How do you solve Maximum 69 Number on LeetCode?

Maximum 69 Number (LeetCode #1323) 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: Changing the first '6' to '9' maximizes the number.

What is the brute force solution for Maximum 69 Number?

The brute force approach for Maximum 69 Number is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every digit in the number. For each digit, we change it from 6 to 9 or from 9 to 6, and then we compare the new number to find the maximum possible value. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum 69 Number?

Maximum 69 Number uses the following concepts: Math, Greedy. The recommended patterns to study are: Greedy, String Manipulation.

What are the interview tips for Maximum 69 Number?

Always consider the simplest solution first. Think about how changing a digit affects the overall number. Practice converting between data types (e.g., int to string) efficiently.

What are common mistakes when solving Maximum 69 Number?

Not stopping after the first change in the optimal solution. Overthinking the problem and trying to change multiple digits.

#1323

Maximum 69 Number

Easy

MathGreedyGreedyString 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 we only need to change the first occurrence of '6' to '9' to maximize the number. This is because changing a '6' to '9' at a higher place value increases the overall number more than changing a '9' to '6'.

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Algorithm

3 steps

1Step 1: Convert the number to a string.
2Step 2: Find the first occurrence of '6'.
3Step 3: Change that '6' to '9' and return the new number.

solution.py7 lines

1def maximum69Number (num):
2    num_str = list(str(num))
3    for i in range(len(num_str)):
4        if num_str[i] == '6':
5            num_str[i] = '9'
6            break
7    return int(''.join(num_str))

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once to find the first '6'. The space complexity is O(n) due to the string conversion.

1Changing the first '6' to '9' maximizes the number.
2Only one change is needed to achieve the maximum.

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