LeetCode #1507 — EASY

Reformat Date

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

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The Problem

Problem Statement

Given a date string in the form Day Month Year, where:

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].

Convert the date string to the format YYYY-MM-DD, where:

YYYY denotes the 4 digit year.
MM denotes the 2 digit month.
DD denotes the 2 digit day.

Example 1:

Input: date = "20th Oct 2052"
Output: "2052-10-20"

Example 2:

Input: date = "6th Jun 1933"
Output: "1933-06-06"

Example 3:

Input: date = "26th May 1960"
Output: "1960-05-26"

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: Given a date string in the form Day Month Year, where: 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]. Convert the date string to the format YYYY-MM-DD, where: YYYY denotes the 4 digit year. MM denotes the 2 digit month. DD denotes the 2 digit day.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"20th Oct 2052"

Example 2

"6th Jun 1933"

Example 3

"26th May 1960"

Step 02

Core Insight

What unlocks the optimal approach

Handle the conversions of day, month and year separately.
Notice that days always have a two-word ending, so if you erase the last two characters of this days you'll get the number.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1507: Reformat Date
class Solution {
    public String reformatDate(String date) {
        var s = date.split(" ");
        String months = " JanFebMarAprMayJunJulAugSepOctNovDec";
        int day = Integer.parseInt(s[0].substring(0, s[0].length() - 2));
        int month = months.indexOf(s[1]) / 3 + 1;
        return String.format("%s-%02d-%02d", s[2], month, day);
    }
}

// Accepted solution for LeetCode #1507: Reformat Date
func reformatDate(date string) string {
	s := strings.Split(date, " ")
	day, _ := strconv.Atoi(s[0][:len(s[0])-2])
	months := " JanFebMarAprMayJunJulAugSepOctNovDec"
	month := strings.Index(months, s[1])/3 + 1
	year, _ := strconv.Atoi(s[2])
	return fmt.Sprintf("%d-%02d-%02d", year, month, day)
}

# Accepted solution for LeetCode #1507: Reformat Date
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)

// Accepted solution for LeetCode #1507: Reformat Date
struct Solution;
use std::collections::HashMap;

impl Solution {
    fn reformat_date(date: String) -> String {
        let month: HashMap<String, i32> = vec![
            ("Jan", 1),
            ("Feb", 2),
            ("Mar", 3),
            ("Apr", 4),
            ("May", 5),
            ("Jun", 6),
            ("Jul", 7),
            ("Aug", 8),
            ("Sep", 9),
            ("Oct", 10),
            ("Nov", 11),
            ("Dec", 12),
        ]
        .into_iter()
        .map(|(m, mm)| (m.to_string(), mm))
        .collect();
        let mut v: Vec<String> = date.split_whitespace().map(|s| s.to_string()).collect();
        let yyyy = v.pop().unwrap();
        let mm = month[&v.pop().unwrap()];
        let mut dd = v.pop().unwrap();
        dd.pop();
        dd.pop();
        let dd = dd.parse::<i32>().unwrap();
        format!("{}-{:02}-{:02}", yyyy, mm, dd)
    }
}

#[test]
fn test() {
    let date = "20th Oct 2052".to_string();
    let res = "2052-10-20".to_string();
    assert_eq!(Solution::reformat_date(date), res);
    let date = "6th Jun 1933".to_string();
    let res = "1933-06-06".to_string();
    assert_eq!(Solution::reformat_date(date), res);
    let date = "26th May 1960".to_string();
    let res = "1960-05-26".to_string();
    assert_eq!(Solution::reformat_date(date), res);
}

// Accepted solution for LeetCode #1507: Reformat Date
function reformatDate(date: string): string {
    const s = date.split(' ');
    const months = ' JanFebMarAprMayJunJulAugSepOctNovDec';
    const day = parseInt(s[0].substring(0, s[0].length - 2));
    const month = Math.floor(months.indexOf(s[1]) / 3) + 1;
    return `${s[2]}-${month.toString().padStart(2, '0')}-${day.toString().padStart(2, '0')}`;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.