Shashank and the Palindromic Strings

Problem Statement :

Shashank loves strings, but he loves palindromic strings the most. He has a list of n strings, A = [a0,a1,...,an-1] , where each string, ai, consists of lowercase English alphabetic letters. Shashank wants to count the number of ways of choosing non-empty subsequences s0,s1,s2,...,sn-1 such that the following conditions are satisfied:

1.s0 is a subsequence of string a0, s1 is a subsequence of string a1, s2 is a subsequence of string a2, ..., and sn-1 is a subsequence of string .
2. s-+s1+s2+ ...+ sn-1 is a palindromic string, where + denotes the string concatenation operator.
You are given q queries where each query consists of some list, A. For each query, find and print the number of ways Shashank can choose n non-empty subsequences satisfying the criteria above, modulo 10^9 + 7, on a new line.

Note: Two subsequences consisting of the same characters are considered to be different if their characters came from different indices in the original string.

Input Format

The first line contains a single integer, q, denoting the number of queries. The subsequent lines describe each query in the following format:

The first line contains an integer, n, denoting the size of the list.
Each line i of the n subsequent lines contains a non-empty string describing ai.
Constraints
1 <= q <= 50
1 <= n <= 50
Σ |ai| <= 1000 over a test case.
For 40% of the maximum score:
1 <= n <= 5
Σ |ai| <= 250 over a test case.
Output Format

For each query, print the number of ways of choosing non-empty subsequences, modulo 10^9 + 7, on a new line.

Solution :

                            Solution in C :

In C++ :





#include <bits/stdc++.h>

using namespace std;

#define vec vector
#define ALL(x) (x).begin(), (x).end()
#define mp make_pair
#define mt make_tuple

typedef pair< int, int > pii;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

int const inf = 1000 * 1000 * 1000;
ll const inf64 = 1ll * inf * inf;

int const mod = inf + 7;

inline int sum(int a, int b) { return (a + b) % mod; }
inline int raz(int a, int b) { return ((a - b) % mod + mod) % mod; }
inline int mul(int a, int b) { return (1ll * a * b) % mod; }

bool solve() {

    int n;
    cin >> n;

    vec< string > a(n);
    string text = "";

    for(int i = 0;i < n;i++) cin >> a[i], text += a[i];

    int sz = (int)text.size();

    vec< int > id(sz);

    for(int j = 0, i = 0;i < n;i++) {
        for(int iter = 0;iter < (int)a[i].size();iter++) {
            id[j++] = i;
        }
    }

    vec< int > le(n, inf);
    vec< int > ri(n,-inf);

    for(int i = 0;i < sz;i++) {
        le[id[i]] = min(le[id[i]], i);
        ri[id[i]] = max(ri[id[i]], i);
    }

    vec< vec< int > > dp(sz, vec< int >(sz));
    vec< vec< int > > sm(sz, vec< int >(sz));

    for(int l = sz - 1;l >= 0;l--) {
        for(int r = l;r < sz;r++) {
            if(r > 0) sm[l][r] = sum(sm[l][r], sm[l][r - 1]);
            if(l + 1 < sz) sm[l][r] = sum(sm[l][r], sm[l + 1][r]);
            if(r > 0 && l + 1 < sz) sm[l][r] = raz(sm[l][r], sm[l + 1][r - 1]);

            if(r < l || text[l] != text[r]) continue;

            dp[l][r] = id[r] - id[l] <= 1;

            int tl1, tr1, tl2, tr2;

            tl1 = l + 1;
            tr1 = min(r - 1, id[l] + 1 < n ? ri[id[l] + 1] : ri[id[l]]);

            tl2 = max(l + 1, id[r] ? le[id[r] - 1] : le[id[r]]);
            tr2 = r - 1;

            if(tl1 <= tr1 && tl2 <= tr2) {
                dp[l][r] = sum(dp[l][r], sm[tl1][tr2]);
                if (tl2 > 0) dp[l][r] = raz(dp[l][r], sm[tl1][tl2 - 1]);
                if (tr1 + 1 < sz) dp[l][r] = raz(dp[l][r], sm[tr1 + 1][tr2]);
                if (tr1 + 1 < sz && tl2 > 0) dp[l][r] = sum(dp[l][r], sm[tr1 + 1][tl2 - 1]);
            }

            sm[l][r] = sum(sm[l][r], dp[l][r]);
        }
    }

    int res = 0;

    for(int l = 0;l < (int)a[0].size();l++) {
        for(int r = sz - 1;r >= l && r >= sz - (int)a[n - 1].size();r--) {
            res = sum(res, dp[l][r]);
        }
    }

    cout << res << "\n";

    return true;
}

int main() {

    int T;
    cin >> T;

    while(T--) solve();

    return 0;
}








In Java : 





import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;

public class Solution {
        final static int MODE = 1000000007;
    /*
     * Complete the palindromicStrings function below.
     */
    static int palindromicStrings(int[] a, char[] cc) {
        final int N = cc.length;
        int[][] dp = new int[6*N][N];
        int[] ids = new int[N];
        int t=a[0]==1?0:1;
        for(int i=1;i<N;++i){
            if(t==0){
                ids[i]=ids[i-1]+1;
            }else{
                ids[i]=ids[i-1];
            }
            ++t;
            if(a[ids[i]]==t)t=0;
        }
        int[] st = new int[6];
        for(int i=1;i<6;++i)st[i]=i*N;
        for(int i=N-1;i>=0;--i){
            dp[st[5]+i][i]=dp[st[4]+i][i]=dp[i][i]=1;
            dp[st[3]+i][i]=dp[st[2]+i][i]=dp[st[1]+i][i]=2;
            for(int j=i+1;j<N;++j){
                if(ids[j]==ids[j-1]){
                    dp[st[5]+i][j]=dp[st[5]+i][j-1];
                    dp[st[4]+i][j]=dp[st[4]+i][j-1];
                }else{
                    dp[st[5]+i][j]=dp[st[4]+i][j-1];
                }
                if(cc[i]==cc[j]){
                    if(i+1==j){
                        ++dp[st[5]+i][j];
                        ++dp[st[4]+i][j];
                    }else{
                        int k=0;
                        if(ids[i]==ids[i+1])k+=2;
                        if(ids[j]==ids[j-1])k+=1;
                        dp[st[5]+i][j]=(dp[st[5]+i][j]+dp[st[k]+i+1][j-1])%MODE;
                        dp[st[4]+i][j]=(dp[st[4]+i][j]+dp[st[k]+i+1][j-1])%MODE;
                    }
                }
                dp[st[1]+i][j]=dp[st[3]+i][j]=dp[st[5]+i][j];
                dp[st[0]+i][j]=dp[st[2]+i][j]=dp[st[4]+i][j];
                if(ids[i]==ids[i+1]){
                    dp[st[3]+i][j]=(dp[st[3]+i][j]+dp[st[3]+i+1][j])%MODE;
                    dp[st[2]+i][j]=(dp[st[2]+i][j]+dp[st[2]+i+1][j])%MODE;
                    dp[st[1]+i][j]=(dp[st[1]+i][j]+dp[st[1]+i+1][j])%MODE;
                    dp[st[0]+i][j]=(dp[st[0]+i][j]+dp[st[0]+i+1][j])%MODE;
                }else{
                    dp[st[3]+i][j]=(dp[st[3]+i][j]+dp[st[1]+i+1][j])%MODE;
                    dp[st[2]+i][j]=(dp[st[2]+i][j]+dp[st[0]+i+1][j])%MODE;
                }
            }
        }
        return dp[0][N-1];
    }

    private static final Scanner scanner = new Scanner(System.in);

    public static void main(String[] args) throws Exception {
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));
        long t = System.currentTimeMillis();
        int q = scanner.nextInt();
        scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])*");

        for (int qItr = 0; qItr < q; qItr++) {
            int aCount = scanner.nextInt();
            scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])*");

            int[] a = new int[aCount];
            StringBuilder sb = new StringBuilder();

            for (int aItr = 0; aItr < aCount; aItr++) {
                String aItem = scanner.nextLine();
                //scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])*");
                a[aItr] = aItem.length();
                sb.append(aItem);
            }
            int result = palindromicStrings(a, sb.toString().toCharArray());

            bufferedWriter.write(String.valueOf(result));
            bufferedWriter.newLine();
        }

        bufferedWriter.close();

        scanner.close();
        //Thread.sleep(3800);
        System.out.println(System.currentTimeMillis()-t);
    }
}








In C :





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

#define md 1000000007

char cv[1001], fv[1000], lv[1000], gv[1000];
unsigned int mv[1000 * 1000 * 4];
int m;

void readInput() {
    int i, j, k, l, n;
    char *s;
    for (i = 0; i < 1000; ++i)
    {
        fv[i] = 0;
        lv[i] = 0;
    }
    s = cv;
    k = 0;
    scanf("%d", &n);
    for (i = 0; i < n; ++i)
    {
        scanf("%s", s);
        l = strlen(s);
        fv[k] = 1;
        lv[k + l - 1] = 1;
        for (j = 0; j < l; ++j)
        {
            gv[k + j] = i;
        }
        k += l;
        s += l;
    }
    m = strlen(cv);
}

int ix(int i, int j, int oi, int oj) {
    return (i * m * 4) + (j * 4) + (oi * 2) + oj;
}

unsigned int f(int i, int j, int oi, int oj) {
    if (i == j)
    {
        return (oi || oj) ? 2 : 1;
    }
    if (j < i)
    {
        return 1;
    }
    if (gv[i] == gv[j])
    {
        oi = oi || oj;
        oj = oi;
    }
    return mv[ix(i, j, oi, oj)];
}

void run() {
    int l, i, j, p;
    int il, jf, oi, oj, oi1, oj1, b1, b2;
    unsigned int c;
    readInput();
    for (l = 2; l <= m; ++l)
    {
        for (i = 0; i <= m - l; ++i)
        {
            j = i + l - 1;
            for (p = 0; p < 4; ++p)
            {
                if (p == 0 || p == 3 || gv[i] < gv[j])
                {
                    il = lv[i];
                    jf = fv[j];
                    oi = p & 1;
                    oj = p >> 1;
                    c = 0;
                    b1 = oi || !il;
                    b2 = oj || !jf;
                    oi1 = !il && oi;
                    oj1 = !jf && oj;
                    if (b1)
                    {
                        c += f(i + 1, j, oi1, oj);
                    }
                    if (b2)
                    {
                        c += f(i, j - 1, oi, oj1);
                    }
                    if (b1 && b2 && (l > 2 || oi || oj))
                    {
                        c += md - f(i + 1, j - 1, oi1, oj1);
                    }
                    if (cv[i] == cv[j])
                    {
                        c += f(i + 1, j - 1, !il, !jf);
                    }
                    //printf("%d %d %d %d %u\n", i, j, oi, oj, c % md);
                    mv[ix(i, j, oi, oj)] = c % md;
                }
            }
        }
    }
    printf("%u\n", mv[ix(0, m - 1, 0, 0)]);
}

int main() {
    int q, i;
    scanf("%d", &q);
    for(i = 0; i < q; ++i) {
        run();
    }
    return 0;
}

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Shashank and the Palindromic Strings

Problem Statement :

Solution :

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