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Programming101
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HackerEarth The Suffix Matches problem solution

YASH PAL, 31 July 2024
In this HackerEarth The Suffix Matches problem solution Rob needs to give a string to each of his students. There is huge string S that is known to all. He doesn’t want to work much, hence decides to give each of them, some prefix of S.
Now, the students consider their strings holy, if their string is very similar to this known string S. Now, this similarity comparison is done from suffix matching.
So, now you are to help in this computation. So, given an integer i, find the longest matching suffix in between S[1…i] and S.
HackerEarth The Suffix Matches problem solution

HackerEarth The Suffix Matches problem solution.

#include<iostream>
#include<bits/stdc++.h>
using namespace std;
#define MOD 1000000007
#define ft first
#define sd second
#define VI vector<int>
#define VLL vector<long long int>
#define PII pair<int,int>
#define pb push_back
#define rsz(v,n) v.resize(n)
#define scan(x) scanf("%d",&x)
#define scanll(x) scanf("%lld",&x)
#define ll long long int
#define rep(i,x,y) for(i=x;i<y;i++)
#define print(x) printf("%dn",x)
#define printll(x) printf("%lldn",x)
#define all(v) v.begin(),v.end()
#define ms(v) memset(v,0,sizeof(v))
#define FOR(i,a,b) for(i=a;i<b;i++)
#define f_in(st) freopen(st,"r",stdin)
#define f_out(st) freopen(st,"w",stdout)
#define PIE 3.14159265358979323846264338327950
#ifdef ONLINE_JUDGE
inline void inp( int &n )
{
n=0;
int ch=getchar_unlocked();int sign=1;
while( ch < '0' || ch > '9' ){if(ch=='-')sign=-1; ch=getchar_unlocked();}

while( ch >= '0' && ch <= '9' )
n = (n<<3)+(n<<1) + ch-'0', ch=getchar_unlocked();
n=n*sign;
}
#else
inline void inp(int &n){
cin>>n;
}
#endif
const int MAX = 200100;
char A[MAX],C[MAX];
string txt; //input
int iSA[MAX], SA[MAX]; //output
int cnt[MAX], next_gen[MAX]; //internal
bool bh[MAX], b2h[MAX],m_arr[MAX];
int len;
bool smaller_first_char(int a, int b){
return txt[a] < txt[b];
}
inline void suffixSort(int n){

for (int i=0; i<n; ++i){
SA[i] = i;
}
sort(SA, SA + n, smaller_first_char);


for (int i=0; i<n; ++i){
bh[i] = i == 0 || txt[SA[i]] != txt[SA[i-1]];
b2h[i] = false;
}

for (int h = 1; h < n; h <<= 1){

int buckets = 0;
for (int i=0, j; i < n; i = j){
j = i + 1;
while (j < n && !bh[j]) j++;
next_gen[i] = j;
buckets++;
}
if (buckets == n) break; // We are done! Lucky bastards!

for (int i = 0; i < n; i = next_gen[i]){
cnt[i] = 0;
for (int j = i; j < next_gen[i]; ++j){
iSA[SA[j]] = i;
}
}

cnt[iSA[n - h]]++;
b2h[iSA[n - h]] = true;
for (int i = 0; i < n; i = next_gen[i]){
for (int j = i; j < next_gen[i]; ++j){
int s = SA[j] - h;
if (s >= 0){
int head = iSA[s];
iSA[s] = head + cnt[head]++;
b2h[iSA[s]] = true;
}
}
for (int j = i; j < next_gen[i]; ++j){
int s = SA[j] - h;
if (s >= 0 && b2h[iSA[s]]){
for (int k = iSA[s]+1; !bh[k] && b2h[k]; k++) b2h[k] = false;
}
}
}
for (int i=0; i<n; ++i){
SA[iSA[i]] = i;
bh[i] |= b2h[i];
}
}
for (int i=0; i<n; ++i){
iSA[SA[i]] = i;
}
}
int lcp[MAX];
inline void getlcp(int n)
{
for (int i=0; i<n; ++i)
iSA[SA[i]] = i;

lcp[0] = 0;

for (int i=0, h=0; i<n; ++i)
{
if (iSA[i] > 0)
{
int j = SA[iSA[i]-1];
while (i + h < n && j + h < n && txt[i+h] == txt[j+h])
h++;
lcp[iSA[i]] = h;

if (h > 0)
h--;
}
}
}
int LCP[MAX][22];

inline void initialize_lcp()
{
for(int i = 0;i<len;++i)
LCP[i][0] = lcp[i];

for(int j = 1;(1<<j)<=len;++j){
for(int i = 0;i+(1<<j)-1<len;++i){
if(LCP[i][j-1]<=LCP[i+ ( 1<<(j-1) )][j-1])
LCP[i][j] = LCP[i][j-1];
else
LCP[i][j] = LCP[i+(1<<(j-1))][j-1];
}
}
}

inline int lcpmain(int x, int y)
{
if(x == y) return len-iSA[x];
if(x > y) swap(x,y);

int log = 0;
while((1<<log)<=(y-x)) ++log;
--log;

return min(LCP[x+1][log],LCP[y-(1<<log)+1][log]);
}
int main()
{
cin>>txt;
len=txt.length();
assert(len<=200000);
reverse(all(txt));
suffixSort(len);
getlcp(len);
initialize_lcp();
int q;
inp(q);
assert(q<=200000);
while(q--)
{
int idx;
inp(idx);
idx--;
idx=len-idx;
idx--;
if(idx==0)
print(len);
else
print(lcpmain(iSA[idx],iSA[0]));

}
return 0;
}
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