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784. Letter Case Permutation

Description

Given a string s, you can transform every letter individually to be lowercase or uppercase to create another string.

Return a list of all possible strings we could create. Return the output in any order.

 

Example 1:

Input: s = "a1b2"
Output: ["a1b2","a1B2","A1b2","A1B2"]

Example 2:

Input: s = "3z4"
Output: ["3z4","3Z4"]

 

Constraints:

  • 1 <= s.length <= 12
  • s consists of lowercase English letters, uppercase English letters, and digits.

Solutions

Solution 1

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class Solution:
    def letterCasePermutation(self, s: str) -> List[str]:
        def dfs(i):
            if i >= len(s):
                ans.append(''.join(t))
                return
            dfs(i + 1)
            if t[i].isalpha():
                t[i] = chr(ord(t[i]) ^ 32)
                dfs(i + 1)

        t = list(s)
        ans = []
        dfs(0)
        return ans

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class Solution {
    private List<String> ans = new ArrayList<>();
    private char[] t;

    public List<String> letterCasePermutation(String s) {
        t = s.toCharArray();
        dfs(0);
        return ans;
    }

    private void dfs(int i) {
        if (i >= t.length) {
            ans.add(String.valueOf(t));
            return;
        }
        dfs(i + 1);
        if (t[i] >= 'A') {
            t[i] ^= 32;
            dfs(i + 1);
        }
    }
}

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class Solution {
public:
    vector<string> letterCasePermutation(string s) {
        vector<string> ans;
        function<void(int)> dfs = [&](int i) {
            if (i >= s.size()) {
                ans.emplace_back(s);
                return;
            }
            dfs(i + 1);
            if (s[i] >= 'A') {
                s[i] ^= 32;
                dfs(i + 1);
            }
        };
        dfs(0);
        return ans;
    }
};

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func letterCasePermutation(s string) (ans []string) {
    t := []byte(s)
    var dfs func(int)
    dfs = func(i int) {
        if i >= len(t) {
            ans = append(ans, string(t))
            return
        }
        dfs(i + 1)
        if t[i] >= 'A' {
            t[i] ^= 32
            dfs(i + 1)
        }
    }

    dfs(0)
    return ans
}

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function letterCasePermutation(s: string): string[] {
    const n = s.length;
    const cs = [...s];
    const res = [];
    const dfs = (i: number) => {
        if (i === n) {
            res.push(cs.join(''));
            return;
        }
        dfs(i + 1);
        if (cs[i] >= 'A') {
            cs[i] = String.fromCharCode(cs[i].charCodeAt(0) ^ 32);
            dfs(i + 1);
        }
    };
    dfs(0);
    return res;
}

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impl Solution {
    fn dfs(i: usize, cs: &mut Vec<char>, res: &mut Vec<String>) {
        if i == cs.len() {
            res.push(cs.iter().collect()