Problem Statement
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.
- Note:
- Digits in the string should remain unchanged.
- The solution should handle both uppercase and lowercase transformations of letters.
Example Inputs
Example 1:
Problem Solution
Approach
To generate all possible letter case permutations of the given string s, we can use a recursive backtracking approach. The idea is to explore all possible combinations by deciding for each character whether to keep it as is (if it's a digit) or toggle its case (if it's a letter).
Steps
-
Initialize:
- Create an empty array
ans to store all valid permutations.
- Start with an empty string
output and begin processing from the first character.
-
Recursive Backtracking Function (backtrack):
-
Parameters:
idx: Current index in the string s.output: The current string being built.
-
Base Case:
- If
idx equals the length of s, it means we have processed all characters. Add output to ans.
-
Recursive Case:
- If the current character is a digit:
- Append it to
output and recurse for the next index.
- If the current character is a letter:
- Choose lowercase:
- Append the lowercase version of the character to
output and recurse for the next index.
- Remove the last character to backtrack.
- Choose uppercase:
- Append the uppercase version of the character to
output and recurse for the next index.
- Remove the last character to backtrack.
-
Start Backtracking:
- Call the
backtrack function starting from index 0 with an empty output string.
-
Return the Result:
- After exploring all possibilities, return the
ans array containing all valid permutations.
Time & Space Complexity
- Time Complexity: O(2^n), where
n is the number of alphabetic characters in s. Each letter has two choices: lowercase or uppercase.
- Space Complexity: O(n), which is the space required for the recursion stack and the
output string.
Code Snippet
Dry Run
Let's perform a dry run with s = "a1B".
-
Initial Call: backtrack(0, [])
-
First Level (idx = 0):
- Character: 'a' (letter)
- Choose lowercase:
current = ['a']- Recurse:
backtrack(1, ['a'])
-
Second Level (idx = 1):
- Character: '1' (digit)
- Add '1':
current = ['a', '1']- Recurse:
backtrack(2, ['a', '1'])
-
Third Level (idx = 2):
- Character: 'B' (letter)
- Choose lowercase:
current = ['a', '1', 'b']- Recurse:
backtrack(3, ['a', '1', 'b'])
-
Fourth Level (idx = 3):
- End of string reached.
- Add
"a1b" to ans.
- Backtrack to
current = ['a', '1']
-
Third Level (idx = 2):
- Choose uppercase:
current = ['a', '1', 'B']- Recurse:
backtrack(3, ['a', '1', 'B'])
-
Fourth Level (idx = 3):
- End of string reached.
- Add
"a1B" to ans.
- Backtrack to
current = ['a', '1']
-
Back to Second Level (idx = 1):
- Backtrack to
current = ['a']
-
First Level (idx = 0):
- Choose uppercase:
current = ['A']- Recurse:
backtrack(1, ['A'])
-
Second Level (idx = 1):
- Character: '1' (digit)
- Add '1':
current = ['A', '1']- Recurse:
backtrack(2, ['A', '1'])
-
Third Level (idx = 2):
- Character: 'B' (letter)
- Choose lowercase:
current = ['A', '1', 'b']- Recurse:
backtrack(3, ['A', '1', 'b'])
-
Fourth Level (idx = 3):
- End of string reached.
- Add
"A1b" to ans.
- Backtrack to
current = ['A', '1']
-
Third Level (idx = 2):
- Choose uppercase:
current = ['A', '1', 'B']- Recurse:
backtrack(3, ['A', '1', 'B'])
-
Fourth Level (idx = 3):
- End of string reached.
- Add
"A1B" to ans.
- Backtrack to
current = ['A', '1']
-
Back to Second Level (idx = 1):
- Backtrack to
current = ['A']
-
Final Result: ["a1b", "a1B", "A1b", "A1B"]
Complexity Analysis
Optimized Time and Space
-
Time Complexity: O(2^n), where n is the number of alphabetic characters in the string s. Each letter has two choices: lowercase or uppercase.
-
Space Complexity: O(n), where n is the length of the string s. This space is used by the recursion stack and the current string being built.
Optimizations:
-
Pruning Paths: Since digits do not have case variations, we only branch when encountering letters. This reduces the number of recursive calls.
-
Efficient String Handling: Using an array current to build the string and joining it only when adding to ans can be more efficient than string concatenation in some programming languages. However, in JavaScript, string concatenation is generally efficient for small to medium-sized strings.
Conclusion
The Letter Case Permutation problem is a classic example of using recursive backtracking to explore all possible combinations that meet specific criteria—in this case, transforming letters to either lowercase or uppercase while keeping digits unchanged. By systematically deciding for each character whether to toggle its case or leave it as is, we can efficiently generate all valid permutations.
Understanding this approach not only aids in solving similar combinatorial problems but also reinforces fundamental concepts in recursion, backtracking, and algorithm optimization. Implementing the solution in a functional style without using classes makes the code more concise and easier to understand, especially for beginners. By carefully managing the state and making strategic choices at each step, complex problems become manageable and solvable with clarity and confidence.