How do you solve Reformat Date on LeetCode?

Reformat Date (LeetCode #1507) can be solved using Brute Force. The optimal approach is Brute Force with undefined time complexity and undefined space complexity. Key insight: The brute force approach involves breaking down the date string into its components: day, month, and year. We then manually convert each component into the desired format, which can be straightforward but inefficient.

What is the time complexity of Reformat Date?

The optimal solution for Reformat Date runs in undefined time and uses undefined space. undefined

What algorithm or data structure is used to solve Reformat Date?

Reformat Date uses the following concepts: String. The recommended patterns to study are: .

#1507

Reformat Date

Easy

String

LeetCode ↗

Approaches

💡

Intuition

Time Space

The brute force approach involves breaking down the date string into its components: day, month, and year. We then manually convert each component into the desired format, which can be straightforward but inefficient.

⚙️

Algorithm

4 steps

1Step 1: Split the input date string into day, month, and year.
2Step 2: Remove the last two characters from the day to get the numeric value.
3Step 3: Map the month abbreviation to its corresponding numeric value.
4Step 4: Format the year, month, and day into 'YYYY-MM-DD'.

solution.py6 lines

1def reformatDate(date):
2    month_map = {'Jan': '01', 'Feb': '02', 'Mar': '03', 'Apr': '04', 'May': '05', 'Jun': '06', 'Jul': '07', 'Aug': '08', 'Sep': '09', 'Oct': '10', 'Nov': '11', 'Dec': '12'}
3    day, month, year = date.split()
4    day = day[:-2].zfill(2)
5    month = month_map[month]
6    return f'{year}-{month}-{day}'

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