How to Solve Generate Parentheses
Generate Parentheses Introduction
Generate Parentheses Introduction
The Generate Parenthesis problem involves generating all valid combinations of parentheses for a given number of pairs of parenthesis. This problem can be solved using both recursive and iterative approaches, and is a great introduction to backtracking style recursion.
Generate Parentheses Problem
Generate Parentheses Problem
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
Example Inputs and Outputs
Example 1
Input: n = 3 Output: ["((()))","(()())","(())()","()(())","()()()"]
Input: n = 2 Output: [“()()”, “(())”]
Constraints
1 <= n <= 10,000
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Generate Parentheses Solutions
Generate Parentheses Solutions
Approach 1: Recursion
We are tasked with generating all combinations of random parenthesis, and the word combinations should serve as a clue that recursion could be useful. Let’s say we are solving this problem by hand for n = 2. Logically, we start with one forward paren (, and from here we have two options: add another forward paren, or add a backward paren.
Note that we will not always be able to add a forward or backward paren. If we are generating all combinations for n = 2 and we already have two forward parens, then we shouldn’t try to add more forward parens. Likewise, if we only have one forward paren in the solution thus far but we already have one backward paren, we can’t add any more backwards parentheses as that would make our solution invalid.
Putting it all together we have a recursive solution where we build up and tear down possible solutions, adding parens as we go. We modify a list containing the forward and backward parens instead of concatenating strings because string concatenation is a linear time complexity operation, while appending to and popping from a list can be done in constant time.
Generate Parentheses Python, JavaScript and Java Solutions - Recursion
class Solution:
def generateParenthesis(self, n: int) -> List[str]:
current_string = []
valid_answers = []
def recurse(forward_parens_needed: int, backward_parens_needed: int) -> None:
nonlocal valid_answers
nonlocal current_string
if forward_parens_needed == 0 and backward_parens_needed == 0:
valid_answers.append(''.join(current_string))
if forward_parens_needed > 0:
current_string.append('(')
recurse(forward_parens_needed - 1, backward_parens_needed)
if backward_parens_needed > 0 and forward_parens_needed < backward_parens_needed:
current_string.append(')')
recurse(forward_parens_needed, backward_parens_needed - 1)
if len(current_string) > 0:
current_string.pop()
recurse(n, n)
return valid_answers
1class Solution:
2 def generateParenthesis(self, n: int) -> List[str]:
3 current_string = []
4 valid_answers = []
5
6 def recurse(forward_parens_needed: int, backward_parens_needed: int) -> None:
7 nonlocal valid_answers
8 nonlocal current_string
9
10 if forward_parens_needed == 0 and backward_parens_needed == 0:
11 valid_answers.append(''.join(current_string))
12
13 if forward_parens_needed > 0:
14 current_string.append('(')
15 recurse(forward_parens_needed - 1, backward_parens_needed)
16
17 if backward_parens_needed > 0 and forward_parens_needed < backward_parens_needed:
18 current_string.append(')')
19 recurse(forward_parens_needed, backward_parens_needed - 1)
20
21 if len(current_string) > 0:
22 current_string.pop()
23
24 recurse(n, n)
25 return valid_answers
26Time/Space Complexity
- Time complexity:
O(2n) - Space complexity:
O(n), as we never have more thann* 2 parentheses in thevalid_answersarray.
Approach 2: Dynamic Programming
When solving an algorithm problem and looking for a more efficient solution, dynamic programming can often be a great tool to improve the time complexity of a solution, though spotting the pattern and implementing the code can be tricky.
To spot a place where dynamic programming could be helpful, we look for situations where solving a subproblem (or a few subproblems) can help solve the main problem at hand. For example, consider a simpler problem in which we are trying to determine the number of ways a robot can traverse a grid from the top left corner to the bottom right corner while only going down and to the right.
To get to a given square, the robot can either move to the right from the square on the left or move down from the square above. Thus, the number of ways the robot can get to square [x][y] = routes to [x - 1][y] + routes to [x][y - 1].
Turning back to generate parentheses, the dynamic programming pattern is more difficult to spot. At a high level, we solve for generateParenthesis(0), generateParenthesis(1) ... generateParentheses(X - 1) and then combine those solutions to solve for generateParentheses(X).
For each value of X, we insert a new pair of parentheses around all possible solutions to generateParentheses(0) ... generateParentheses(X - 1), and then combining each of those solutions with however many parentheses we still need in order to generate X parentheses total. By the time we get to n we have solved this problem for every number between 0 and n, with the answers stored in a zero-indexed array, so we simply return the nth index in the array.
This particular problem is a fairly difficult dynamic programming problem, so if you are new to algorithms and/or dynamic programming I would suggest focussing on simpler problems such as the robot in grid problem mentioned above first, then coming back to this problem once solidifying your understanding of the basics.
Generate Parentheses Python, JavaScript and Java Solutions - Dynamic Programming
class Solution:
def generateParenthesis(self, n: int) -> List[str]:
dp = [[] for i in range(n + 1)]
dp[0].append('')
# iterate over all f(0) ... f(n) parenthesis pair targets
for i in range(n + 1):
# iterate over all solutions generated thus far --> f(0) ... f(i)
for j in range(i):
to_be_added = []
# iterate over all solutions for f(j)
for x in dp[j]:
# iterate over all solutions for f(n - j)
for y in dp[i - j - 1]:
# insert solution for f(j) between parens
# --> and add all solutions for f(n - j) to back
to_be_added.append('(' + x + ')' + y)
dp[i] += to_be_added
return dp[n]1class Solution:
2 def generateParenthesis(self, n: int) -> List[str]:
3 dp = [[] for i in range(n + 1)]
4 dp[0].append('')
5
6 # iterate over all f(0) ... f(n) parenthesis pair targets
7 for i in range(n + 1):
8 # iterate over all solutions generated thus far --> f(0) ... f(i)
9 for j in range(i):
10 to_be_added = []
11 # iterate over all solutions for f(j)
12 for x in dp[j]:
13 # iterate over all solutions for f(n - j)
14 for y in dp[i - j - 1]:
15 # insert solution for f(j) between parens
16 # --> and add all solutions for f(n - j) to back
17 to_be_added.append('(' + x + ')' + y)
18
19 dp[i] += to_be_added
20
21 return dp[n]Time/Space Complexity
- Time complexity:
O(n * Cat(n)), whereCatrepresents the time complexity of Catalan numbers (out of scope for this article and likely for an interview). - Space complexity:
O(n * Cat(n)).
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