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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

DFS.

  • 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);
            }
        }
    }
    
  • 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;
        }
    };
    
  • 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
    
    
  • 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
    }
    
  • 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;
    }
    
    
  • 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
        }
    }
    
    
  • 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;
        }
    }
    
  • class Solution {
    public:
        vector<string> letterCasePermutation(string s) {
            int n = count_if(s.begin(), s.end(), [](char c) { return isalpha(c); });
            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;
        }
    };
    
    
  • 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
    
    
  • 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
    }
    
  • function letterCasePermutation(s: string): string[] {
        const ans: string[] = [];
        const n: number = Array.from(s).filter(c => /[a-zA-Z]/.test(c)).length;
        for (let i = 0; i < 1 << n; ++i) {
            let j = 0;
            const t: string[] = [];
            for (let c of s) {
                if (/[a-zA-Z]/.test(c)) {
                    t.push((i >> j) & 1 ? c.toLowerCase() : c.toUpperCase());
                    j++;
                } else {
                    t.push(c);
                }
            }
            ans.push(t.join(''));
        }
        return ans;
    }
    
    
  • impl Solution {
        pub fn letter_case_permutation(s: String) -> Vec<String> {
            let n = s.chars().filter(|&c| c.is_alphabetic()).count();
            let mut ans = Vec::new();
            for i in 0..(1 << n) {
                let mut j = 0;
                let mut t = String::new();
                for c in s.chars() {
                    if c.is_alphabetic() {
                        if (i >> j) & 1 == 1 {
                            t.push(c.to_lowercase().next().unwrap());
                        } else {
                            t.push(c.to_uppercase().next().unwrap());
                        }
                        j += 1;
                    } else {
                        t.push(c);
                    }
                }
                ans.push(t);
            }
            ans
        }
    }
    
    

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