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#include "string/z_algorithm.hpp"
#pragma once #include "../template.hpp" /* * z[i] <- LCP(s[0..], s[i..]) * * Finding z[0..N-1] quickly relies on finding an estimation for z[i] (that is less than the actual value), and * increasing it until it reaches the correct value. This works because z[i] stores the LCP of some strings, so if z[i] * is a common prefix, then clearly z[i]-1 is one as well. * * To do this, we maintain the rightmost interval [l = i, r = i+z[i]-1] and solve for i-s sequentially. At any point, * we have a few cases: * * (1) i > r : We start our estimate of z[i] as 0, and extend * (2) i <= r : Also note that in this case, l<=i<=r as we process i-s in order. We know that s[l..r] == s[0..r-l] as * it is an interval that is defined by [k, k+z[k]-1] (for some k). Thus, we use one of our previously computed values * to initialize z[i]. In this case, we use min(z[i-l], r-i+1), which is when we shift the interval [k, k+z[k]-1] to * [0, z[k]-1]. * * Complexity Proof: * * (1) : We are always extending r and since r<N, we will extend at most N times * (2) : * (2a) z[i-l] == r-i+1 : Again, we are extending r so this will happen at most N times (combined with case 1) * (2b) z[i-l] < r-i+1 : We will extend 0 times, as if we could extend z[i], we could also extend z[i-l] as we know * that s[l..r] == s[0..r-l] (and we are still within that range, so the extended character will be the same * at both locations) * * Note: while z[0] is technically undefined, we'll define it as N for now */ template <typename Container> vector<int> z_algorithm(int N, const Container &s) { vector<int> z(N); z[0] = N; int l = 0, r = -1; for (int i = 1; i < N; i++) { if (i <= r) z[i] = min(r-i+1, z[i-l]); while (i + z[i] < N && s[z[i]] == s[i + z[i]]) z[i]++; if (i+z[i]-1 > r) { r = i+z[i]-1; l = i; } } return z; }
#line 2 "template.hpp" #include <bits/stdc++.h> #define DEBUG 1 using namespace std; // Defines #define fs first #define sn second #define pb push_back #define eb emplace_back #define mpr make_pair #define mtp make_tuple #define all(x) (x).begin(), (x).end() // Basic type definitions #if __cplusplus == 201703L // CPP17 only things template <typename T> using opt_ref = optional<reference_wrapper<T>>; // for some templates #endif using ll = long long; using ull = unsigned long long; using ld = long double; using pii = pair<int, int>; using pll = pair<long long, long long>; #ifdef __GNUG__ // PBDS order statistic tree #include <ext/pb_ds/assoc_container.hpp> // Common file #include <ext/pb_ds/tree_policy.hpp> using namespace __gnu_pbds; template <typename T, class comp = less<T>> using os_tree = tree<T, null_type, comp, rb_tree_tag, tree_order_statistics_node_update>; template <typename K, typename V, class comp = less<K>> using treemap = tree<K, V, comp, rb_tree_tag, tree_order_statistics_node_update>; // HashSet #include <ext/pb_ds/assoc_container.hpp> template <typename T, class Hash> using hashset = gp_hash_table<T, null_type, Hash>; template <typename K, typename V, class Hash> using hashmap = gp_hash_table<K, V, Hash>; const ll RANDOM = chrono::high_resolution_clock::now().time_since_epoch().count(); struct chash { ll operator()(ll x) const { return x ^ RANDOM; } }; #endif // More utilities int SZ(string &v) { return v.length(); } template <typename C> int SZ(C &v) { return v.size(); } template <typename C> void UNIQUE(vector<C> &v) { sort(v.begin(), v.end()); v.resize(unique(v.begin(), v.end()) - v.begin()); } template <typename T, typename U> void maxa(T &a, U b) { a = max(a, b); } template <typename T, typename U> void mina(T &a, U b) { a = min(a, b); } const ll INF = 0x3f3f3f3f, LLINF = 0x3f3f3f3f3f3f3f3f; #line 3 "string/z_algorithm.hpp" /* * z[i] <- LCP(s[0..], s[i..]) * * Finding z[0..N-1] quickly relies on finding an estimation for z[i] (that is less than the actual value), and * increasing it until it reaches the correct value. This works because z[i] stores the LCP of some strings, so if z[i] * is a common prefix, then clearly z[i]-1 is one as well. * * To do this, we maintain the rightmost interval [l = i, r = i+z[i]-1] and solve for i-s sequentially. At any point, * we have a few cases: * * (1) i > r : We start our estimate of z[i] as 0, and extend * (2) i <= r : Also note that in this case, l<=i<=r as we process i-s in order. We know that s[l..r] == s[0..r-l] as * it is an interval that is defined by [k, k+z[k]-1] (for some k). Thus, we use one of our previously computed values * to initialize z[i]. In this case, we use min(z[i-l], r-i+1), which is when we shift the interval [k, k+z[k]-1] to * [0, z[k]-1]. * * Complexity Proof: * * (1) : We are always extending r and since r<N, we will extend at most N times * (2) : * (2a) z[i-l] == r-i+1 : Again, we are extending r so this will happen at most N times (combined with case 1) * (2b) z[i-l] < r-i+1 : We will extend 0 times, as if we could extend z[i], we could also extend z[i-l] as we know * that s[l..r] == s[0..r-l] (and we are still within that range, so the extended character will be the same * at both locations) * * Note: while z[0] is technically undefined, we'll define it as N for now */ template <typename Container> vector<int> z_algorithm(int N, const Container &s) { vector<int> z(N); z[0] = N; int l = 0, r = -1; for (int i = 1; i < N; i++) { if (i <= r) z[i] = min(r-i+1, z[i-l]); while (i + z[i] < N && s[z[i]] == s[i + z[i]]) z[i]++; if (i+z[i]-1 > r) { r = i+z[i]-1; l = i; } } return z; }