Print all combinations of balanced parentheses

Try it on GfG Practice

Given a number n, print all combinations of balanced parentheses of length n.

Note: A sequence of parentheses is balanced if every opening bracket ( has a corresponding closing bracket ) in the correct order.

Examples: 

Input: n = 2
Output: ["( )"]
Explanation: Only one valid sequence can be formed using 1 pair of parentheses, and it is balanced.

Input : n = 4
Output: [ "( ( ) )", "( )( )"]
Explanation: Two valid balanced sequences are possible with 2 pairs: one with nested parentheses, and one with pairs side by side.

[Naive Approach] Generating All Parentheses - O(2^n * n) Time and O(n) Space

The main idea is to generate all possible combinations of parentheses and check if it is balance or not. If a sequence is balanced, we store it as a valid result.

[Expected Approach] Using BackTracking

We recursively construct the sequence by considering these two conditions.

• The i-th character can be '(' if the number of opening '(' brackets used so far is less than n. This ensures we don’t add more opening brackets than are needed, keeping the sequence valid.
• The i-th character can be ')' if the number of closing ')' brackets used so far is less than the number of opening '(' brackets. This ensures that each closing bracket has a matching opening bracket before it, maintaining the balance.

// c++ program to generate all the combinations of balanced
// parenthesis.

#include <iostream>
#include <vector>
using namespace std;

// Function to generate valid parentheses sequences
void validParentheses(int n, int open, string curr, vector<string>& res) {
    
    // If the current sequence has reached
    // the length of 2 * n, add it to the result
    if (curr.length() == 2 * n) {
        res.push_back(curr);
        return;
    }

    // Add opening parenthesis if we haven't used all n opening parentheses
    if (open < n) validParentheses(n, open + 1, curr + '(', res);

    // Add closing parenthesis if the number of closing
    // parentheses is less than the number of opening ones
    if (curr.length() - open < open) validParentheses(n, open, curr + ')', res);
}

//function to return all valid parentheses sequences
vector<string> generateParentheses(int n) {
    vector<string> res;
    
     // Starting with 0 open parentheses and an empty string
    validParentheses(n/2, 0, "", res); 
    return res;
}

int main() {
    int n = 4; 
    vector<string> res = generateParentheses(n);

    for (const string& seq : res) {
        cout << seq << endl;
    }

    return 0;
}

// java program to generate all the combinations of balanced
// parenthesis.
import java.util.ArrayList;
class GfG {

    // Function to generate valid parentheses sequences
   static void validParentheses(int n, int open, String curr,
                                ArrayList<String> res) {
                                    
        // If the current sequence has reached the length of 2 * n, 
        // add it to the result
        if (curr.length() == 2 * n) {
            res.add(curr);
            return;
        }

        // Add opening parenthesis if we haven't used all n opening
        // parentheses
        if (open < n)
        validParentheses(n, open + 1, curr + "(", res);

        // Add closing parenthesis if the number of
        // closing parentheses is less than the number of opening ones
        if (curr.length() - open < open) 
        validParentheses(n, open, curr + ")", res);
    }

    // Function to return all valid parentheses sequences
    static ArrayList<String> generateParentheses(int n) {
        ArrayList<String> res = new ArrayList<>();
        
        // Start recursion with 0 open parentheses and an empty string
        validParentheses(n/2, 0, "", res);  
        return res;
    }

    public static void main(String[] args) {
        int n = 4; 
        ArrayList<String> res = generateParentheses(n);

        for (String seq : res) {
            System.out.println(seq);
        }
    }
}

# Python program to print all the combinations of balanced
# parentheses.

# Function to generate valid parentheses sequences
def validParentheses(n, openCount, curr, res):
   
    # If the current sequence has reached
    # the length of 2 * n, add it to the result
    if len(curr) == 2 * n:
        res.append(curr)
        return

    # Add opening parenthesis if we haven't used all n opening parentheses
    if openCount < n:
        validParentheses(n, openCount + 1, curr + '(', res)

    # Add closing parenthesis if the number of
    # closing parentheses is less than the number of opening ones
    if len(curr) - openCount < openCount:
        validParentheses(n, openCount, curr + ')', res)

# Function to return all valid parentheses sequences
def generateParentheses(n):
    res = []
    
    # Start recursion with 0 open parentheses and an empty string
    validParentheses(n/2, 0, '', res)  
    return res

if __name__ == "__main__":
    n = 4 
    res = generateParentheses(n)

    for seq in res:
        print(seq)

// C# program to print all the combinations of balanced
// parentheses.

using System;
using System.Collections.Generic;

class GfG {
    // Function to generate valid parentheses sequences
    static void validParentheses(int n, int open, string curr, List<string> res) {
        if (curr.Length == 2 * n) {
            res.Add(curr);
            return;
        }

        if (open < n)
            validParentheses(n, open + 1, curr + "(", res);

        if (curr.Length - open < open)
            validParentheses(n, open, curr + ")", res);
    }

    // Function to return all valid parentheses sequences
    static List<string> generateParentheses(int n) {
        var res = new List<string>();
        validParentheses(n, 0, "", res);
        return res;
    }

    static void Main() {
        int n = 2;
        List<string> res = generateParentheses(n);

        foreach (string seq in res) {
            Console.WriteLine(seq);
        }
    }
}

// JavaScript program to generate all the combinations of balanced
// parenthesis.

// Function to generate valid parentheses sequences
function validParentheses(n, open, curr, res) {
    
    // If the current sequence has reached
    // the length of 2 * n, add it to the result
    if (curr.length === 2 * n) {
        res.push(curr);
        return;
    }

    // Add opening parenthesis if we haven't used all n opening parentheses
    if (open < n) validParentheses(n, open + 1, curr + '(', res);

    // Add closing parenthesis if the number of 
    // closing parentheses is less than the number of opening ones
    if (curr.length - open < open) validParentheses(n, open, curr + ')', res);
}

// Function to return all valid parentheses sequences
function generateParentheses(n) {
    let res = [];
    
    // Start recursion with 0 open parentheses and an empty string
    validParentheses(n/2, 0, '', res);  
    return res;
}

// Driver code
let n = 4; 
let res = generateParentheses(n);

res.forEach(seq => {
    console.log(seq);
});

(())
()()

Time complexity: O(C_n​ × n), where C_n​ is the nth Catalan number.

C_n = \frac{(2n)!}{(n+1)!n!}

• Every valid sequence must start with an opening '('.
• That '(' has to match some closing ')' later. Suppose it matches the very next character then you’ve enclosed 0 other pairs inside, and the rest of the string (after that match) is a valid sequence of n − 1 pairs.
• Or, it might match after one full pair is inside—so you’ve enclosed 1 pair inside " (( … )) ", and the remainder after is a valid sequence of n−2 pairs.
• Or you could enclose 2 pairs, 3 pairs, … up to n − 1 pairs.

If you let C_k​ be the number of sequences of k pairs, then for each choice of how many pairs you “enclose” inside the first match (call that k), you get

C_k ×  C_n−1−k

Summing over k = 0 to k = n-1 gives the Catalan recurrence

C_n = (k = 0 to n - 1) ∑ C_k ×  C_n−1−k

Auxiliary space: O(n), for the recursion stack depth and the temporary string storing the current sequence.