Hello Programmers, In this post, you will Know how to solve HackerRank Maximum Palindromes Solution. This problem is a part of the HackerRank Algorithms Series.
One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.
HackerRank Maximum Palindromes Solution
Task
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 s consisting of lowercase English letters. Every day, for q days, she would select two integers l and r, take the substring s1 . . . r (the substring of s from index l to index r), and ask the following question:
Consider all the palindromes that can be constructed from some of the letters from s1 . . . r. 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?
For example, if s = madamimadam, l = 4 and r = 7, then we have,
Your job as the head of the marketing team is to answer all the queries
Complete the functions initialize and answerQuery and return the number of maximum–length palindromes modulo 109 + 7.
Input Format
The first line contains the string s.
The second line contains a single integer q.
The ith of the next q lines contains two space-separated integers li , ri denoting the l and r values Anna selected on the ith day.
Constraints
Here, |s| denotes the length of s.
- 1 <= |s| <= 105
- 1 <= q <= 105
- 1 <= li <= ri <= |s|
Subtasks
For 30% of the total score:
- 1 <= |s| <= 100
- 1 <= q <= 1000
- ri – li <= 3
For 60% of the total score:
- 1 <= |s| <= 100
- 1 <= q <= 1000
Output Format
For each query, print a single line containing a single integer denoting the answer.
Sample Input 0
week 2 1 4 2 3
Sample Output 0
2 1
Explanation 0
On the first day, l = 1 and r = 4. The maximum-length palindromes are “ewe” and “eke”.
On the second day, l = 2 and r = 3. The maximum-length palindrome is “ee”.
Sample Input 1
abab 1 1 4
Sample Output 1
2
Explanation 1
Here, the maximum-length palindromes are “abba” and “baab”.
HackerRank Maximum Palindromes Solution
Maximum Palindromes Solution 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;
}Maximum Palindromes Solution in Cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1000000007;
const int Maxn = 100005;
const int Maxl = 26;
int fac[Maxn], inv[Maxn];
int freq[Maxn][Maxl];
int Inv(int x)
{
int res = 1;
int p = mod - 2;
while (p) {
if (p & 1) res = ll(res) * x % mod;
p >>= 1; x = ll(x) * x % mod;
}
return res;
}
int C(int n, int k)
{
if (n < 0 || k < 0 || k > n) return 0;
return ll(fac[n]) * inv[k] % mod * inv[n - k] % mod;
}
void initialize(string s) {
for (int i = 1; i <= s.length(); i++) {
for (int j = 0; j < Maxl; j++)
freq[i][j] = freq[i - 1][j];
freq[i][s[i - 1] - 'a']++;
}
}
int answerQuery(int l, int r) {
int odd = 0, tot = 0;
for (int i = 0; i < Maxl; i++) {
tot += (freq[r][i] - freq[l - 1][i]) / 2;
odd += (freq[r][i] - freq[l - 1][i]) % 2;
}
int res = 1;
if (odd > 0) res = ll(res) * odd % mod;
for (int i = 0; i < Maxl; i++) {
int my = (freq[r][i] - freq[l - 1][i]) / 2;
res = ll(res) * C(tot, my) % mod; tot -= my;
}
return res;
}
int main() {
fac[0] = inv[0] = 1;
for (int i = 1; i < Maxn; i++) {
fac[i] = ll(fac[i - 1]) * i % mod;
inv[i] = Inv(fac[i]);
}
string s;
cin >> s;
initialize(s);
int q;
cin >> q;
for(int a0 = 0; a0 < q; a0++){
int l;
int r;
cin >> l >> r;
int result = answerQuery(l, r);
cout << result << endl;
}
return 0;
}Maximum Palindromes Solution in Java
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
static int R = 26;
static int n;
static int[][] csum;
static long[] fac;
static long[] ifac;
static long MOD = 1000000007;
static void initialize(String s) {
// This function is called once before all queries.
csum = new int[R][n+1];
for (int i = 1; i <= n; i++) {
int id = s.charAt(i-1) - 'a';
for (int j = 0; j < R; j++)
csum[j][i] = csum[j][i-1];
csum[id][i] = csum[id][i-1] + 1;
}
fac = new long[n+1];
ifac = new long[n+1];
fac[0] = 1;
ifac[0] = 1;
for (int i = 1; i <= n; i++) {
fac[i] = fac[i-1] * i % MOD;
ifac[i] = ifac[i-1] * inv(i, MOD) % MOD;
}
}
static private long inv(long v, long m) {
return (extendedGCD(v, m)[0] % m + m) % m;
}
static private long[] extendedGCD(long x, long y) {
long[] ans = new long[3];
if (y == 0) {
ans[0] = 1;
ans[2] = x;
return ans;
}
long q = x / y;
long r = x % y;
long[] t = extendedGCD(y, r);
ans[0] = t[1];
ans[2] = t[2];
ans[1] = t[0] - ans[0] * q;
return ans;
}
static int answerQuery(int l, int r) {
// Return the answer for this query modulo 1000000007.
int nmid = 0;
int nside = 0;
long ans = 1;
for (int i = 0; i < R; i++) {
int cnt = csum[i][r] - csum[i][l-1];
int left = cnt / 2;
nside += left;
ans = ans * ifac[left] % MOD;
if (cnt % 2 == 1)
nmid++;
}
ans = ans * fac[nside] % MOD;
if (nmid > 1)
ans = ans * nmid % MOD;
return (int)ans;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String s = in.next();
n = s.length();
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();
}
}
Maximum Palindromes Solution in Python
#!/bin/python
import sys
mod = (10**9)+7
fact = []
infact =[]
S = []
def exp(a,b,mod):
ans = 1
while(b!=0):
if((b%2)== 1):
ans = (ans*a)%mod
a = (a*a)%mod
b>>=1
return ans
def initialize(s):
# This function is called once before all queries.
global mod,fact,infact,S
fact = [1]*((10**5) +1)
for i in xrange(1,(10**5) + 1):
fact[i]= (i*fact[i-1])%mod
infact = [0]*((10**5) +1)
infact[-1]= exp(fact[-1],mod-2,mod)
for i in xrange((10**5)-1,-1,-1):
infact[i] = ((i+1)*infact[i+1])%mod
l = len(s)
for i in s:
S += [[0]*26]
S[-1][ord(i)-97]=1
for i in xrange(1,l):
for j in xrange(26):
S[i][j]+=S[i-1][j]
def answerQuery(s,l, r):
global mod
temp = [0]*26
for i in xrange(26):
temp[i]=S[r][i]-S[l][i]
temp[ord(s[l])-97]+=1
tot = 0
one = 0
ans = 1
for i in temp:
one += i%2
tot += i/2
ans = (ans*infact[i/2])%mod
ans = (ans*fact[tot])%mod
if(one!=0):
ans = (ans * one)%mod
return ans
if __name__ == "__main__":
s = raw_input().strip()
initialize(s)
q = int(raw_input().strip())
for a0 in xrange(q):
l, r = raw_input().strip().split(' ')
l, r = [int(l), int(r)]
result = answerQuery(s,l-1, r-1)
print resultMaximum Palindromes Solution using JavaScript
Maximum Palindromes Solution in Scala
Maximum Palindromes Solution in Pascal
uses math;
const base=round(1e9+7);
var s:ansistring; i,q,l,r:longint; cnt1,cnt2,cnt3:int64; c:char;
cnt:array[0..100000,'a'..'z']of int64;
factorial:array[0..100000]of int64;
function power(a,b:int64):int64;
begin
if b=0 then exit(1);
power:=power(a,b>>1);
power:=power*power mod base;
if odd(b) then power:=power*a mod base
end;
begin
read(s,q);
factorial[0]:=1;
for i:=1 to length(s) do
factorial[i]:=factorial[i-1]*i mod base;
for c:='a' to 'z' do
for i:=1 to length(s) do
cnt[i][c]:=cnt[i-1][c] + ord(s[i]=c);
for q:=1 to q do
begin
read(l,r);
cnt1:=0; cnt2:=0; cnt3:=1;
for c:='a' to 'z' do
begin
i:=cnt[r][c]-cnt[l-1][c];
cnt1:=cnt1+(i and 1);
cnt2:=(cnt2+i>>1) mod base;
cnt3:=cnt3*factorial[i>>1] mod base
end;
cnt1:=max(cnt1,1);
writeln((cnt1*factorial[cnt2] mod base)*power(cnt3,base-2) mod base)
end
end.
Disclaimer: This problem (Maximum Palindromes) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.