# TLE on USACO "Subset Equality"

For Silver Problem 2 on the 2022 USACO Open (http://www.usaco.org/index.php?page=viewproblem2&cpid=1231), my solution times out even though the number of operations it does seem to fit within the time limit.

I was testing my solution, and I found that this piece of code is what causes it to time out:

for (int i=0; i<18; i++){
for (int j=i+1; j<18; j++){
if (check(i, j)) eq[(1<<i)+(1<<j)]=1;
else eq[(1<<i)+(1<<j)]=0;
}
}


where I defined the function check as :

 bool check(int i, int j){
if (posS[i].size()!=posT[i].size() || posS[j].size()!=posT[j].size()) return false;
string s1="", t1="";
int pi=0, pj=0;
for (int k=0; k<posS[i].size()+posT[j].size(); k++){
if (pi<posS[i].size() && pj<posS[j].size()){
if (posS[i][pi]<posS[j][pj]) s1=s1+let[i], pi++;
else s1=s1+let[j], pj++;
}
else if (pi>=posS[i].size()) s1=s1+let[j], pj++;
else s1=s1+let[i], pi++;
}
pi=0, pj=0;
for (int k=0; k<posT[i].size()+posT[j].size(); k++){
if (pi<posT[i].size() && pj<posT[j].size()){
if (posT[i][pi]<posT[j][pj]) t1=t1+let[i], pi++;
else t1=t1+let[j], pj++;
}
else if (pi>=posT[i].size()) t1=t1+let[j], pj++;
else t1=t1+let[i], pi++;
}
return(s1==t1);
}


Basically, at the beginning I iterated through both strings s and t once, and kept track of the positions of each letter (with let[0]=βaβ, let[1]=βbβ, β¦, let[17]=βrβ). This takes O(|s|+|t|).

Then, I iterated through all pairs of letters: the check function uses two-pointers to reconstruct s, t when restricted to that pair of letters. But since there are 153 pairs, doing the check function on each pair should take a total of at most O(153*(|s|+|t|)) since there are that many pairs (and actually it should take a lot less time than that due to amortization). Could someone please help me figure out why the solution is timing out?

(The full code is below if anyone needs it)

#include <iostream>
#include <bits/stdc++.h>
#include <fstream>
#include <string>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <math.h>
#include <numeric>
using namespace std;
using ll=long long;

int eq[1<<18];
vector<int> posS[18], posT[18];
char let[18]={'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r'};

bool check(int i, int j){
if (posS[i].size()!=posT[i].size() || posS[j].size()!=posT[j].size()) return false;
string s1="", t1="";
int pi=0, pj=0;
for (int k=0; k<posS[i].size()+posT[j].size(); k++){
if (pi<posS[i].size() && pj<posS[j].size()){
if (posS[i][pi]<posS[j][pj]) s1=s1+let[i], pi++;
else s1=s1+let[j], pj++;
}
else if (pi>=posS[i].size()) s1=s1+let[j], pj++;
else s1=s1+let[i], pi++;
}
pi=0, pj=0;
for (int k=0; k<posT[i].size()+posT[j].size(); k++){
if (pi<posT[i].size() && pj<posT[j].size()){
if (posT[i][pi]<posT[j][pj]) t1=t1+let[i], pi++;
else t1=t1+let[j], pj++;
}
else if (pi>=posT[i].size()) t1=t1+let[j], pj++;
else t1=t1+let[i], pi++;
}
// cout<<s1<<" "<<t1<<"\n";
return(s1==t1);
}

int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
// ifstream cin("file.in");
string s, t;
int Q;
cin>>s>>t>>Q;
eq[0]=0;

for (int i=0; i<s.size(); i++) posS[(int)(s[i]-'a')].push_back(i);
for (int i=0; i<t.size(); i++) posT[(int)(t[i]-'a')].push_back(i);
for (int i=0; i<18; i++){
if (posS[i].size()==posT[i].size()) eq[1<<i]=1;
else eq[1<<i]=0;
}
for (int i=0; i<18; i++){
for (int j=i+1; j<18; j++){
if (check(i, j)) eq[(1<<i)+(1<<j)]=1;
else eq[(1<<i)+(1<<j)]=0;
}
}
vector<int> subs[19];
vector<int> v;
for (int i=0; i<18; i++){
}
}
for (int i=3; i<=18; i++){
vector<int> v;
for (int j=0; j<18; j++){
}
int works=1;
for (int i: v){
}
}
}
for (int i=0; i<Q; i++){
string s0;
cin>>s0;
}