string/manachers.hpp

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Last update: 2021-09-12 23:20:08-04:00
Include: #include "string/manachers.hpp"

Depends on

template.hpp

Verified with

tests/string/manachers.test.cpp

Code

#pragma once
#include "../template.hpp"

/*
 * d[i] = Longest odd-length palindrome centered at i.  Formally, largest value d[i] s.t. s[i-k] == s[i+k] for all
 * 0<=k<d[i].
 *
 * This algorithm works almost identically to z-algorithm.  For more information and explanation, see
 * string/z_algorithm.hpp
 *
 * The key differences between the algorithms are: instead of 'shifting' from i -> i-l when using precomputed values,
 * we instead flip across the midpoint of rightmost palindrome substring we found to find our precomputed value.
 * Additionally, we extend out in both directions instead of only forwards as we are looking for palindromes.
 *
 * To find the longest palindromes for even locations, add placeholders between each character.
 */
template <typename Container> vector<int> manachers(int N, const Container &s) {
    vector<int> d(N);
    int l = 0, r = -1;
    for (int i = 0; i < N; i++) {
        if (i <= r) d[i] = min(r-i+1, d[l + r - i]); // flip across (l+r)/2
        while (0 <= i-d[i] && i+d[i] < N && s[i-d[i]] == s[i+d[i]]) d[i]++;
        if (i+d[i]-1 > r) {
            l = i-d[i]+1;
            r = i+d[i]-1;
        }
    }
    return d;
}

#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/manachers.hpp"

/*
 * d[i] = Longest odd-length palindrome centered at i.  Formally, largest value d[i] s.t. s[i-k] == s[i+k] for all
 * 0<=k<d[i].
 *
 * This algorithm works almost identically to z-algorithm.  For more information and explanation, see
 * string/z_algorithm.hpp
 *
 * The key differences between the algorithms are: instead of 'shifting' from i -> i-l when using precomputed values,
 * we instead flip across the midpoint of rightmost palindrome substring we found to find our precomputed value.
 * Additionally, we extend out in both directions instead of only forwards as we are looking for palindromes.
 *
 * To find the longest palindromes for even locations, add placeholders between each character.
 */
template <typename Container> vector<int> manachers(int N, const Container &s) {
    vector<int> d(N);
    int l = 0, r = -1;
    for (int i = 0; i < N; i++) {
        if (i <= r) d[i] = min(r-i+1, d[l + r - i]); // flip across (l+r)/2
        while (0 <= i-d[i] && i+d[i] < N && s[i-d[i]] == s[i+d[i]]) d[i]++;
        if (i+d[i]-1 > r) {
            l = i-d[i]+1;
            r = i+d[i]-1;
        }
    }
    return d;
}