How do you solve Maximum Odd Binary Number on LeetCode?

Maximum Odd Binary Number (LeetCode #2864) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: To form the maximum odd binary number, the last digit must be '1'.

What is the brute force solution for Maximum Odd Binary Number?

The brute force approach for Maximum Odd Binary Number is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible permutations of the binary string and checking which one forms the maximum odd binary number. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Odd Binary Number?

Maximum Odd Binary Number uses the following concepts: Math, String, Greedy. The recommended patterns to study are: Greedy, String Manipulation.

What are the interview tips for Maximum Odd Binary Number?

Always clarify the requirements, especially about odd/even conditions. Think about edge cases, such as strings with only one '1'. Practice counting and arranging elements efficiently.

What are common mistakes when solving Maximum Odd Binary Number?

Not ensuring the last digit is '1' when forming the number. Forgetting to count '1's and '0's correctly.

#2864

Maximum Odd Binary Number

Easy

MathStringGreedyGreedyString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages the fact that to create the largest odd binary number, we should place all '1's at the front, except for one '1' which must be at the end to ensure the number is odd.

⚙️

Algorithm

3 steps

1Step 1: Count the number of '1's and '0's in the string.
2Step 2: If there's at least one '1', place all '1's except one at the front.
3Step 3: Place one '1' at the end and fill the rest with '0's.

solution.py4 lines

1def maxOddBinary(s):
2    count1 = s.count('1')
3    count0 = s.count('0')
4    return '1' * (count1 - 1) + '0' * count0 + '1' if count1 > 0 else ''

ℹ

Complexity note: The time complexity is O(n) because we traverse the string once to count '1's and '0's. The space complexity is O(n) due to the construction of the result string.

1To form the maximum odd binary number, the last digit must be '1'.
2The arrangement of '1's and '0's affects the overall value of the binary number.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.