Number of balanced bracket expressions that can be formed from a string

Given a string str comprising of characters (, ), {, }, [, ] and ?. The task is to find the total number of balanced bracket expressions formed when ? can be replaced with any of the bracket characters. 
Here are some examples of balanced bracket expressions: {([])}, {()}[{}] etc. 
And, unbalanced bracket expressions: {[}, {()], {()}[) etc.

Examples:

Input: str = “(?([?)]?}?” 
Output: 3 
({([()]]}), ()([()]{}) and ([([])]{}) are the only possible balanced expressions that can be generated from the input.

Input: str = “???[???????]????” 
Output: 392202 

Approach:

1. If n is odd, then the result will always be 0 as there will be no balanced expression possible.
2. If n id even, then create a dp array for storing precomputations.
3. Call a recursive function with the following operations: 
   • If starting index > ending index then you return 1.
   • If dp[start][end] is already computed, return dp[start][end].
   • Run a loop from start + 1 till end incrementing by 2 to find its pair bracket or ‘?’.
   • Then divide the string into two halves to check to proper subsequent bracket expressions from start + 1 till i – 1 and i + 1 till the end by calling the recursive function.
   • If st[start] = ’?’ and st[i] = ’?’ then a total of 3 combinations of bracket pairs can be formed, thus multiplying the result of the recursive function by 3.
4. Return dp[start][end]

Below is the implementation of the above approach: 

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>

Output:

3

Time Complexity: O(N^3)
Auxiliary Space: O(MAX^2)