String Function Calculation
Problem Statement :
Jane loves strings more than anything. She has a string with her, and value of string over function can be calculated as given below: Jane wants to know the maximum value of among all the substrings of string . Can you help her? Input Format A single line containing string . Output Format Print the maximum value of f(s) among all the substrings ( s ) of string t. Constraints 1 <= | t | <= 10^5 The string consists of lowercase English alphabets. Sample Input 0 aaaaaa Sample Output 0 12
Solution :
Solution in C :
In C++ :
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <cmath>
#include <memory.h>
using namespace std;
typedef long long ll;
const int N = 3e5+6;
#define link ololink
int link[N], len[N], to[N][26], cnt[N], d;
vector<int> v[N];
char s[N];
int main(){
//freopen("input.txt","r",stdin);// freopen("output.txt","w",stdout);
int n, m;
gets(s);
n = strlen(s);
d = 0;
int l = d++;
link[l] = -1;
for(int i=0;i<n;++i){
int c = s[i]-'a';
int x = d++;
len[x] = len[l]+1;
cnt[x] = 1;
for(;l!=-1 && !to[l][c]; l=link[l]) to[l][c]=x;
if(~l){
int p = to[l][c];
if(len[p]==len[l]+1) link[x] = p; else{
int q = d++;
memcpy(to[q], to[p], sizeof(to[p]));
link[q] = link[p];
len[q] = len[l]+1;
for(;l!=-1 && to[l][c]==p; l=link[l]) to[l][c] = q;
link[x] = link[p] = q;
}
}
l = x;
}
ll ans = 0;
for(int i=0;i<d;++i) v[len[i]].push_back(i);
for(l=n;l;--l)
for(int k=0;k<v[l].size();++k){
int i = v[l][k];
int j = link[i];
ans = max(ans, 1LL*len[i]*cnt[i]);
cnt[j]+=cnt[i];
}
cout<<ans<<endl;
return 0;
}
In Java :
import java.io.*;
import java.util.ArrayList;
import java.util.List;
public class Solution {
static class SuffixAutomata {
static class Vertex {
Vertex suffixLink = null;
Vertex[] edges;
int log = 0;
int terminals;
boolean visited;
public Vertex(Vertex o, int log) {
edges = o.edges.clone();
this.log = log;
}
public Vertex(int log) {
edges = new Vertex[26];
this.log = log;
}
long dp() {
if (visited) {
return 0;
}
visited = true;
long r = 0;
for (Vertex v : edges) {
if (v != null) {
r = Math.max(r, v.dp());
terminals += v.terminals;
}
}
return Math.max(r, 1L * log * terminals);
}
}
Vertex root, last;
public SuffixAutomata(String str) {
last = root = new Vertex(0);
for (int i = 0; i < str.length(); i++) {
addChar(str.charAt(i));
}
addTerm();
}
private void addChar(char c) {
Vertex cur = last;
last = new Vertex(cur.log + 1);
while (cur != null && cur.edges[c - 'a'] == null) {
cur.edges[c - 'a'] = last;
cur = cur.suffixLink;
}
if (cur != null) {
Vertex q = cur.edges[c - 'a'];
if (q.log == cur.log + 1) {
last.suffixLink = q;
} else {
Vertex r = new Vertex(q, cur.log + 1);
r.suffixLink = q.suffixLink;
q.suffixLink = r;
last.suffixLink = r;
while (cur != null) {
if (cur.edges[c - 'a'] == q) {
cur.edges[c - 'a'] = r;
} else {
break;
}
cur = cur.suffixLink;
}
}
} else {
last.suffixLink = root;
}
}
private void addTerm() {
Vertex cur = last;
while (cur != null) {
cur.terminals++;
cur = cur.suffixLink;
}
}
}
public static void solve(Input in, PrintWriter out) throws IOException {
String s = in.next();
SuffixAutomata a = new SuffixAutomata(s);
out.println(a.root.dp());
}
public static void main(String[] args) throws IOException {
PrintWriter out = new PrintWriter(System.out);
solve(new Input(new BufferedReader(new InputStreamReader(System.in))), out);
out.close();
}
static class Input {
BufferedReader in;
StringBuilder sb = new StringBuilder();
public Input(BufferedReader in) {
this.in = in;
}
public Input(String s) {
this.in = new BufferedReader(new StringReader(s));
}
public String next() throws IOException {
sb.setLength(0);
while (true) {
int c = in.read();
if (c == -1) {
return null;
}
if (" \n\r\t".indexOf(c) == -1) {
sb.append((char)c);
break;
}
}
while (true) {
int c = in.read();
if (c == -1 || " \n\r\t".indexOf(c) != -1) {
break;
}
sb.append((char)c);
}
return sb.toString();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
}
}
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define MAXN 100000+2
char str[MAXN];
int sa[MAXN];
int rank[MAXN];
int cnt[MAXN];
int wb[MAXN];
int wv[MAXN];
int height[MAXN];
int stack[MAXN];
inline int max(int a, int b) {
return a > b? a : b;
}
int cmp(int *r, int a, int b, int k) {
return r[a] == r[b] && r[a+k] == r[b+k];
}
void gen_sa(char *str, int n, int *sa, int *rank) {
int m = 128, p;
int i, j, k;
int *x, *y, *t;
x = rank; y = wb;
memset(cnt, 0, sizeof(int) * m);
for (i = 0; i < n; ++ i) ++ cnt[x[i] = str[i]];
for (i = 1; i < m; ++ i) cnt[i] += cnt[i-1];
for (i = n-1; i >= 0; -- i) sa[--cnt[x[i]]] = i;
for (k = 1; k <= n; k = k << 1) {
for (p = 0, i = n-k; i < n; ++ i) y[p++] = i;
for (i = 0; i < n; ++ i) if (sa[i] >= k) y[p++] = sa[i] - k;
memset(cnt, 0, sizeof(int) * m);
for (i = 0; i < n; ++ i) {
wv[i] = x[y[i]];
++ cnt[wv[i]];
}
for (i = 1; i < m; ++ i) cnt[i] += cnt[i-1];
for (i = n-1; i >= 0; -- i) sa[--cnt[wv[i]]] = y[i];
t = x; x = y; y = t;
x[sa[0]] = 0;
for (p = 1, i = 0; i < n; ++ i) {
x[sa[i]] = cmp(y, sa[i], sa[i-1], k) ? p-1: p++;
}
m = p;
}
if (x != rank) memcpy(rank, x, sizeof(int)*n);
}
void gen_height(char *str, int n, int *sa, int *rank, int *height) {
int i, j, k;
height[0] = 0;
k = 0;
for (i = 0; i < n-1; ++ i) {
if (k) -- k;
j = rank[i]-2;
if (j == -1) continue;
for (j = sa[j]; str[i+k] == str[j+k]; ) {
++ k;
}
height[rank[i]-1] = k;
}
}
int max_rectangle(int *height, int n) {
int i, j, left, right, cur, top = -1;
int result = 0;
height[n] = 0;
stack[++top] = 0;
for (i = 0; i <= n; ++ i) {
while (top > -1 && height[i] < height[stack[top]]) {
cur = stack[top--];
left = (top > -1? cur-stack[top]: cur+1) * height[cur];
right = (i - cur - 1) * height[cur];
result = max(result, left+right+height[cur]);
}
stack[++top] = i;
}
return max(result, n-1);
}
int main() {
int n, result;
scanf("%s", str);
n = strlen(str);
gen_sa(str, n+1, sa, rank);
gen_height(str, n+1, sa, rank, height);
result = max_rectangle(height, n+1);
printf("%d\n", result);
return 0;
}
In Python3 :
#!/bin/python3
import os
from itertools import zip_longest, islice
def suffix_array_best(s):
"""
suffix array of s
O(n * log(n)^2)
"""
n = len(s)
k = 1
line = to_int_keys_best(s)
while max(line) < n - 1:
line = to_int_keys_best(
[a * (n + 1) + b + 1
for (a, b) in
zip_longest(line, islice(line, k, None),
fillvalue=-1)])
k <<= 1
return line
def inverse_array(l):
n = len(l)
ans = [0] * n
for i in range(n):
ans[l[i]] = i
return ans
def to_int_keys_best(l):
"""
l: iterable of keys
returns: a list with integer keys
"""
seen = set()
ls = []
for e in l:
if not e in seen:
ls.append(e)
seen.add(e)
ls.sort()
index = {v: i for i, v in enumerate(ls)}
return [index[v] for v in l]
def suffix_matrix_best(s):
"""
suffix matrix of s
O(n * log(n)^2)
"""
n = len(s)
k = 1
line = to_int_keys_best(s)
ans = [line]
while max(line) < n - 1:
line = to_int_keys_best(
[a * (n + 1) + b + 1
for (a, b) in
zip_longest(line, islice(line, k, None), fillvalue=-1)])
ans.append(line)
k <<= 1
return ans
def suffix_array_alternative_naive(s):
return [rank for suffix, rank in sorted((s[i:], i) for i in range(len(s)))]
def lcp(sm, i, j):
"""
longest common prefix
O(log(n))
sm: suffix matrix
"""
n = len(sm[-1])
if i == j:
return n - i
k = 1 << (len(sm) - 2)
ans = 0
for line in sm[-2::-1]:
if i >= n or j >= n:
break
if line[i] == line[j]:
ans ^= k
i += k
j += k
k >>= 1
return ans
def maxValue(a):
# print()
# print(a)
# print()
res = inverse_array(suffix_array_best(a))
# res = suffix_array_alternative_naive(a)
mtx = suffix_matrix_best(a)
lcp_res = []
for n in range(len(res) - 1):
lcp_res.append(lcp(mtx, res[n], res[n+1]))
lcp_res.append(0)
# само слово набирает столько баллов, сколько в нем символов
max_score = len(a)
lcp_res_len = len(lcp_res)
for i, num in enumerate(res):
if lcp_res[i] <= 1:
continue
lcp_res_i = lcp_res[i]
cnt = 2
for ii in range(i+1, lcp_res_len):
if lcp_res[ii] >= lcp_res_i:
cnt += 1
else:
break
for ii in range(i-1, -1, -1):
if lcp_res[ii] >= lcp_res_i:
cnt += 1
else:
break
max_score = max(cnt * lcp_res_i, max_score)
return max_score
if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')
t = input()
result = maxValue(t)
fptr.write(str(result) + '\n')
fptr.close()
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