22. Generate Parentheses

LeetCode
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Recursion (My Solution)

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class Solution {
    public List<String> generateParenthesis(int n) {
        List<String> result = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            result = addParenthesis(result);
        }

        return result;
        
    }

    private List<String> addParenthesis(List<String> s) {
        if (s.isEmpty()) {
            return new ArrayList<>(Arrays.asList("()"));
        }

        Set<String> newSet = new HashSet<>();

        for (String current : s) {
            for (int i = 0; i <= current.length(); i++) {
                String newStr = current.substring(0, i) + "()" + current.substring(i);
                newSet.add(newStr);
            }
        }

        return new ArrayList<>(newSet);
    }
}

Remarks:

  1. TC $O(n!\times n)$
  2. Add new () to each gaps; will have duplication, use Set to remove.

Traceback

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class Solution {
    public List<String> generateParenthesis(int n) {
        List<String> result = new ArrayList<>();
        backtrack(result, "", 0, 0, n);
        return result;
    }

    private void backtrack(List<String> result, String current, int open, int close, int max) {
        if (current.length() == max * 2) {
            result.add(current);
            return;
        }

        if (open < max) {
            backtrack(result, current + "(", open + 1, close, max);
        }
        if (close < open) {
            backtrack(result, current + ")", open, close + 1, max);
        }
    }
}

Remarks:

  1. TC $O(\cfrac{4^n}{\sqrt{n}})$
  2. From left to right, first add ( , then add )
  3. ( -> (( -> (() -> (()) -> ( -> ()()