Reformat Date - Practice Coding Problems

Reformat Date - Problem

String Easy

Date String Reformatter

You're working on a legacy data migration project where you need to convert human-readable dates into standardized format for database storage.

Given a date string in the friendly format "Day Month Year", convert it to the ISO standard format "YYYY-MM-DD".

Input Format:
• Day: Ordinal numbers like "1st", "2nd", "3rd", "4th", ..., "30th", "31st"
• Month: Three-letter abbreviations "Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"
• Year: Four-digit year between 1900 and 2100

Output Format:
• YYYY-MM-DD where each component is zero-padded to required length

Example: "20th Oct 2052" → "2052-10-20"

Input & Output

example_1.py — Standard Date

$ Input: date = "20th Oct 2052"

› Output: "2052-10-20"

💡 Note: Day '20th' becomes '20' (remove 'th' suffix), 'Oct' maps to '10', year '2052' stays the same. Format as YYYY-MM-DD with zero-padding.

example_2.py — Single Digit Day

$ Input: date = "6th Jun 1933"

› Output: "1933-06-06"

💡 Note: Day '6th' becomes '6', then zero-padded to '06'. 'Jun' maps to '06', resulting in '1933-06-06'.

example_3.py — Different Ordinal Suffix

$ Input: date = "22nd May 2025"

› Output: "2025-05-22"

💡 Note: Day '22nd' becomes '22' (remove 'nd' suffix), 'May' maps to '05'. All ordinal suffixes (st, nd, rd, th) are handled the same way.

Constraints

The given dates are guaranteed to be valid
Day is in the set {"1st", "2nd", "3rd", "4th", ..., "30th", "31st"}
Month is in the set {"Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"}
Year is in the range [1900, 2100]
Input format is always "Day Month Year" with spaces as separators

Visualization

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Understanding the Visualization

Parse Components

Split the friendly date into its three parts: day, month, year

Clean Day

Remove the ordinal suffix (st/nd/rd/th) to get the numeric day

Convert Month

Use hash map lookup to convert 3-letter month to 2-digit number

Format Output

Apply zero-padding and combine in YYYY-MM-DD format

Key Takeaway

🎯 Key Insight: Hash maps provide O(1) lookup time, making month conversion instant and eliminating the need for lengthy conditional chains.

Asked in

G Google 45 a Amazon 38 f Meta 28 ⊞ Microsoft 25

The optimal solution uses a hash map to store month abbreviations for O(1) lookup time. Simply split the input string, remove the ordinal suffix from the day using string slicing, lookup the month number, and format with zero-padding. This approach is clean, efficient, and maintainable compared to lengthy if-else chains.

Common Approaches

Approach	Time	Space	Notes
✓ Hash Map Lookup (Optimal)	O(1)	O(1)	Use hash map for instant month conversion and simple string operations
Brute Force (Multiple Conditionals)	O(1)	O(1)	Use long if-else chains to parse and convert each component

Hash Map Lookup (Optimal) — Algorithm Steps

Create a hash map mapping month abbreviations to their numeric strings
Split the input string into day, month, and year components
Remove ordinal suffix from day using string slicing (last 2 characters)
Look up month number using hash map in O(1) time
Format day and month with zero-padding and concatenate result

Visualization

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Step-by-Step Walkthrough

Initialize Hash Map

Create month mapping: {'Jan':'01', 'Feb':'02', ...}

Split and Parse

Split '20th Oct 2052' and remove '20th'[:-2] = '20'

Hash Lookup

O(1) lookup: months['Oct'] = '10'

Format Result

Combine with padding: '2052-10-20'

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

char* getMonthNum(char* month) {
    if (strcmp(month, "Jan") == 0) return "01";
    if (strcmp(month, "Feb") == 0) return "02";
    if (strcmp(month, "Mar") == 0) return "03";
    if (strcmp(month, "Apr") == 0) return "04";
    if (strcmp(month, "May") == 0) return "05";
    if (strcmp(month, "Jun") == 0) return "06";
    if (strcmp(month, "Jul") == 0) return "07";
    if (strcmp(month, "Aug") == 0) return "08";
    if (strcmp(month, "Sep") == 0) return "09";
    if (strcmp(month, "Oct") == 0) return "10";
    if (strcmp(month, "Nov") == 0) return "11";
    return "12";
}

char* reformatDate(char* date) {
    char* result = (char*)malloc(11 * sizeof(char));
    char day[4], month[4], year[5];
    
    sscanf(date, "%s %s %s", day, month, year);
    
    // Remove ordinal suffix from day (last 2 characters)
    int dayLen = strlen(day);
    day[dayLen-2] = '\0';
    
    // Get month number using lookup function
    char* monthNum = getMonthNum(month);
    
    // Format result with zero-padding
    sprintf(result, "%s-%s-%02d", year, monthNum, atoi(day));
    return result;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Constant time operations: string split, hash map lookup, string formatting

✓ Linear Growth

Space Complexity

O(1)

Fixed-size hash map with 12 entries, constant space regardless of input

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Hash Map Lookup (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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