LeetCode #3798 — EASY

Largest Even Number

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

The Problem

Problem Statement

You are given a string s consisting only of the characters '1' and '2'.

You may delete any number of characters from s without changing the order of the remaining characters.

Return the largest possible resultant string that represents an even integer. If there is no such string, return the empty string "".

Example 1:

Input: s = "1112"

Output: "1112"

Explanation:

The string already represents the largest possible even number, so no deletions are needed.

Example 2:

Input: s = "221"

Output: "22"

Explanation:

Deleting '1' results in the largest possible even number which is equal to 22.

Example 3:

Input: s = "1"

Output: ""

Explanation:

There is no way to get an even number.

Constraints:

1 <= s.length <= 100
s consists only of the characters '1' and '2'.

Step 01

Brute Force Baseline

Problem summary: You are given a string s consisting only of the characters '1' and '2'. You may delete any number of characters from s without changing the order of the remaining characters. Return the largest possible resultant string that represents an even integer. If there is no such string, return the empty string "".

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"1112"

Example 2

"221"

Example 3

"1"

Step 02

Core Insight

What unlocks the optimal approach

A number ending with <code>'1'</code> is odd.
Find the last <code>'2'</code> in <code>s</code>. If none, return <code>""</code>. Otherwise return the prefix up to and including that <code>'2'</code>.

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 #3798: Largest Even Number
class Solution {
    public String largestEven(String s) {
        int i = s.length();
        while (i > 0 && s.charAt(i - 1) == '1') {
            i--;
        }
        return s.substring(0, i);
    }
}

// Accepted solution for LeetCode #3798: Largest Even Number
func largestEven(s string) string {
	return strings.TrimRight(s, "1")
}

# Accepted solution for LeetCode #3798: Largest Even Number
class Solution:
    def largestEven(self, s: str) -> str:
        return s.rstrip("1")

// Accepted solution for LeetCode #3798: Largest Even Number
fn largest_even(s: String) -> String {
    let cs: Vec<_> = s.chars().collect();
    if let Some(p) = cs.iter().rposition(|c| *c == '2') {
        cs.into_iter().take(p + 1).collect()
    } else {
        String::new()
    }
}

fn main() {
    let ret = largest_even("1112".to_string());
    println!("ret={ret}");
}

#[test]
fn test() {
    assert_eq!(largest_even("1112".to_string()), "1112");
    assert_eq!(largest_even("221".to_string()), "22");
    assert_eq!(largest_even("1".to_string()), "");
}

// Accepted solution for LeetCode #3798: Largest Even Number
function largestEven(s: string): string {
    return s.replace(/1+$/, '');
}

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.