Square Subsequences


Problem Statement :


Square Subsequences

A string is called a square string if it can be obtained by concatenating two copies of the same string. For example, "abab", "aa" are square strings, while "aaa", "abba" are not. Given a string, how many (non-empty) subsequences of the string are square strings? A subsequence of a string can be obtained by deleting zero or more characters from it, and maintaining the relative order of the remaining characters.

Input Format

The first line contains the number of test cases, T.
T test cases follow. Each case contains a string, S.

Output Format

Output T lines, one for each test case, containing the required answer modulo 1000000007.

Constraints:
1 <= T <= 20
S will have at most 200 lowercase characters ('a' - 'z').


Solution :



title-img 


                            Solution in C :

In C++ :






#include<iostream>
#include<set>
#include<map>
#include<string>
#include<stdio.h>
#include<sstream>
#include<algorithm>
#include<queue>
#include<cmath>
#include<string.h>
using namespace std ;
#define MAXN 305
#define INF (int)1e9
#define MOD 1000000007

int brute(string s)
{
 int n = s.size() ;
 int ret = 0 ;
 for(int i = 1;i < 1 << n;i++)
 {
  string t ;
  for(int j = 0;j < n;j++)
   if(i & 1 << j)
    t.push_back(s[j]) ;
  if(t.size() % 2 == 1) continue ;
  int tt = t.size() ;
  bool valid = true ;
  for(int j = 0;j < tt / 2;j++)
   if(t[j] != t[tt / 2 + j])
    valid = false ;
  if(valid) ret++ ;
 }
 return ret ;
}

string s ;
int n ;
int start1,start2 ;

int memo[MAXN][MAXN] ;
int solve(int k1,int k2)
{
 if(k1 == start2 || k2 == n) return 0 ;
 int ret = s[k1] == s[k2] ? 1 : 0 ;
 if(memo[k1][k2] != -1) return memo[k1][k2] ;
 ret += solve(k1 + 1,k2) ;
 ret += solve(k1,k2 + 1) ;
 if(ret >= MOD) ret -= MOD ; 
 ret -= solve(k1 + 1,k2 + 1) ;
 if(ret < 0) ret += MOD ;
 if(s[k1] == s[k2]) ret += solve(k1 + 1,k2 + 1) ;
 if(ret >= MOD) ret -= MOD ; 
 return memo[k1][k2] = ret ;
}

int solve(string _s)
{
 s = _s ;
 n = s.size() ;
 int ret = 0 ;
 for(start2 = 0;start2 < n;start2++)
 {
  memset(memo,255,sizeof memo) ;
  for(start1 = 0;start1 < start2;start1++)
   if(s[start1] == s[start2])
   {
    int cret = 1 + solve(start1 + 1,start2 + 1) ;
    ret += cret ;
    if(ret >= MOD) ret -= MOD ;
   }
 }
 return ret ;
}

void test()
{
 for(int test = 0;test < 10000;test++)
 {
  int n = rand() % 10 + 1 ;
  string t ;
  for(int j = 0;j < n;j++) t.push_back('a' + rand() % 2) ;
  int ret1 = brute(t) ;
  int ret2 = solve(t) ;
  cout << ret1 << " " << ret2 << endl ;
  if(ret1 != ret2)
  {
   cout << "Failed on: " << test << endl ;
   cout << t << endl ;
   while(1) ;
  }
 }
}

void generate()
{
 char in[10] = "in .txt" ;
 for(int test = 0;test < 10;test++)
 {
  in[2] = test + '0' ;
  FILE * fout = fopen(in,"w") ;
  
  int runs = 20 ;
  fprintf(fout,"%d\n",runs) ;
  for(int t = 0;t < runs;t++)
  {
   string c ;
   int n = rand() % 200 + 1 ;
   if(test < 2) n = rand() % 20 + 1 ;
   
   if(test < 4) for(int i = 0;i < n;i++) c.push_back(rand() % 26 + 'a') ;
   else if(test < 7) for(int i = 0;i < n;i++) c.push_back(rand() % 3 + 'a') ;
   else if(test < 10) for(int i = 0;i < n;i++) c.push_back(rand() % 2 + 'a') ;

   fprintf(fout,"%s\n",c.c_str()) ;
  }
 }
}


int main()
{
// test() ; return 0 ;
// generate() ; return 0 ;

 int runs ;
 cin >> runs ;
 while(runs--)
 {
  string s ;
  cin >> s ;
  int ret = solve(s) ;
  cout << ret << endl ;
 }
 return 0 ;
}








In Java :





import java.io.*;
import java.util.*;

public class Solution implements Runnable {
    final static int MOD = 1000000007;

    int count(String a, String b) {
        int n = a.length();
        int m = b.length();
        int dp[][] = new int[n + 1][m + 1];
        int sum[][] = new int[n + 1][m + 1];
        for (int i = 0; i <= n; ++ i) {
            sum[i][m] = 1;
        }
        for (int j = 0; j <= m; ++ j) {
            sum[n][j] = 1;
        }
        for (int i = n - 1; i >= 0; -- i) {
            for (int j = m - 1; j >= 0; -- j) {
                if (a.charAt(i) == b.charAt(j)) {
                    dp[i][j] = sum[i + 1][j + 1];
                } 
                sum[i][j] = (sum[i + 1][j] + sum[i][j + 1]
                    - sum[i + 1][j + 1] + dp[i][j]) % MOD;
            }
        }
        int result = 0;
        for (int i = 0; i < n; ++ i) {
            result += dp[i][0];
            result %= MOD;
        }
        return result;
    }

    int count(String s) {
        int n = s.length();
        int result = 0;
        for (int i = 1; i < n; ++ i) {
            result += count(s.substring(0, i), s.substring(i, n));
            result %= MOD;
        }
        return (result + MOD) % MOD;
    }

    public void run() {
        try {
            BufferedReader reader = new BufferedReader(
                    new InputStreamReader(System.in));
            int testCount = Integer.parseInt(reader.readLine());
            while (testCount > 0) {
                testCount --;
                System.out.println(count(reader.readLine()));
            }
        } catch (Exception e) {
        }
    }

    public static void main(String args[]) {
        new Thread(new Solution()).run();
    }
}








In C :





#include <stdio.h>

#define P 1000000007

char s[500];
long long i,j,k,l,m,n,t,kk;
//a[210][210][210],
long long b[210][210][210];

int main()
{

scanf("%lld\n",&t);
while(t--)
{
scanf("%s\n",s);

//printf("%s %lld",s,n);

n=0;
while(s[n]) n++;

//printf("%s %lld\n",s,n);

for(j=0;j<=n;j++)
 for(i=j;i<=n;i++) 
  for(k=i;k<=n;k++) b[j][i][k]=0;

m=0;

for(k=n-1;k>=0;k--)
 for(i=k-1;i>=0;i--)
  for(j=i;j>=0;j--)
   {
     b[j][i][k] = 0;
     //b[j+1][i][k];
     
     if(s[j]==s[k])
           {
          // a[j][i][k] = (1+b[j+1][i][k+1])%P; 
           if(i+1==k) m = (m+1+b[j+1][i][k+1])%P;
           
           b[j][i][k] = (b[j][i][k] +1 + b[j+1][i][k+1])%P;
//           printf("%c %lld %lld b=%lld\n",s[j],j,k,b[j][i][k]);
           } 
    b[j][i][k] = (b[j][i][k]+b[j][i][k+1]+b[j+1][i][k]-b[j+1][i][k+1]);
   
   l++;
   }

//printf("%lld=l\n",l);

//m=0;

//for(i=0;i<n;i++) 
// for(k=i+1;k<n;k++) m+= a[i][k-1][k];

printf("%lld\n",m);

//printf("%lld %lld=n\n",b[0][13][14],n);
//printf("%lld %lld=n\n",b[4][14][15],n);

}


return 0;
}








In Python3 :





from sys import stderr

def dp(a,b):
    c = [[0 for j in range(len(b))] for i in range(len(a))]
    for i in range(len(b)):
        if a[0] == b[i]:
            c[0][i] = 1
        if i:
            c[0][i] += c[0][i-1]
            c[0][i] %= mod
    for i in range(1,len(a)):
        c[i][0] = c[i-1][0]
        for j in range(1,len(b)):
            c[i][j] = c[i-1][j] + c[i][j-1] - c[i-1][j-1]
            if a[i] == b[j]:
                c[i][j] += c[i-1][j-1]
                c[i][j] %= mod
    return c[len(a)-1][len(b)-1]
    

def backtrack(c,a,b,i,j):
    if i == 0 or j == 0:
        return ''
    elif a[i-1] == b[j-1]:
        return backtrack(c,a,b,i-1,j-1) + a[i-1]
    else:
        if c[i][j-1] > c[i-1][j]:
            return backtrack(c,a,b,i,j-1)
        else:
            return backtrack(c,a,b,i-1,j)

mod = 1000000007
    
t = int(input().strip())

for _ in ' '*t:
    s = list(input().strip())
    s = ''.join(x for x in s if s.count(x) > 1)
    c = 0
    
    for k in range(1,len(s)):
        d = dp(s[k:],s[:k])
        print(s[:k],s[k:],file=stderr)
        print(d,file=stderr)
        c += d
        c %= mod
            
    print(c)








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