Power of String
Problem Statement: here
Solution:
We have to find the longest substring occurring in the string at least K times.
The golden point is:
EVRY SUBSTRING IS A PREFIX OF SOME SUFFIX.
So, we can quickly calculate the suffix array and LCP array of the given string.
Once, we have calculated the suffix array, we now want the substring of the maximum common prefix, but it should be present at least k times. This means, in LCP array, it should be present in at least k-1 consecutive elements. So, we can use the sliding window technique to find out the minimum in a sliding window of length k-1.
Why does the longest substring need to be present in consecutive suffix array indices??
Since we know that suffix array is nothing but lexicographically shortest suffixes array. Let's say the length of the longest substring is x, and let us represent is by L. Then our string can be represented as:
S = A₁LA₂L....Aₚ, where L is the longest substring repeating at least K times. So, when we look at the suffix array, all the suffixes which start with this L, they have to be present together in our suffix array, because a large part of it is common. But, it might be possible that we have a suffix that starts from L and shares prefix length much larger than just L. Thus, we consider the minimum of the entire window because we know in minimum, they will be sharing exactly L length, if they would have been sharing more, that would have already been included in counting for L.
Complexities:
Building Suffix Array - O(NlogN)
Building LCP Table - O(N)
Computing minimum in a sliding window - O(N)
Thus, the overall complexity is O(NlogN). We could have even used any RMQ data structure like segment trees to answer the minimum in the sliding window, but we have used the most optimal way of doing it - using a monotonic queue. I will make a separate blog post (maybe) discussing the monotonic queue and its wide applications.
The code:
#include<bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define M1 1000000007
#define M2 998244353
#define ll long long int
#define pll pair<ll,ll>
#define mll map<ll,ll>
#define F first
#define S second
#define PB push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define V(a) vector<a>
#define endl '\n'
#define test(t) while(t--)
#define PI acos(-1.0)
#define rep(i,a,b) for(ll i=a;i<b;i++)
#define repp(i,b,a) for(ll i=b-1;i>=a;i--)
#define clr(ar, val) memset(ar, val, sizeof(ar))
#define setbits(x) __builtin_popcountll(x)
#define zrobits(x) __builtin_ctzll(x)
#define ps(y) fixed << setprecision(y)
#define all(x) begin(x),end(x)
#define allr(x) rbegin(x),rend(x)
const int inf= 0x3f3f3f3f;
const ll INF= 0x3f3f3f3f3f3f3f3f;
const int dx[4]= { 0, -1, 0, 1 };
const int dy[4]= { -1, 0, 1, 0 };
ll floor_div(ll a, ll b) {
return a / b - (((a ^ b) < 0) and a % b);
}
ll ceil_div(ll a, ll b){
return a / b + (((a ^ b) >= 0) and a % b);
}
vector<int> suffix_array_cyclic(string s){
int n = s.size(), N = n + 256;
vector<int> sa(n), ra(n);
for(int i = 0; i < n; i++) sa[i] = i, ra[i] = s[i];
for(int k = 0; k < n; k ? k *= 2 : k++)
{
vector<int> nsa(sa), nra(n), cnt(N);
for(int i = 0; i < n; i++) nsa[i] = (nsa[i] - k + n) % n;
for(int i = 0; i < n; i++) cnt[ra[i]]++;
for(int i = 1; i < N; i++) cnt[i] += cnt[i - 1];
for(int i = n - 1; i >= 0; i--) sa[--cnt[ra[nsa[i]]]] = nsa[i];
int r = 0;
for(int i = 1; i < n; i++)
{
if(ra[sa[i]] != ra[sa[i - 1]]) r++;
else if(ra[(sa[i] + k) % n] != ra[(sa[i - 1] + k) % n]) r++;
nra[sa[i]] = r;
}
ra = nra;
}
return sa;
}
vector<int> kasai(string s, vector<int> sa)
{
int n = s.size(), k = 0;
vector<int> ra(n), lcp(n);
for(int i = 0; i < n; i++) ra[sa[i]] = i;
for(int i = 0; i < n; i++)
{
if(k) k--;
if(ra[i] == n - 1) {k = 0; continue;}
int j = sa[ra[i] + 1];
while(k < n && s[(i + k) % n] == s[(j + k) % n]) k++;
lcp[ra[i]] = k;
if(ra[(sa[ra[i]] + 1) % n] > ra[(sa[ra[j]] + 1) % n]) k = 0;
}
return lcp;
}
inline void INP()
{
#ifndef ONLINE_JUDGE
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
}
void solve()
{
ll n,k;
cin >> k >> n;
if (k == 1){
cout << n << endl;
return;
}
string s;
cin >> s;
s += '#';
vector<int> sa = suffix_array_cyclic(s);
vector<int> lcp = kasai(s,sa);
lcp.erase(lcp.begin());
// for(int i: lcp) cout << i << " ";cout << endl;
deque<pair<int,int>> dq;
int gmax = INT_MIN;
rep(i,0,k-1){
while(!dq.empty() and dq.back().F >= lcp[i]){
dq.pop_back();
}
dq.push_back({lcp[i],i});
}
gmax = max(gmax,dq.front().F);
rep(i,k-1,lcp.size()-1){
while(!dq.empty() and dq.back().F >= lcp[i]){
dq.pop_back();
}
dq.push_back({lcp[i],i});
while(dq.front().S <= i-(k-1)) dq.pop_front();
gmax = max(gmax,dq.front().F);
}
cout << gmax << endl;
}
int32_t main(){
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
int t=1;
// cin>>t;
test(t){
solve();
}
return 0;
}
