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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());
            return;
        }
        Self::dfs(i + 1, cs, res);
        if cs[i] >= 'A' {
            cs[i] = char::from((cs[i] as u8) ^ 32);
            Self::dfs(i + 1, cs, res);
        }
    }

    pub fn letter_case_permutation(s: String) -> Vec<String> {
        let mut res = Vec::new();
        let mut cs = s.chars().collect::<Vec<char>>();
        Self::dfs(0, &mut cs, &mut res);
        res
    }
}

Solution 2

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class Solution:
    def letterCasePermutation(self, s: str) -> List[str]:
        ans = []
        n = sum(c.isalpha() for c in s)
        for i in range(1 << n):
            j, t = 0, []
            for c in s:
                if c.isalpha():
                    c = c.lower() if (i >> j) & 1 else c.upper()
                    j += 1
                t.append(c)
            ans.append(''.join(t))
        return ans

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class Solution {
    public List<String> letterCasePermutation(String s) {
        int n = 0;
        for (int i = 0; i < s.length(); ++i) {
            if (s.charAt(i) >= 'A') {
                ++n;
            }
        }
        List<String> ans = new ArrayList<>();
        for (int i = 0; i < 1 << n; ++i) {
            int j = 0;
            StringBuilder t = new StringBuilder();
            for (int k = 0; k < s.length(); ++k) {
                char x = s.charAt(k);
                if (x >= 'A') {
                    x = ((i >> j) & 1) == 1 ? Character.toLowerCase(x) : Character.toUpperCase(x);
                    ++j;
                }
                t.append(x);
            }
            ans.add(t.toString());
        }
        return ans;
    }
}

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class Solution {
public:
    vector<string> letterCasePermutation(string s) {
        int n = 0;
        for (char c : s)
            if (isalpha(c)) ++n;
        vector<string> ans;
        for (int i = 0; i < 1 << n; ++i) {
            int j = 0;
            string t;
            for (char c : s) {
                if (isalpha(c)) {
                    c = (i >> j & 1) ? tolower(c) : toupper(c);
                    ++j;
                }
                t += c;
            }
            ans.emplace_back(t);
        }
        return ans;
    }
};

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func letterCasePermutation(s string) (ans []string) {
    n := 0
    for _, c := range s {
        if c >= 'A' {
            n++
        }
    }
    for i := 0; i < 1<<n; i++ {
        j := 0
        t := []rune{}
        for _, c := range s {
            if c >= 'A' {
                if ((i >> j) & 1) == 1 {
                    c = unicode.ToLower(c)
                } else {
                    c = unicode.ToUpper(c)
                }
                j++
            }
            t = append(t, c)
        }
        ans = append(ans, string(t))
    }
    return ans
}

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