Maximum Palindromes
Problem Statement :
Madam Hannah Otto, the CEO of Reviver Corp., is fond of palindromes, or words that read the same forwards or backwards. She thinks palindromic brand names are appealing to millennials. As part of the marketing campaign for the company's new juicer called the Rotator™, Hannah decided to push the marketing team's palindrome-searching skills to a new level with a new challenge. In this challenge, Hannah provides a string consisting of lowercase English letters. Every day, for days, she would select two integers and , take the substring (the substring of from index to index ), and ask the following question: Consider all the palindromes that can be constructed from some of the letters from . You can reorder the letters as you need. Some of these palindromes have the maximum length among all these palindromes. How many maximum-length palindromes are there? Input Format The first line contains the string . The second line contains a single integer . The of the next lines contains two space-separated integers , denoting the and values Anna selected on the Output Format For each query, print a single line containing a single integer denoting the answer.
Solution :
Solution in C :
In C++ :
#include <cstdio>
#include <cmath>
#include <iostream>
#include <set>
#include <algorithm>
#include <vector>
#include <map>
#include <cassert>
#include <string>
#include <cstring>
#include <queue>
using namespace std;
#define rep(i,a,b) for(int i = a; i < b; i++)
#define S(x) scanf("%d",&x)
#define S2(x,y) scanf("%d%d",&x,&y)
#define P(x) printf("%d\n",x)
#define all(v) v.begin(),v.end()
#define FF first
#define SS second
#define pb push_back
#define mp make_pair
typedef long long int LL;
typedef pair<int, int > pii;
typedef vector<int > vi;
const int mod = 1000000007;
const int N = 100005;
LL F[N], IF[N];
LL _pow(LL a, LL b) {
if(!b) return 1;
if(b == 1) return a;
if(b == 2) return a * a % mod;
if(b & 1) {
return a * _pow(a,b-1) % mod;
}
return _pow(_pow(a,b/2),2);
}
void pre() {
F[0] = 1;
rep(i,1,N) {
F[i] = i * F[i-1] % mod;
}
IF[N - 1] = _pow(F[N - 1], mod - 2);
for(int i = N - 2; i >= 0; i--) {
IF[i] = IF[i + 1] * (i + 1) % mod;
}
}
int X[26][N];
int main() {
pre();
string s;
cin >> s;
int n = s.size();
rep(i,0,n) {
rep(j,0,26) X[j][i+1] = X[j][i];
X[s[i]-'a'][i+1]++;
}
int q;
S(q);
while(q--) {
int l,r;
S2(l,r);
int tot = 0;
LL ans = 1;
int odd = 0;
rep(i,0,26) {
int x = X[i][r] - X[i][l-1];
odd += (x&1);
ans *= IF[x / 2];
ans %= mod;
tot += x / 2;
}
ans *= F[tot];
ans %= mod;
if(odd) ans = ans * odd % mod;
printf("%lld\n",ans);
}
return 0;
}
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
static int[][] count;
static long[] f;
static long mod = 1000000007;
static void initialize(String s)
{
count = new int[s.length()][26];
for (int i = 0; i < s.length(); i++)
{
if (i > 0)
{
for (int j = 0; j < 26; j++) count[i][j] = count[i-1][j];
}
count[i][s.charAt(i) - 'a']++;
}
f = new long[100000];
f[0] = 1;
for (int i = 1; i < 100000; i++) f[i] = f[i - 1] * i % mod;
}
static int c[] = new int[26];
static int answerQuery(int l, int r) {
// Return the answer for this query modulo 1000000007.
l--;
r--;
for (int i = 0; i < 26; i++)
{
c[i] = count[r][i];
if (l > 0)
{
c[i] -= count[l - 1][i];
}
}
int len = 0;
int oddCount = 0;
for (int i = 0; i < 26; i++)
{
len += c[i] / 2;
if (c[i] % 2 == 1) oddCount++;
}
long ans = f[len];
long div = 1;
for (int i = 0; i < 26; i++)
{
if (c[i] >= 4)
{
div *= f[c[i] / 2];
div %= mod;
}
}
if (div != 1) ans = solveCongruence(div, ans, mod);
if (oddCount > 0)
{
ans *= oddCount;
ans %= mod;
}
return (int) ans;
}
static long solveCongruence(long a, long b, long mod)
{
if (a == 1)
{
return b;
}
long b1 = -b%a;
if (b1 != 0)
{
b1 += a;
}
long y = solveCongruence(mod % a, b1, a);
return (mod * y + b) / a % mod;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String s = in.next();
initialize(s);
int q = in.nextInt();
for(int a0 = 0; a0 < q; a0++){
int l = in.nextInt();
int r = in.nextInt();
int result = answerQuery(l, r);
System.out.println(result);
}
in.close();
}
}
In C :
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
#define MOD 1000000007
#define MAX 100000
#define MAXP 100000
int a[MAX+1][26];
long fact[MAX+1];
void initialize(char* s) {
for(int j=0; j<26; j++) {
a[0][j] = 0;
}
for(int i=0; s[i]; i++) {
for(int j=0; j<26; j++) {
a[i+1][j] = a[i][j];
}
a[i+1][s[i]-'a'] ++;
/*for(int j=0; j<26; j++) {
printf("a[until(inc) %c][for %c] = %d\n", s[i], j+'a', a[i+1][j]);
}*/
}
fact[0] = 1L;
for(int i=0; s[i]; i++) {
fact[i+1] = (fact[i]*(i+1))%MOD;
}
}
long dinv(long x) {
int i;
static long r[MAXP], s[MAXP], t[MAXP], q[MAXP];
r[0] = MOD; r[1] = x;
s[0] = 1; s[1] = 0;
t[0] = 0; t[1] = 1;
i = 1;
while(r[i] > 0) {
q[i] = r[i-1]/r[i];
r[i+1] = r[i-1] - q[i]*r[i];
s[i+1] = s[i-1] - q[i]*s[i];
t[i+1] = t[i-1] - q[i]*t[i];
//printf("%ld %ld %ld\n", r[i+1], s[i+1], t[i+1]);
i ++;
}
return (t[i-1]+MOD)%MOD;
}
int answerQuery(char* s, int l, int r) {
int v[26];
long res;
for(int i=0; i<26; i++) {
v[i] = a[r][i] - a[l-1][i];
}
/*for(int i=0; i<26; i++) {
printf("v[%c] = %d\n", i+'a', v[i]);
}
printf("\n");*/
int oddcount = 0;
int eventotal = 0;
for(int i=0; i<26; i++) {
oddcount += v[i]%2;
eventotal += v[i]/2;
}
res = fact[eventotal];
if(oddcount > 0) {
res = (res*oddcount)%MOD;
}
for(int i=0; i<26; i++) {
if(v[i]/2 > 0) {
res = (res*dinv(fact[v[i]/2]))%MOD;
}
}
return (int)res;
}
int main() {
char* s = (char *)malloc(512000 * sizeof(char));
scanf("%s", s);
initialize(s);
int q;
scanf("%i", &q);
for(int a0 = 0; a0 < q; a0++){
int l;
int r;
scanf("%i %i", &l, &r);
int result = answerQuery(s, l, r);
printf("%d\n", result);
}
return 0;
}
In Python3 :
from collections import defaultdict
from copy import copy
M = int(1e9) + 7
fact = [1]
for i in range(1,int(1e5) + 10):
fact.append(fact[-1]*i % M)
def egcd(a, b):
if a == 0:
return (b, 0, 1)
else:
g, y, x = egcd(b % a, a)
return (g, x - (b // a) * y, y)
memo = {}
def modinv(a, m):
if a in memo:
return memo[a]
while a < 0:
a += m
g, x, y = egcd(a, m)
if g != 1:
raise Exception('Modular inverse does not exist')
else:
memo[a] = x%m
return x % m
s = input().strip()
occ = [defaultdict(int)]
for ch in s:
newd = copy(occ[-1])
newd[ch] += 1
occ.append(newd)
def query(l, r):
d = defaultdict(int)
for ch in occ[r]:
d[ch] += occ[r][ch]
for ch in occ[l-1]:
d[ch] -= occ[l-1][ch]
odds = 0
for k, v in copy(d).items():
if v&1:
odds += 1
d[k] = v - (v&1)
res = 1
total = 0
for k, v in d.items():
res *= modinv(fact[v//2], M)
total += v//2
res %= M
return (max(1, odds)*res*fact[total])%M
for _ in range(int(input())):
l, r = map(int, input().split())
print(query(l, r))
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