Longest Palindromic Subsequence

Problem Statement :

```Steve loves playing with palindromes. He has a string, s, consisting of n lowercase English alphabetic characters (i.e., a through z). He wants to calculate the number of ways to insert exactly 1 lowercase character into string s such that the length of the longest palindromic subsequence of s increases by at least k. Two ways are considered to be different if either of the following conditions are satisfied:

The positions of insertion are different.
The inserted characters are different.
This means there are at most 26*(n+1) different ways to insert exactly  1character into a string of length n.

Given q queries consisting of n, k, and s, print the number of different ways of inserting exactly 1 new lowercase letter into string s such that the length of the longest palindromic subsequence of s increases by at least k.

Input Format

The first line contains a single integer, q, denoting the number of queries. The 2q subsequent lines describe each query over two lines:

1.The first line of a query contains two space-separated integers denoting the respective values of n and k.
2.The second line contains a single string denoting s.
Constraints
1 <= q <= 10
1 <= n <= 3000
0 <= k <= 50
It is guaranteed that s consists of lowercase English alphabetic letters (i.e., a to z) only.```

Solution :

```                            ```Solution in C :

In C++ :

#include <bits/stdc++.h>
#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define REP(I, N) for (int I = 0; I < (N); ++I)
#define REPP(I, A, B) for (int I = (A); I < (B); ++I)
#define RI(X) scanf("%d", &(X))
#define RII(X, Y) scanf("%d%d", &(X), &(Y))
#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))
#define DRI(X) int (X); scanf("%d", &X)
#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define RS(X) scanf("%s", (X))
#define CASET int ___T, case_n = 1; scanf("%d ", &___T); while (___T-- > 0)
#define MP make_pair
#define PB push_back
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define PII pair<int,int>
#define VI vector<int>
#define VPII vector<pair<int,int> >
#define PLL pair<long long,long long>
#define VPLL vector<pair<long long,long long> >
#define F first
#define S second
typedef long long LL;
using namespace std;
const int MOD = 1e9+7;
const int SIZE = 3005;
int dp[SIZE][SIZE];
int dp2[SIZE][SIZE];
char s[SIZE];
int main(){
CASET{
DRII(n,K);
RS(s+1);
if(K>2)puts("0");
else if(K==0){
printf("%d\n",n*26+26);
}
else{
MS0(dp);
MS0(dp2);
REPP(i,1,n+1)dp2[i][i]=1;
REPP(j,1,n){
for(int k=1;k+j<=n;k++){
int ll=k,rr=k+j;
if(s[ll]==s[rr])dp2[ll][rr]=max(dp2[ll][rr],dp2[ll+1][rr-1]+2);
dp2[ll][rr]=max(dp2[ll][rr],dp2[ll+1][rr]);
dp2[ll][rr]=max(dp2[ll][rr],dp2[ll][rr-1]);
}
}
int ma=0;
for(int i=1;i<n;i++){
for(int j=n;j>i;j--){
if(s[i]==s[j])dp[i][j]=dp[i-1][j+1]+2;
else dp[i][j]=max(dp[i-1][j],dp[i][j+1]);
ma=max(ma,dp[i][j]);
}
}
REPP(i,1,n+1)ma=max(ma,dp[i-1][i+1]+1);
int an=0;
REP(i,n+1){
int me=0;
me=dp[i][i+1]+1;
if(me>=ma+K){
an+=26;
continue;
}
bool used[26]={};
REPP(j,1,n+1){
if(j<=i){
if(dp2[j+1][i]+dp[j-1][i+1]+2>=ma+K)used[s[j]-'a']=1;
}
else{
if(dp2[i+1][j-1]+dp[i][j+1]+2>=ma+K)used[s[j]-'a']=1;
}
}
REP(j,26)
if(used[j]){
an++;
}
}
printf("%d\n",an);
}
}
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 longestPalindromicSubsequence(String s, int k)
{
int n = s.length();

if (k == 0)
{
return (n + 1) * 26;
}

if (k > 2)
{
return 0;
}

if (n == 1)
{
if (k == 1)
{
return 2;
}

return 0;
}

short pal[][] = computePal(s);
short rightEnds[][] = computeRightEnds(s);
short leftEnds[][] = computeLeftEnds(s);
int sMax = pal[0][n - 1];

boolean ok[][] = new boolean[n + 1][26];

for (int i = 0; i <= n; i++)
{
for (int j = 0; j < n; j++)
{
char c = s.charAt(j);
int min;
int max;
int middle = 0;

if (i <= j)
{
if (i < j)
{
middle = pal[i][j - 1];
}

min = i;
max = j;
}
else
{
if (j + 1 <= i - 1)
{
middle = pal[j + 1][i - 1];
}

min = j;
max = i - 1;
}

int need = sMax + k - middle - 2;

if (need % 2 == 1)
{
need++;
}

if (rightEnds[min][need / 2] > max + 1 ||
leftEnds[max + 1][need / 2] <  min)
{
ok[i][c - 'a'] = true;
//System.out.println("ok " + i + " " + c);
}
}
}

/*for (int i = 0; i <= n; i++)
{
for (int j = n - 1; j > i; j--)
{
int midLength = sMax - pal[i][j - 1];

if (midLength % 2 == 0 &&
rightEnds[i][midLength / 2] > j)
{
ok[i][s.charAt(j) - 'a'] = true;
}
}
}

for (int i = n; i >= 0; i--)
{
for (int j = 0; j < i - 1; j++)
{
int midLength = sMax - pal[j + 1][i - 1];

if (midLength % 2 == 0 &&
leftEnds[i][midLength / 2] < j)
{
ok[i][s.charAt(j) - 'a'] = true;
}
}
}*/

if (k == 1)
{
if (sMax % 2 == 0)
{
for (int i = sMax / 2; i < n; i++)
{
if (rightEnds[i][sMax / 2] >= i)
{
for (int j = 0; j < 26; j++)
{
ok[i][j] = true;
}
}
}
}
else
{
/*int half = sMax / 2;

for (int i = 0; i < n; i++)
{
int ch = s.charAt(i) - 'a';

for (int j = i; j >= half; j--)
{
if (rightEnds[j][half] > i)
{
ok[j][ch] = true;
}
else
{
break;
}
}

for (int j = i + 1; j <= n - half; j++)
{
if (leftEnds[j][half] < i)
{
ok[j][ch] = true;
}
else
{
break;
}
}
}*/
}
}

int total = 0;

for (int i = 0; i <= n; i++)
{
for (int j = 0; j < 26; j++)
{
if (ok[i][j])
{
total++;
}
}
}

}

private static short[][] computeRightEnds(String s)
{
short n = (short) s.length();
short ends[][] = new short[n + 1][n / 2 + 1];

for (int i = 0; i < ends.length; i++)
{
for (int j = 0; j < ends[i].length; j++)
{
ends[i][j] = -1;
}
}

ends[0][0] = n;

for (int len = 1; len <= n; len++)
{
ends[len][0] = n;

int i = n - 1;

while (i >= 0 && s.charAt(i) != s.charAt(len - 1))
{
i--;
}

for (int c = 1; c <= n / 2 && c <= len; c++)
{
ends[len][c] = (short) Math.max(-1, ends[len - 1][c]);

while (i >= ends[len - 1][c - 1])
{
i--;

while (i >= 0 && s.charAt(i) != s.charAt(len - 1))
{
i--;
}
}

if (i >= len)
{
ends[len][c] = (short) Math.max(ends[len][c], i);
}
}
}

return ends;
}

private static short[][] computeLeftEnds(String s)
{
short n = (short) s.length();
short ends[][] = new short[n + 1][n / 2 + 1];

for (int i = 0; i < ends.length; i++)
{
for (int j = 0; j < ends[i].length; j++)
{
ends[i][j] = n;
}
}

ends[n][0] = -1;

for (int k = n - 1; k >= 0; k--)
{
ends[k][0] = -1;

int i = 0;

while (i < n && s.charAt(i) != s.charAt(k))
{
i++;
}

for (int c = 1; c <= n / 2 && c <= n - k; c++)
{
ends[k][c] = (short) Math.min(n, ends[k + 1][c]);

while (i <= ends[k + 1][c - 1])
{
i++;

while (i < n && s.charAt(i) != s.charAt(k))
{
i++;
}
}

if (i < k)
{
ends[k][c] = (short) Math.min(ends[k][c], i);
}
}
}

return ends;
}

private static short[][] computePal(String s)
{
int n = s.length();
short pal[][] = new short[n][n];

for (int i = 0; i < n; i++)
{
pal[i][i] = 1;
}

for (int d = 1; d < n; d++)
{
for (int i = 0, j = d; j < n; i++, j++)
{
if (s.charAt(i) == s.charAt(j))
{
if (d == 1)
{
pal[i][j] = 2;
}
else
{
pal[i][j] = (short) (pal[i + 1][j - 1] + 2);
}
}
else
{
pal[i][j] = (short) Math.max(pal[i][j - 1], pal[i + 1][j]);
}
}
}

return pal;
}

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int q = in.nextInt();
for(int a0 = 0; a0 < q; a0++){
int n = in.nextInt();
int k = in.nextInt();
String s = in.next();
int result = longestPalindromicSubsequence(s, k);
System.out.println(result);
}
in.close();
}
}

In C :

#include<stdio.h>
#define M 3005
int q, n, k;
char s[M];
int in[M][M], out[M][M];
int max(int a, int b)
{
return a > b ? a : b;
}
int main()
{
scanf("%d", &q);
while(q--)
{
scanf("%d %d", &n, &k);
scanf("%s", s);
if( k == 0 )
{
printf("%d\n", ( n + 1 ) * 26);
continue;
}
if( k > 2 )
{
printf("0\n");
continue;
}
for( int l = 0 ; l < n ; l++ )
{
for( int i = 0 ; i + l < n ; i++ )
{
int j = i + l;
if( i == j )
{
in[i][j] = 1;
}
else if( s[i] == s[j] )
{
if( i + 1 < j )
{
in[i][j] = 2 + in[i+1][j-1];
}
else
{
in[i][j] = 2;
}
}
else
{
in[i][j] = max(in[i+1][j], in[i][j-1]);
}
}
}
for( int l = n - 1 ; l >= 0 ; l-- )
{
for( int i = 0 ; i + l < n ; i++ )
{
int j = i + l;
if( i == j )
{
if( 0 < i && j + 1 < n )
{
out[i][j] = 1 + out[i-1][j+1];
}
else
{
out[i][j] = 1;
}
}
else if( s[i] == s[j] )
{
if( 0 < i && j + 1 < n )
{
out[i][j] = 2 + out[i-1][j+1];
}
else
{
out[i][j] = 2;
}
}
else
{
out[i][j] = 0;
if( 0 < i )
{
out[i][j] = max(out[i][j], out[i-1][j]);
}
if( j + 1 < n )
{
out[i][j] = max(out[i][j], out[i][j+1]);
}
}
}
}
int cur = in[0][n-1], res = 0;
for( int i = 0 ; i <= n ; i++ )
{
for( char ch = 'a' ; ch <= 'z' ; ch++ )
{
int my = ( i == 0 || i == n ) ? 1 : 1 + out[i-1][i];
for( int j = 0 ; j < i && my < cur + k ; j++ )
{
if( s[j] == ch )
{
int cand = 2;
if( 0 < j && i < n )
{
cand += out[j-1][i];
}
if( j + 1 <= i - 1 )
{
cand += in[j+1][i-1];
}
my = max(my, cand);
}
}
for( int j = i ; j < n && my < cur + k ; j++ )
{
if( s[j] == ch )
{
int cand = 2;
if( 0 < i && j + 1 < n )
{
cand += out[i-1][j+1];
}
if( i <= j - 1 )
{
cand += in[i][j-1];
}
my = max(my, cand);
}
}
if( my >= cur + k )
{
res++;
}
}
}
printf("%d\n", res);
}
return 0;
}```
```

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