
Problem Statement
The task is to create a function that generates all possible well-formed combinations of parentheses for a given number of pairs. A combination is considered well-formed if every opening parenthesis "(" has a corresponding closing parenthesis ")" and the pairs are correctly nested. For instance, with three pairs (n = 3
), some valid combinations include "((()))", "(()())", and "()(())", among others. The goal is to generate all such possible combinations for any given n
within the constraints.
Examples
Example 1
Input:
n = 3
Output:
["((()))","(()())","(())()","()(())","()()()"]
Example 2
Input:
n = 1
Output:
["()"]
Constraints
1 <= n <= 8
Approach and Intuition
The challenge of generating well-formed parentheses can be approached recursively or using iterative methods such as backtracking, which is well-suited for this kind of problem where we need to explore all possible combinations under certain constraints. Here’s the intuitive breakdown of this process:
Initialization:
- Start with an empty string and initiate the process with the total number of opening and closing brackets set to
n
.
- Start with an empty string and initiate the process with the total number of opening and closing brackets set to
Recursive Backtracking Function:
The recursive function takes the current string in construction, the remaining count of opening brackets, and closing brackets.
Base Case:
- When both the remaining opening and closing brackets count reaches zero, a valid combination has been formed. This combination is added to the result list.
Recursive Case:
If the count of remaining opening brackets > 0, a new combination can be formed by adding an opening bracket and recursively calling the function with decremented count of opening brackets.
If the count of remaining closing brackets is greater than the count of remaining opening brackets, a closing bracket can be added. This ensures that no closing bracket is added without a corresponding opening bracket before it, maintaining the well-formed criterion.
Backtracking:
- This is inherent in the recursive approach as it explores one branch (like adding an opening bracket), reaches an end (either finding a valid combination or violating the well-formed rule), and then returns to explore the other branch (adding a closing bracket).
This approach effectively builds all combinations by adding parentheses step-by-step, while the recursive structure ensures that only valid sequences are created and added to the list. The constraints of n
being between 1 and 8 inclusively guarantee that the recursion depth and computational cost remain feasible.
Solutions
- C++
- Java
- C
- JavaScript
- Python
class Solution {
public:
std::vector<std::string> createParentheses(int count) {
if (count == 0) {
return std::vector<std::string>{""};
}
std::vector<std::string> result;
for (int lc = 0; lc < count; ++lc) {
for (std::string lPart : createParentheses(lc)) {
for (std::string rPart : createParentheses(count - 1 - lc)) {
result.push_back("(" + lPart + ")" + rPart);
}
}
}
return result;
}
};
The provided C++ code defines a class named Solution
which includes a method called createParentheses
. This method generates all valid combinations of count
pairs of parentheses.
Explanation of the Implementation:
- The method
createParentheses
takes an integercount
which specifies the number of pairs of parentheses. - It starts with a base case: if
count
is 0, it returns a vector containing an empty string. This represents the base scenario where no parentheses are needed. - For generating combinations, the method uses a recursive approach:
- It initializes an empty vector of strings named
result
to store the combinations of parentheses. - It iterates through possible splits of the
count
using a variablelc
, wherelc
indicates the number of left parentheses. - Inside the outer loop, it calls itself recursively to generate the left part of the parentheses string (
lPart
) usinglc
as the argument. - It then calls itself again to generate the right part of the parentheses string (
rPart
), usingcount - 1 - lc
as the argument. - For each pair of left and right parts obtained from recursive calls, the method constructs a valid set of parentheses by wrapping
lPart
in a single pair of parentheses and then concatenating it withrPart
. This constructed string is then added to theresult
vector.
- It initializes an empty vector of strings named
- Finally, after iterating through all possibilities, the
result
vector containing all valid combinations of parentheses is returned.
How to Utilize This Method:
- Create an instance of the
Solution
class. - Call the
createParentheses
method with a specific number of pairs of parentheses you wish to generate. - The method returns a vector of strings, where each string represents a valid permutation of the parentheses.
class Solution {
public List<String> generateBalancedParenthesis(int count) {
if (count == 0) {
return new ArrayList(Arrays.asList(""));
}
List<String> result = new ArrayList();
for (
int openBrace = 0;
openBrace < count;
++openBrace
) for (String leftPart : generateBalancedParenthesis(
openBrace
)) for (String rightPart : generateBalancedParenthesis(
count - 1 - openBrace
)) result.add("(" + leftPart + ")" + rightPart);
return result;
}
}
The given Java solution is designed to generate all combinations of well-formed parentheses for a given number count
. Here are the core components and the functionality of the code:
The
generateBalancedParenthesis(int count)
method is part of aSolution
class. It uses recursion to create all possible combinations of balanced parentheses.An
ArrayList
namedresult
is created to store the combinations.If
count
is 0, the method returns an ArrayList with an empty string. This represents the base case of the recursion, indicating there are no more parentheses to balance.For the non-base cases, the method iterates through possible splits between the left and right parts of the parentheses string. It makes a recursive call for each possible number of opening braces
openBrace
and combines the results.The loop combines the left and right parts of the parentheses: for each combination from the recursive call, it adds an opening parenthesis
(
then theleftPart
, a closing parenthesis)
, and then therightPart
.Finally, the method returns the
result
list containing all possible well-formed parentheses combinations.
For example, when count
is 3, some of the strings returned would be ((()))
, (()())
, (())()
, ()(())
, and ()()()
.
This approach ensures that all potential combinations are covered by exploring every possible distribution of balanced pairs of parentheses through recursion, guaranteeing the correct format and balance for each combination before adding it to the result list.
char **generateCombinations(int count, int *sizeOut) {
if (count == 0) {
char **output = malloc(sizeof(char *));
output[0] = malloc(sizeof(char));
output[0][0] = '\0';
*sizeOut = 1;
return output;
}
int totalSize = 0, currIndex = 0;
char **combinations = NULL;
for (int parLeft = 0; parLeft < count; ++parLeft) {
int countLeft = 0;
int countRight = 0;
char **leftParts = generateCombinations(parLeft, &countLeft);
char **rightParts =
generateCombinations(count - 1 - parLeft, &countRight);
for (int i = 0; i < countLeft; ++i) {
for (int j = 0; j < countRight; ++j) {
totalSize +=
2 + strlen(leftParts[i]) + strlen(rightParts[j]);
combinations = realloc(combinations, totalSize * sizeof(char *));
combinations[currIndex] = malloc(2 + strlen(leftParts[i]) +
strlen(rightParts[j]) + 1);
sprintf(combinations[currIndex], "(%s)%s", leftParts[i],
rightParts[j]);
++currIndex;
}
}
for (int i = 0; i < countLeft; ++i) free(leftParts[i]);
free(leftParts);
for (int j = 0; j < countRight; ++j) free(rightParts[j]);
free(rightParts);
}
*sizeOut = currIndex;
return combinations;
}
The solution provided is a C program that generates all possible valid parentheses combinations for a given number of pairs count
. The function generateCombinations
utilizes recursion and dynamic memory allocation to construct each combination.
Here's a breakdown of how the solution works:
- The function accepts an integer
count
, representing the number of pairs of parentheses, and a pointersizeOut
to store the total number of combinations generated. - If
count
is zero (base case), the function initializes an output array with an empty string and setssizeOut
to 1. - Two integer variables
totalSize
andcurrIndex
track the total size of thecombinations
array and the current index, respectively. - The function iterates through all possible splits of parentheses pairs, recursively generating combinations for the left part (
parLeft
) and the right part (count - 1 - parLeft
). - For each left and right combination, the function concatenates them into new strings surrounded by a pair of parentheses and stores them in the
combinations
array, adjusting memory allocation accordingly. - Memory allocated for individual parts is freed after adding the new combination to avoid memory leaks.
- Finally, the function updates
*sizeOut
with the count of unique combinations generated, returning thecombinations
array.
This efficient recursive approach ensures all combinations of valid parentheses are generated properly, respecting scoping rules by ensuring each left parenthesis has a corresponding right parenthesis. Memory management is crucial, as all dynamically allocated space is properly cleaned up to prevent memory leaks.
var makeParentheses = function (count) {
if (count === 0) {
return [""];
}
let result = [];
for (let i = 0; i < count; ++i) {
for (let left of makeParentheses(i)) {
for (let right of makeParentheses(count - 1 - i)) {
result.push("(" + left + ")" + right);
}
}
}
return result;
};
The function makeParentheses
effectively generates all possible combinations of well-formed parentheses given a specific number count
representing pairs of parentheses. Implement this solution in JavaScript using a recursive approach.
- Define a recursive function named
makeParentheses
which accepts an integercount
. - Use a base case to return a list containing an empty string if
count
is 0, indicating no parentheses are needed. - Initialize an empty list
result
to store the combinations of valid parentheses. - Use a loop to iterate through integer values from 0 up to
count
. - For each iteration, recursively invoke
makeParentheses
for the left part for each valid left parentheses combination and the remaining right part for each valid right parentheses combination. - Concatenate the left and right combinations in the format
("(" + left + ")" + right)
to ensure the parentheses are validly nested, and append this to theresult
list. - Finally, return the
result
list containing all the valid combinations.
This approach ensures that for each position in the sequence, a valid configuration of nested parentheses is created, leveraging the recursive structure to explore every possible valid configuration and storing them efficiently.
class Solution:
def generateBrackets(self, count: int) -> List[str]:
if count == 0:
return [""]
results = []
for open_count in range(count):
for open_bracket_set in self.generateBrackets(open_count):
for close_bracket_set in self.generateBrackets(
count - 1 - open_count
):
results.append("(" + open_bracket_set + ")" + close_bracket_set)
return results
This Python solution effectively generates all possible combinations of well-formed parentheses given a number. The function generateBrackets
uses recursion to build these combinations, forming the key algorithmic approach that ensures every generated string of parentheses is well-formed.
The base case of the recursion addresses situations where no parentheses are needed, simply returning a list containing an empty string. For cases where count
is greater than zero, the program iteratively builds the combinations:
- Iteratively increment the number of opening brackets.
- Recursively call
generateBrackets
for subsets of the count, ensuring different configurations of open and closed brackets are considered. - Concatenate the results in the form "(
open_bracket_set
)close_bracket_set
" to maintain proper nesting and balance of the parentheses.
By iterating over all possible counts of opening brackets and recursively solving for each configuration, this solution efficiently constructs each unique sequence of well-formed parentheses, accumulating them into the results
list which is returned at the end. This approach ensures comprehensive coverage of all potential combinations, leveraging recursion effectively for combinatorial generation.
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