Generate Parentheses
Problem Highlights
- 🔗 Leetcode Link: https://leetcode.com/problems/generate-parentheses/
- 💡 Difficulty: Medium
- ⏰ Time to complete: __ mins
- 🛠️ Topics: Backtracking
- 🗒️ Similar Questions: Letter Combinations of a Phone Number
1: U-nderstand
Understand what the interviewer is asking for by using test cases and questions about the problem.
- Established a set (2-3) of test cases to verify their own solution later.
- Established a set (1-2) of edge cases to verify their solution handles complexities.
- Have fully understood the problem and have no clarifying questions.
- Have you verified any Time/Space Constraints for this problem?
- If solved recursively, what is the time complexity?
- For input
n(the number of pairs), the generated strings have length2n. A naive brute-force that enumerates every length-2nstring over(and)and writes each one out runs in O(2^(2n) · n) = O(4^n · n) time (2^(2n) candidate strings, O(n) to build each). The number of valid sequences is the n-th Catalan number C_n, which grows asymptotically as O(4^n / n^(3/2)). Since the backtracking solution only constructs the valid sequences (each costing O(n) to write), the total enumeration time is O(C_n · n) = O(4^n / sqrt(n)).
- For input
- How do we apply the divide and conquer principle to this problem?
- The combinations can be obtained by dividing the string into two parts, the left part would be embraced by a pair of parenthesis, and then appended with the right without parenthesis, we get a combination.
Run through a set of example cases:
HAPPY CASE
Input: n = 3
Output: ["((()))","(()())","(())()","()(())","()()()"]
Input: n = 1
Output: ["()"]
EDGE CASE
Input: n = 0
Output: []
2: M-atch
Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.
- Can we generate only permutations which we know for sure will be valid? Backtracking can be used and it should reduce the time considerably. Given n number of left parentheses and right parentheses, try generate all possible combinations. If the combination turns to be not valid, backtracking can discard it and then try another combination. Let’s consider the base case when the number of opening and closing parentheses is equal to n, the number of opening parentheses should be less than n, and a closing parenthesis cannot occur before the open parenthesis.
3: P-lan
Plan the solution with appropriate visualizations and pseudocode.
General Idea: Instead of adding ‘(’ or ‘)’ every time, let’s only add them when we know it will remain a valid sequence. We can do this by keeping track of the number of opening and closing brackets we have placed so far.
1) create a list that will store the result
2) call our backtracking function with empty string and initial number of opening and closing parentheses
3) check the base case. If number of opening and closing parentheses are equal to n then we will add the string to the list and return.
4) if the base case does not meet then we will check if number of opening parentheses is less than n, If true, then we will add ( to the current string and increment the count of opening parenthesis.
5) check if number of closing parentheses is less than open parentheses then we will add ) to the current string and increment the count of closing parentheses.
⚠️ Common Mistakes
- What are some common pitfalls students might have when implementing this solution?
From the problem’s description, recursion seems like the way to go. However, this naive approach is to generate all the permutations. All sequence of length n is ( plus all sequences of length n - 1. The time complexity of this will be O(2 to the nth power) which is quite large.
4: I-mplement
Implement the code to solve the algorithm.
class Solution:
def generateParenthesis(self, n):
"""
:type n: int
:rtype: List[str]
"""
OPEN = '('
CLOSE = ')'
def gen(combo, open_count, close_count, n, res):
# we have generated all open and close parens, add combo to result
if open_count == close_count == n:
# when n == 0
if combo != '':
res.append(combo)
# used all opens, must close
elif open_count == n:
gen(combo + CLOSE, open_count, close_count + 1, n, res)
# current pairs all closed, must open before another close
elif open_count == close_count:
gen(combo + OPEN, open_count + 1, close_count, n, res)
# otherwise, free to open or close
else:
gen(combo + OPEN, open_count + 1, close_count, n, res)
gen(combo + CLOSE, open_count, close_count + 1, n, res)
res = []
gen('', 0, 0, n, res)
return res
class Solution {
public List<String> generateParenthesis(int n) {
List<String> result = new ArrayList<String>();
generateParenthesis(n, n, result, "");
return result;
}
private void generateParenthesis(int left, int right, List<String> result,
String currString){
// left and right are the counts of "(" and ")" still available to place
if(left == 0 && right == 0){
result.add(currString);
return;
}
if(left > 0){
// we open a bracket
generateParenthesis(left - 1, right, result, currString + "(");
}
if(right > left){
// more "(" have been placed than ")", so we may close a bracket
generateParenthesis(left, right - 1, result, currString + ")");
}
}
}
5: R-eview
Review the code by running specific example(s) and recording values (watchlist) of your code’s variables along the way.
- Trace through your code with an input to check for the expected output
- Catch possible edge cases and off-by-one errors
6: E-valuate
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
The complexity analysis based on how many elements there are in generateParenthesis(n).
- Time Complexity: This is the n-th Catalan number, where the first Catalan numbers for n = 0, 1, 2, 3, … are 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786,…
- Space Complexity: n-th Catalan Number