# Play with words

### Problem Statement :

```Shaka and his brother have created a boring game which is played like this:

They take a word composed of lowercase English letters and try to get the maximum possible score by building exactly 2 palindromic subsequences. The score obtained is the product of the length of these 2 subsequences.

Let's say A and B are two subsequences from the initial string. If Ai & Aj are the smallest and the largest positions (from the initial word) respectively in A; and Bi & Bj are the smallest and the largest positions (from the initial word) respectively in B, then the following statements hold true:
Ai <= Aj,
Bi <= Bj, &
Aj < Bi.
i.e., the positions of the subsequences should not cross over each other.

Hence the score obtained is the product of lengths of subsequences A & B. Such subsequences can be numerous for a larger initial word, and hence it becomes harder to find out the maximum possible score. Can you help Shaka and his brother find this out?

Input Format

Input contains a word S composed of lowercase English letters in a single line.

Constraints
1 < |S| <= 3000

each character will be a lower case english alphabet.

Output Format

Output the maximum score the boys can get from S.```

### Solution :

```                            ```Solution in C :

In C++ :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

char a[3333];
int n;
int f[3033][3033];

int get(int l, int r) {
if (l > r) return 0;
if (l == r) {
f[l][r] = 1;
}
if (f[l][r] != 0) {
return f[l][r];
}
f[l][r] = max(get(l + 1, r), get(l, r - 1));
if (a[l] == a[r]) {
f[l][r] = max(f[l][r], get(l + 1, r - 1) + 2);
}
return f[l][r];
}

int main() {
string s;
cin >> s;
n = s.size();
for (int i = 0; i < n; i++) {
a[i + 1] = s[i];
}
int ans = 0;
for (int i = 1; i < n; i++) {
ans = max(ans, get(1, i) * get(i + 1, n));
}
cout << ans << endl;
return 0;
}

In Java :

/**
* Created by zhengf1 on 12/18/2014.
*/
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Dec101_C {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
String str = sc.nextLine();
int n = str.length();
int[][] L = new int[n][n];
for (int i = 0; i < n; i++)
L[i][i] = 1;

int i = 0;
int j = 0;
int cl = 0;
for (cl=2; cl<=n; cl++)
{
for (i=0; i<n-cl+1; i++)
{
j = i+cl-1;
if (str.charAt(i) == str.charAt(j) && cl == 2)
L[i][j] = 2;
else if (str.charAt(i) == str.charAt(j))
L[i][j] = L[i+1][j-1] + 2;
else
L[i][j] = Math.max(L[i][j - 1], L[i + 1][j]);
}
}
int res = 0;
for(i = 1; i < n; i++){
int v1 = L[0][i - 1];
int v2 = L[i][n - 1];
res = Math.max(res, v1 * v2);

}
System.out.println(res);
}
}

In C :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int L[3100][3100],cntt[30][3100];

int max(int a,int b)
{
if(a<b)
{
return b;
}
return a;
}

int main() {
char str[4000],c;
int i,j,t,l,cl;
scanf("%s",str);
l=strlen(str);
for(i=0;i<l;i++)
{
L[i][i]=1;
}
for (cl=2; cl<=l; cl++)
{
for (i=0; i<l-cl+1; i++)
{
j = i+cl-1;
if (str[i] == str[j] && cl == 2)
L[i][j] = 2;
else if (str[i] == str[j])
L[i][j] = L[i+1][j-1] + 2;
else
L[i][j] = max(L[i][j-1], L[i+1][j]);
}
}
int mx=0;
for(i=0;i<l;i++)
{
if((L[0][i]*L[i+1][l-1])>mx)
{
mx=(L[0][i]*L[i+1][l-1]);
}
}
printf("%d\n",mx);
return 0;
}

In Python3 :

s = input()
p = [[0] * len(s) for _ in range(len(s))]
for i in range(len(s)): p[i][i] = 1
for k in range(2, len(s) + 1):
for i in range(len(s) - k + 1):
j = i + k - 1
if s[i] == s[j]:
p[i][j] = 2 + (0 if k == 2 else p[i + 1][j - 1])
else:
p[i][j] = max(p[i][j - 1], p[i + 1][j])
print(max(p[0][i] * p[i + 1][-1] for i in range(len(s) - 1)))```
```

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