#22

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Generate Parentheses

Generate Parentheses requires generating all valid combinations of parentheses with given pairs using backtracking and state transition.

string dynamic programming backtracking

Problem Statement

Given an integer n, representing the number of pairs of parentheses, the task is to generate all combinations of well-formed parentheses. Each combination must use exactly n pairs of parentheses, forming a valid sequence where every opening parenthesis has a corresponding closing parenthesis. The order of parentheses in the output doesn't matter, but each combination must be unique.

For example, when n equals 3, valid combinations include '((()))', '(()())', '(())()', '()(())', and '()()()'. The goal is to explore the number of combinations possible with a given n and ensure that the parentheses are well-formed and valid in each sequence generated.

Examples

Example 1

Input: n = 3

Output: ["((()))","(()())","(())()","()(())","()()()"]

Example details omitted.

Example 2

Input: n = 1

Output: ["()"]

Example details omitted.

Constraints

1 <= n <= 8

Solution Approach

Backtracking

Backtracking is the primary approach to solve this problem. By maintaining a count of open and close parentheses, we can incrementally build valid combinations. At each step, we either add an open or a close parenthesis. We stop the recursive call when we've used n open and n close parentheses. This approach avoids invalid combinations and ensures that we generate all possible valid sequences.

State Transition Dynamic Programming

Using dynamic programming, we can optimize the backtracking approach. At each state, we transition from one configuration of parentheses to another by adding either an open or a closing parenthesis. Memoization or caching of intermediate results helps in avoiding redundant calculations and speeds up the generation process. This technique is beneficial for solving subproblems of generating parentheses efficiently.

Efficient Backtracking with Pruning

When building a valid combination, pruning helps to improve efficiency. If the number of closing parentheses exceeds the number of opening parentheses at any point, we stop the recursion for that branch. This ensures we avoid invalid sequences early on and reduces unnecessary computations, keeping the solution efficient while generating all valid parentheses combinations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity for generating all valid parentheses combinations is O(4^n / sqrt(n)), which is derived from Catalan numbers. The space complexity is O(n), where n is the depth of recursion or the number of parentheses. Memoization or dynamic programming optimizes space by reducing redundant calculations.

What Interviewers Usually Probe

Do you understand how to use backtracking for generating all combinations of valid parentheses?
Can you explain how state transition dynamic programming can optimize the generation of valid parentheses combinations?
Will you be able to discuss pruning techniques and their impact on performance when solving this problem?

Common Pitfalls or Variants

Common pitfalls

Failing to properly balance the number of opening and closing parentheses during backtracking leads to invalid sequences.
Not pruning invalid sequences early in the recursion can lead to unnecessary calls and performance issues.
Using an inefficient approach without dynamic programming or caching can result in excessive computation time.

Follow-up variants

Generate Parentheses for k pairs instead of n.
Generate Parentheses with additional constraints such as balanced and non-repeating sequences.
Generate Parentheses but with nesting depth restrictions.

FAQ

What is the time complexity of the Generate Parentheses problem?

The time complexity is O(4^n / sqrt(n)), derived from Catalan numbers, representing the number of valid parentheses sequences for n pairs.

How does backtracking solve the Generate Parentheses problem?

Backtracking incrementally builds valid combinations of parentheses by adding open and close parentheses and ensuring the sequence is valid at each step.

What is state transition dynamic programming in the context of this problem?

State transition dynamic programming optimizes backtracking by caching intermediate results, reducing redundant calculations, and transitioning between valid states of parentheses.

How does pruning improve the backtracking solution for this problem?

Pruning avoids unnecessary recursive calls by stopping when a sequence exceeds valid parentheses constraints, thus improving the overall efficiency of the solution.

What are common pitfalls in solving the Generate Parentheses problem?

Common pitfalls include not properly balancing parentheses, failing to prune invalid sequences, and using inefficient solutions without dynamic programming or memoization.

terminal

Solution

Solution 1: DFS + Pruning

The range of $n$ in the problem is $[1, 8]$, so we can directly solve this problem through "brute force search + pruning".

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class Solution:
    def generateParenthesis(self, n: int) -> List[str]:
        def dfs(l: int, r: int, t: str):
            if l > n or r > n or l < r:
                return
            if l == n and r == n:
                ans.append(t)
                return
            dfs(l + 1, r, t + "(")
            dfs(l, r + 1, t + ")")

        ans = []
        dfs(0, 0, "")
        return ans

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