Número de expresiones de paréntesis equilibradas que se pueden formar a partir de una string

Posted on by Rudeus Greyrat

Dada una string str que consta de caracteres ( , ) , { , } , [ , ] y ? . La tarea es encontrar el número total de expresiones de paréntesis equilibrados formadas cuando ? se puede reemplazar con cualquiera de los caracteres de paréntesis. 
Estos son algunos ejemplos de expresiones de paréntesis equilibrados: {([])} , {()}[{}] etc. 
Y expresiones de paréntesis no equilibrados: {[} , {()] , {()}[) etc.

Ejemplos: 

Entrada: str = “(?([?)]?}?” 
Salida: 3 
({([()]]}),()([()]{}) y ([([])]{} ) son las únicas expresiones balanceadas posibles que se pueden generar a partir de la entrada.

Entrada: str = “???[???????]????” 
Salida: 392202 

Acercarse: 

  1. Si n es impar, entonces el resultado siempre será 0 ya que no habrá expresión balanceada posible.
  2. Si n id es par, cree una array dp para almacenar cálculos previos.
  3. Llame a una función recursiva con las siguientes operaciones: 
    • Si el índice inicial> el índice final , devuelve 1 .
    • Si dp[start][end] ya se calculó, devuelva dp[start][end] .
    • Ejecute un ciclo desde el inicio + 1 hasta el final incrementando en 2 para encontrar su par de corchetes o ‘?’.
    • Luego divida la string en dos mitades para verificar las expresiones de paréntesis posteriores adecuadas desde el inicio + 1 hasta i – 1 e i + 1 hasta el final llamando a la función recursiva.
    • Si st[inicio] = ‘?’ y st[i] = ‘?’ entonces se pueden formar un total de 3 combinaciones de pares de corchetes, multiplicando así el resultado de la función recursiva por 3 .
  4. Regresar dp[inicio][fin]

A continuación se muestra la implementación del enfoque anterior: 

C++

// C++ program to find number of balanced
// bracket expressions possible
#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
 
// Max string length
const int MAX = 300;
 
// Function to check whether index start
// and end can form a bracket pair or not
int checkFunc(int i, int j, string st)
{
    // Check for brackets ( )
    if (st[i] == '(' && st[j] == ')')
        return 1;
    if (st[i] == '(' && st[j] == '?')
        return 1;
    if (st[i] == '?' && st[j] == ')')
        return 1;
 
    // Check for brackets [ ]
    if (st[i] == '[' && st[j] == ']')
        return 1;
    if (st[i] == '[' && st[j] == '?')
        return 1;
    if (st[i] == '?' && st[j] == ']')
        return 1;
 
    // Check for brackets { }
    if (st[i] == '{' && st[j] == '}')
        return 1;
    if (st[i] == '{' && st[j] == '?')
        return 1;
    if (st[i] == '?' && st[j] == '}')
        return 1;
 
    return 0;
}
 
// Function to find number of
// proper bracket expressions
int countRec(int start, int end, int dp[][MAX],
             string st)
{
    int sum = 0;
 
    // If starting index is greater
    // than ending index
    if (start > end)
        return 1;
 
    // If dp[start][end] has already been computed
    if (dp[start][end] != -1)
        return dp[start][end];
 
    lli i, r = 0;
 
    // Search for the bracket in from next index
    for (i = start + 1; i <= end; i += 2) {
 
        // If bracket pair is formed,
        // add number of combination
        if (checkFunc(start, i, st)) {
 
            sum = sum
                  + countRec(start + 1, i - 1, dp, st)
                        * countRec(i + 1, end, dp, st);
        }
 
        // If ? comes then all three bracket
        // expressions are possible
        else if (st[start] == '?' && st[i] == '?') {
 
            sum = sum
                  + countRec(start + 1, i - 1, dp, st)
                        * countRec(i + 1, end, dp, st)
                        * 3;
        }
    }
 
    // Return answer
    return dp[start][end] = sum;
}
 
int countWays(string st)
{
    int n = st.length();
 
    // If n is odd, string cannot be balanced
    if (n % 2 == 1)
        return 0;
    int dp[MAX][MAX];
    memset(dp, -1, sizeof(dp));
    return countRec(0, n - 1, dp, st);
}
 
// Driving function
int main()
{
    string st = "(?([?)]?}?";
    cout << countWays(st);
    return 0;
}

Java

// Java program to find number of balanced
// bracket expressions possible
 
class GFG {
    // Max string length
    static int MAX = 300;
 
    // Function to check whether index start
    // and end can form a bracket pair or not
    static int checkFunc(int i, int j, String st)
    {
        // Check for brackets ( )
        if (st.charAt(i) == '(' && st.charAt(j) == ')')
            return 1;
        if (st.charAt(i) == '(' && st.charAt(j) == '?')
            return 1;
        if (st.charAt(i) == '?' && st.charAt(j) == ')')
            return 1;
 
        // Check for brackets [ ]
        if (st.charAt(i) == '[' && st.charAt(j) == ']')
            return 1;
        if (st.charAt(i) == '[' && st.charAt(j) == '?')
            return 1;
        if (st.charAt(i) == '?' && st.charAt(j) == ']')
            return 1;
 
        // Check for brackets { }
        if (st.charAt(i) == '{' && st.charAt(j) == '}')
            return 1;
        if (st.charAt(i) == '{' && st.charAt(j) == '?')
            return 1;
        if (st.charAt(i) == '?' && st.charAt(j) == '}')
            return 1;
 
        return 0;
    }
 
    // Function to find number of
    // proper bracket expressions
    static int countRec(int start, int end, int dp[][],
                        String st)
    {
        int sum = 0;
 
        // If starting index is greater
        // than ending index
        if (start > end)
            return 1;
 
        // If dp[start][end] has already been computed
        if (dp[start][end] != -1)
            return dp[start][end];
 
        int i, r = 0;
 
        // Search for the bracket in from next index
        for (i = start + 1; i <= end; i += 2) {
 
            // If bracket pair is formed,
            // add number of combination
            if (checkFunc(start, i, st) == 1) {
 
                sum = sum
                      + countRec(start + 1, i - 1, dp, st)
                            * countRec(i + 1, end, dp, st);
            }
 
            // If ? comes then all three bracket
            // expressions are possible
            else if (st.charAt(start) == '?' && st.charAt(i) == '?') {
 
                sum = sum
                      + countRec(start + 1, i - 1, dp, st)
                            * countRec(i + 1, end, dp, st)
                            * 3;
            }
        }
 
        // Return answer
        return dp[start][end] = sum;
    }
 
    static int countWays(String st)
    {
        int n = st.length();
 
        // If n is odd, string cannot be balanced
        if (n % 2 == 1)
            return 0;
 
        int dp[][] = new int[MAX][MAX];
 
        for (int i = 0; i < MAX; i++)
            for (int j = 0; j < MAX; j++)
                dp[i][j] = -1;
 
        return countRec(0, n - 1, dp, st);
    }
 
    // Driving function
    public static void main(String[] args)
    {
 
        String st = "(?([?)]?}?";
        System.out.println(countWays(st));
    }
}
 
// This code is contributed by ihritik

Python 3

# Python 3 program to find number of balanced
# bracket expressions possible
 
# Max string length
MAX = 300
 
# Function to check whether index start
# and end can form a bracket pair or not
def checkFunc(i, j, st):
 
    # Check for brackets ( )
    if (st[i] == '(' and st[j] == ')'):
        return 1
    if (st[i] == '(' and st[j] == '?'):
        return 1
    if (st[i] == '?' and st[j] == ')'):
        return 1
 
    # Check for brackets [ ]
    if (st[i] == '[' and st[j] == ']'):
        return 1
    if (st[i] == '[' and st[j] == '?'):
        return 1
    if (st[i] == '?' and st[j] == ']'):
        return 1
 
    # Check for brackets { }
    if (st[i] == '{' and st[j] == '}'):
        return 1
    if (st[i] == '{' and st[j] == '?'):
        return 1
    if (st[i] == '?' and st[j] == '}'):
        return 1
 
    return 0
 
# Function to find number of
# proper bracket expressions
def countRec(start, end, dp, st):
 
    sum = 0
 
    # If starting index is greater
    # than ending index
    if (start > end):
        return 1
 
    # If dp[start][end] has already
    # been computed
    if (dp[start][end] != -1):
        return dp[start][end]
 
    r = 0
 
    # Search for the bracket in from next index
    for i in range(start + 1, end + 1, 2):
 
        # If bracket pair is formed,
        # add number of combination
        if (checkFunc(start, i, st)):
 
            sum = (sum + countRec(start + 1, i - 1, dp, st) *
                         countRec(i + 1, end, dp, st))
 
        # If ? comes then all three bracket
        # expressions are possible
        elif (st[start] == '?' and st[i] == '?'):
 
            sum = (sum + countRec(start + 1, i - 1, dp, st) *
                         countRec(i + 1, end, dp, st) * 3)
 
    # Return answer
    dp[start][end] = sum
    return dp[start][end]
 
def countWays( st):
 
    n = len(st)
 
    # If n is odd, string cannot be balanced
    if (n % 2 == 1):
        return 0
    dp = [[-1 for i in range(MAX)]
              for i in range(MAX)]
    return countRec(0, n - 1, dp, st)
 
# Driver Code
if __name__ =="__main__":
     
    st = "(?([?)]?}?"
    print(countWays(st))
 
# This code is contributed by ita_c

C#

// C# program to find number of balanced
// bracket expressions possible
 
using System;
class GFG {
    // Max string length
    static int MAX = 300;
 
    // Function to check whether index start
    // and end can form a bracket pair or not
    static int checkFunc(int i, int j, string st)
    {
        // Check for brackets ( )
        if (st[i] == '(' && st[j] == ')')
            return 1;
        if (st[i] == '(' && st[j] == '?')
            return 1;
        if (st[i] == '?' && st[j] == ')')
            return 1;
 
        // Check for brackets [ ]
        if (st[i] == '[' && st[j] == ']')
            return 1;
        if (st[i] == '[' && st[j] == '?')
            return 1;
        if (st[i] == '?' && st[j] == ']')
            return 1;
 
        // Check for brackets { }
        if (st[i] == '{' && st[j] == '}')
            return 1;
        if (st[i] == '{' && st[j] == '?')
            return 1;
        if (st[i] == '?' && st[j] == '}')
            return 1;
 
        return 0;
    }
 
    // Function to find number of
    // proper bracket expressions
    static int countRec(int start, int end, int[, ] dp,
                        string st)
    {
        int sum = 0;
 
        // If starting index is greater
        // than ending index
        if (start > end)
            return 1;
 
        // If dp[start, end] has already been computed
        if (dp[start, end] != -1)
            return dp[start, end];
 
        int i;
 
        // Search for the bracket in from next index
        for (i = start + 1; i <= end; i += 2) {
 
            // If bracket pair is formed,
            // add number of combination
            if (checkFunc(start, i, st) == 1) {
 
                sum = sum
                      + countRec(start + 1, i - 1, dp, st)
                            * countRec(i + 1, end, dp, st);
            }
 
            // If ? comes then all three bracket
            // expressions are possible
            else if (st[start] == '?' && st[i] == '?') {
 
                sum = sum
                      + countRec(start + 1, i - 1, dp, st)
                            * countRec(i + 1, end, dp, st)
                            * 3;
            }
        }
 
        // Return answer
        return dp[start, end] = sum;
    }
 
    static int countWays(string st)
    {
        int n = st.Length;
 
        // If n is odd, string cannot be balanced
        if (n % 2 == 1)
            return 0;
 
        int[, ] dp = new int[MAX, MAX];
 
        for (int i = 0; i < MAX; i++)
            for (int j = 0; j < MAX; j++)
                dp[i, j] = -1;
 
        return countRec(0, n - 1, dp, st);
    }
 
    // Driving function
    public static void Main()
    {
 
        string st = "(?([?)]?}?";
        Console.WriteLine(countWays(st));
    }
}
 
// This code is contributed by ihritik

Javascript

<script>
// Javascript program to find number of balanced
// bracket expressions possible
     
    // Max string length
    let MAX = 300;
    // Function to check whether index start
    // and end can form a bracket pair or not
    function checkFunc(i,j,st)
    {
        // Check for brackets ( )
        if (st[i] == '(' && st[j] == ')')
            return 1;
        if (st[i] == '(' && st[j] == '?')
            return 1;
        if (st[i] == '?' && st[j] == ')')
            return 1;
   
        // Check for brackets [ ]
        if (st[i] == '[' && st[j] == ']')
            return 1;
        if (st[i] == '[' && st[j] == '?')
            return 1;
        if (st[i] == '?' && st[j] == ']')
            return 1;
   
        // Check for brackets { }
        if (st[i] == '{' && st[j] == '}')
            return 1;
        if (st[i] == '{' && st[j] == '?')
            return 1;
        if (st[i] == '?' && st[j] == '}')
            return 1;
   
        return 0;
    }
     
    // Function to find number of
    // proper bracket expressions
    function countRec(start,end,dp,st)
    {
        let sum = 0;
   
        // If starting index is greater
        // than ending index
        if (start > end)
            return 1;
   
        // If dp[start][end] has already been computed
        if (dp[start][end] != -1)
            return dp[start][end];
   
        let i, r = 0;
   
        // Search for the bracket in from next index
        for (i = start + 1; i <= end; i += 2) {
   
            // If bracket pair is formed,
            // add number of combination
            if (checkFunc(start, i, st) == 1) {
   
                sum = sum
                      + countRec(start + 1, i - 1, dp, st)
                            * countRec(i + 1, end, dp, st);
            }
   
            // If ? comes then all three bracket
            // expressions are possible
            else if (st[start] == '?' && st[i] == '?') {
   
                sum = sum
                      + countRec(start + 1, i - 1, dp, st)
                            * countRec(i + 1, end, dp, st)
                            * 3;
            }
        }
   
        // Return answer
        return dp[start][end] = sum;
    }
     
    function countWays(st)
    {
        let n = st.length;
   
        // If n is odd, string cannot be balanced
        if (n % 2 == 1)
            return 0;
   
        let dp = new Array(MAX);
   
        for (let i = 0; i < MAX; i++)
        {   
            dp[i]=new Array(MAX);
            for (let j = 0; j < MAX; j++)
                dp[i][j] = -1;
          }
        return countRec(0, n - 1, dp, st);
    }
     
    // Driving function
    let st = "(?([?)]?}?";
    document.write(countWays(st));
     
     
// This code is contributed by rag2127
</script>
Producción: 

3

 

Publicación traducida automáticamente

Artículo escrito por atulim y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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