# HackerRank Maximum Palindromes Solution

#### ByBrokenprogrammers

Dec 13, 2022

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

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 = madamimadaml = 4 and r = 7, then we have,

Complete the functions `initialize` and `answerQuery` and return the number of maximumlength 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|

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 “abbaand “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;
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();
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]
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)]
print result```

### 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
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
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.