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You are given a binary string s (such as “010”). You need to rearrange the characters of the string in such a way that it forms the largest possible odd binary number. A binary number is odd if its last digit is 1.

Clarifying Questions:

  1. Input Constraints:
    • Is the input string always guaranteed to contain at least one ‘1’ (since a binary number must end in ‘1’ to be odd)?
    • What is the maximum length of the binary string?
  2. Output Requirements:
    • Should the output be the rearranged binary string itself or a specific format?

Assuming the input will always contain at least one ‘1’ and the output should be the binary string.

Strategy:

  1. Understanding Odd Binary Number:
    • A binary number is odd if its last digit is ‘1’.
  2. Objective:
    • To form the largest possible binary number with the given characters, while ensuring that the last digit is ‘1’.
  3. Procedure:
    • Count the number of ‘1’s and ‘0’s in the input string.
    • Place all ‘1’s except one (which will be at the end to ensure the number is odd) at the beginning for the largest magnitude.
    • Follow with all the ‘0’s.
    • Place the last ‘1’ at the end.

Code:

def maximumOddBinaryNumber(s: str) -> str:
    # Count the number of '1's and '0's
    ones = s.count('1')
    zeros = s.count('0')
    
    # Form the largest number by putting (ones-1) '1's, then all '0's, and end with one '1'.
    result = '1' * (ones - 1) + '0' * zeros + '1'
    
    return result

# Example Usage
s = "010"
print(maximumOddBinaryNumber(s))  # Output should be "101"

Time Complexity:

Thus, the overall time complexity of the solution is (O(n)).

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