#1507

Easy

auto_awesomeString-driven solution strategy

LeetCode Problem Workspace

Reformat Date

Reformat Date requires transforming a date string from human-readable form to YYYY-MM-DD using a string-driven strategy efficiently.

string

Problem Statement

You are given a date string in the format "Day Month Year" where Day is 1st to 31st with suffixes, Month is a three-letter abbreviation, and Year is four digits. Your task is to convert this string into the format YYYY-MM-DD, standardizing day and month with leading zeros where needed.

For example, given date = "20th Oct 2052", the output should be "2052-10-20". Similarly, date = "6th Jun 1933" should return "1933-06-06". All input dates are valid and do not require error handling.

Examples

Example 1

Input: date = "20th Oct 2052"

Output: "2052-10-20"

Example details omitted.

Example 2

Input: date = "6th Jun 1933"

Output: "1933-06-06"

Example details omitted.

Example 3

Input: date = "26th May 1960"

Output: "1960-05-26"

Example details omitted.

Constraints

The given dates are guaranteed to be valid, so no error handling is necessary.

Solution Approach

Split and Normalize Components

Separate the input string into day, month, and year. Remove ordinal suffixes like 'st', 'nd', 'rd', 'th' from the day and ensure it is two digits by padding with zero if necessary.

Map Month Abbreviations to Numbers

Use a dictionary mapping from 'Jan' to '01', 'Feb' to '02', and so on, to convert the three-letter month to its numeric representation, handling all twelve months accurately.

Combine into Standard Format

Concatenate the year, month, and normalized day into the final YYYY-MM-DD string. Return this string directly, ensuring consistent formatting for all inputs.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(1) because the input size is fixed and splitting plus mapping operations are constant. Space complexity is O(1) as only a few variables and a fixed-size dictionary are used.

What Interviewers Usually Probe

Look for explicit handling of day suffixes and zero-padding in the solution.
Verify that month abbreviation mapping covers all months and handles case consistently.
Expect a simple string-based solution without relying on built-in date libraries.

Common Pitfalls or Variants

Common pitfalls

Forgetting to remove day suffixes before padding, leading to incorrect day formatting.
Incorrect mapping of month abbreviations or typos in the dictionary.
Failing to pad single-digit months or days, resulting in invalid YYYY-MM-DD format.

Follow-up variants

Input dates could include varying day suffixes or whitespace requiring careful trimming.
Month names could appear in full form rather than three-letter abbreviations.
Extensions could involve converting multiple date strings in a list efficiently.

FAQ

What is the recommended approach to remove day suffixes in Reformat Date?

Split the day from the string and strip suffixes like 'st', 'nd', 'rd', 'th' before padding it to two digits.

How do I convert month abbreviations to numbers efficiently?

Use a fixed dictionary mapping each three-letter month abbreviation to its corresponding two-digit number.

Is it necessary to handle invalid dates in this problem?

No, all input dates are guaranteed valid, so no additional error checking is required.

Can I solve this without using any date libraries?

Yes, a string-driven approach is sufficient and preferred to convert day, month, and year into the correct format.

Why is zero-padding important in Reformat Date?

Zero-padding ensures single-digit days and months comply with the YYYY-MM-DD standard, preventing format errors.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def reformatDate(self, date: str) -> str:
        s = date.split()
        s.reverse()
        months = " JanFebMarAprMayJunJulAugSepOctNovDec"
        s[1] = str(months.index(s[1]) // 3 + 1).zfill(2)
        s[2] = s[2][:-2].zfill(2)
        return "-".join(s)

Continue Practicing

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