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easyBasicsSometimes asked

Longest Word

Basics · 6 min

Given a sentence s, return its longest word. If several tie, return the first.

In plain English

Split into words, then track the longest seen, keeping the earliest on ties.

Examples

Example 1

Input: s = "the quick brown fox"
Output: "quick"
Note: 5 letters, first of the 5-letter words.

Example 2

Input: s = "a bb ccc"
Output: "ccc"
Note: Longest.

Example 3

Input: s = "hi"
Output: "hi"
Note: Only word.

Constraints

1 ≤ s.length ≤ 10^4

Clarifying questions1First question: On a tie, first or last?Explore questions

Q1.On a tie, first or last?
First — so I use strict >.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Split on spaces and fold a running longest, updating only when a word is strictly longer (so the first of a tie wins).

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you map this to a running max and handle the tie rule with strictness.

Where candidates slip

`>=` returning the wrong tie, or empty tokens from a bad split.

How to stand out

Identify it as array-maximum keyed by word length.

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Running longest

TimeO(n)SpaceO(n)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Split into words, then it's a running max over lengths.

2Hint 2Explore hint

Which comparison keeps the FIRST word on a length tie?

3Hint 3Explore hint

Use strict > so later equal-length words don't replace the earlier one.

A · Core idea

Tokenise the sentence into words, then apply the running-maximum pattern over word lengths, keeping the first word on a tie.

Why strict `>` for the tie rule. Updating best only when a word is strictly longer means a later word of equal length never replaces the earlier one — so the first of several same-length words wins, as required. Using >= would keep the last instead.

Why tokenise robustly. Split on runs of whitespace so extra spaces don't produce empty 'words' that could masquerade as length-0 entries.

This is the same fold as array-maximum, just with the comparison key being length(word) instead of the value itself.

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

s = "the quick brown fox"

the

len 3 > 0

Statebest=the

Final result"quick"

C · Code with teaching comments

solution.javaJava

Running longest — optimal

1 teaching note

1
public String longestWord(String s) {
2
    String best = "";
3
    for (String w : s.split(" ")) {
4
        // Strict > keeps the first word on a length tie.
5
        if (w.length() > best.length()) best = w;
6
    }
7
    return best;
8
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.8 lines

D · Why it works and complexityO(n) timeO(n) spaceOpen the invariant and proofExplore the proof

ComplexityO(n) time, O(n) space — one pass over the words comparing lengths; tokenising is linear in the input.

E · Edge cases and pitfalls5s = "a bb ccc" — Longest is "ccc".Explore edge cases

Edge cases handled

s = "a bb ccc"Longest is "ccc".
s = "hi"Only word → "hi".
s = "to be or"All len 2; first "to" wins.

Pitfalls

Using `>=` and returning the last tie instead of the first.
Splitting incorrectly and getting empty tokens.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Editable code previewpublic String longestWord(String s) { String best = ""; for (String w : s.split(" ")) { if (w.length() > best.length()) best = w; }Explore the playground

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Using `>=` and returning the last tie instead of the first.
Splitting incorrectly and getting empty tokens.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Running maximum (by length)

Code pattern to memorize

best=''; for w in words: if len(w)>len(best): best=w

Rules to remember

1Split into words.
2Strict > for first-on-tie.
3Track the best.

Use this pattern when

Max-by-key over tokens
First-on-tie selection

Interview takeaway

Use strict > in running-max when ties should keep the earlier item.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Running maximum (by length). Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

easyBasicsSometimes asked

Longest Word

Basics · 6 min

Given a sentence s, return its longest word. If several tie, return the first.

In plain English

Split into words, then track the longest seen, keeping the earliest on ties.

Examples

Example 1

Input: s = "the quick brown fox"
Output: "quick"
Note: 5 letters, first of the 5-letter words.

Example 2

Input: s = "a bb ccc"
Output: "ccc"
Note: Longest.

Example 3

Input: s = "hi"
Output: "hi"
Note: Only word.

Constraints

1 ≤ s.length ≤ 10^4

Clarifying questions1First question: On a tie, first or last?Explore questions

Q1.On a tie, first or last?
First — so I use strict >.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Split on spaces and fold a running longest, updating only when a word is strictly longer (so the first of a tie wins).

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you map this to a running max and handle the tie rule with strictness.

Where candidates slip

`>=` returning the wrong tie, or empty tokens from a bad split.

How to stand out

Identify it as array-maximum keyed by word length.

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Running longest

TimeO(n)SpaceO(n)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Split into words, then it's a running max over lengths.

2Hint 2Explore hint

Which comparison keeps the FIRST word on a length tie?

3Hint 3Explore hint

Use strict > so later equal-length words don't replace the earlier one.

A · Core idea

Tokenise the sentence into words, then apply the running-maximum pattern over word lengths, keeping the first word on a tie.

Why tokenise robustly. Split on runs of whitespace so extra spaces don't produce empty 'words' that could masquerade as length-0 entries.

This is the same fold as array-maximum, just with the comparison key being length(word) instead of the value itself.

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

s = "the quick brown fox"

the

len 3 > 0

Statebest=the

Final result"quick"

C · Code with teaching comments

solution.javaJava

Running longest — optimal

1 teaching note

1
public String longestWord(String s) {
2
    String best = "";
3
    for (String w : s.split(" ")) {
4
        // Strict > keeps the first word on a length tie.
5
        if (w.length() > best.length()) best = w;
6
    }
7
    return best;
8
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.8 lines

D · Why it works and complexityO(n) timeO(n) spaceOpen the invariant and proofExplore the proof

ComplexityO(n) time, O(n) space — one pass over the words comparing lengths; tokenising is linear in the input.

E · Edge cases and pitfalls5s = "a bb ccc" — Longest is "ccc".Explore edge cases

Edge cases handled

s = "a bb ccc"Longest is "ccc".
s = "hi"Only word → "hi".
s = "to be or"All len 2; first "to" wins.

Pitfalls

Using `>=` and returning the last tie instead of the first.
Splitting incorrectly and getting empty tokens.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Editable code previewpublic String longestWord(String s) { String best = ""; for (String w : s.split(" ")) { if (w.length() > best.length()) best = w; }Explore the playground

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Using `>=` and returning the last tie instead of the first.
Splitting incorrectly and getting empty tokens.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Running maximum (by length)

Code pattern to memorize

best=''; for w in words: if len(w)>len(best): best=w

Rules to remember

1Split into words.
2Strict > for first-on-tie.
3Track the best.

Use this pattern when

Max-by-key over tokens
First-on-tie selection

Interview takeaway

Use strict > in running-max when ties should keep the earlier item.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Running maximum (by length). Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups