]!=-1)foo(start[i],end[i],ind+1);
}
void toposort()
{
int count[26]={};
vector < int > cur;
for(int i=0; i<26; i++)
{
int flag=0;
for(int j=0; j<26; j++)
if(graph[j][i])flag=1,count[i]++;
if(flag==0)
cur.pb(i);
}
while(1)
{
vector < int > ncur;
if(cur.empty())break;
ans.pb(cur);
for(int i=0; i<cur.size(); i++)
for(int j=0; j<26; j++)
if(graph[cur[i]][j])
{
graph[cur[i]][j]=0,count[j]--;
if(count[j]==0)ncur.pb(j);
}
cur.clear();
for(int i=0; i<ncur.size(); i++)
cur.pb(ncur[i]);
}
}
int main()
{
int flag[26]={};
cin >> n;
for(int i=0; i<n; i++)
{
cin >> str[i];
for(int j=0; j<str[i].length(); j++)
flag[str[i][j]-'a']++;
}
foo(0,n-1,0);
/* for(int i=0; i<26; i++)
for(int j=0; j<26; j++)
if(graph[i][j])
cout << (char)(i+'a') << " -> "<< (char)(j+'a') << endl; */
toposort();
for(int i=0; i<ans.size(); i++,cout<<endl)
{
sort(ans[i].begin(),ans[i].end());
for(int j=0; j<ans[i].size(); j++)
if(flag[ans[i][j]])
cout << (char)(ans[i][j]+'a');
}
return 0;
}