LeetCode 22: Generate Parentheses Solution
Master LeetCode problem 22 (Generate Parentheses), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
22. Generate Parentheses
Problem Explanation
Explanation
To generate all combinations of well-formed parentheses, we can use a backtracking approach. At each step, we have two choices: either add an open parenthesis '(', or add a close parenthesis ')'. We need to ensure that the parentheses are well-formed by keeping track of the number of open and close parentheses used.
We can start with an empty string and add parentheses recursively while keeping track of the number of open and close parentheses we have used. We can backtrack if we reach a state where the number of open parentheses is greater than n or the number of close parentheses is greater than the number of open parentheses.
The time complexity of this approach is O(4^n / sqrt(n)), where n is the number of pairs of parentheses. The space complexity is O(n) to store the current combination of parentheses.
Solution Code
class Solution {
public List<String> generateParenthesis(int n) {
List<String> result = new ArrayList<>();
generateParenthesisHelper(result, "", 0, 0, n);
return result;
}
private void generateParenthesisHelper(List<String> result, String current, int open, int close, int max) {
if (current.length() == max * 2) {
result.add(current);
return;
}
if (open < max) {
generateParenthesisHelper(result, current + "(", open + 1, close, max);
}
if (close < open) {
generateParenthesisHelper(result, current + ")", open, close + 1, max);
}
}
}Try It Yourself
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