algoadvance

Given an alphanumeric string s, return the string after reformating it so that no two adjacent characters are of the same type. Assume that to reformat the string, you may only use the string’s existing characters. If it’s impossible to reformat the string, return an empty string.

Constraints:

Example:

Clarifying Questions

  1. Can the input string contain spaces or special characters?
    • No, the input only consists of lowercase English letters and digits.
  2. What should be done in case the string can’t be reformatted?
    • Return an empty string.
  3. Is there a preferred order for letters and digits in the result?
    • No specific order is required as long as no two adjacent characters are of the same type. However, a consistent alternating pattern is needed.

Strategy

  1. Separate Characters:
    • Traverse the string and separate letters and digits into two lists.
  2. Check Possibility:
    • If the absolute difference between the counts of the two lists is greater than 1, return an empty string because it’s impossible to reformat.
  3. Merging:
    • Start merging the characters by alternating between letters and digits, starting with the one with the higher count, to ensure a valid alternating pattern.
  4. Edge Cases:
    • If the input string has only one character, it’s already reformatted.

Code

def reformat(s: str) -> str:
    letters = []
    digits = []
    
    # Separate characters
    for char in s:
        if char.isdigit():
            digits.append(char)
        else:
            letters.append(char)
    
    # Check for the possibility of reformat
    if abs(len(letters) - len(digits)) > 1:
        return ""
    
    # Decide the order of alternating based on count
    if len(letters) > len(digits):
        longer, shorter = letters, digits
    else:
        longer, shorter = digits, letters
    
    result = []
    
    # Merge the characters
    for i in range(len(shorter)):
        result.append(longer[i])
        result.append(shorter[i])
        
    if len(longer) > len(shorter):
        result.append(longer[-1])
    
    return "".join(result)

# Example usage
print(reformat("a0b1c2"))  # Output: "0a1b2c" or "a0b1c2" etc.

Time Complexity

Overall, the solution runs in linear time, O(n), because both the operations of separating and merging are linear in nature. The space complexity is also O(n) due to the storage of characters in separate lists.

Try our interview co-pilot at AlgoAdvance.com