LeetCode #3582 — EASY

Generate Tag for Video Caption

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

Solve on LeetCode

The Problem

Problem Statement

You are given a string caption representing the caption for a video.

The following actions must be performed in order to generate a valid tag for the video:

Combine all words in the string into a single camelCase string prefixed with '#'. A camelCase string is one where the first letter of all words except the first one is capitalized. All characters after the first character in each word must be lowercase.
Remove all characters that are not an English letter, except the first '#'.
Truncate the result to a maximum of 100 characters.

Return the tag after performing the actions on caption.

Example 1:

Input: caption = "Leetcode daily streak achieved"

Output: "#leetcodeDailyStreakAchieved"

Explanation:

The first letter for all words except "leetcode" should be capitalized.

Example 2:

Input: caption = "can I Go There"

Output: "#canIGoThere"

Explanation:

The first letter for all words except "can" should be capitalized.

Example 3:

Input: caption = "hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh"

Output: "#hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh"

Explanation:

Since the first word has length 101, we need to truncate the last two letters from the word.

Constraints:

1 <= caption.length <= 150
caption consists only of English letters and ' '.

Step 01

Brute Force Baseline

Problem summary: You are given a string caption representing the caption for a video. The following actions must be performed in order to generate a valid tag for the video: Combine all words in the string into a single camelCase string prefixed with '#'. A camelCase string is one where the first letter of all words except the first one is capitalized. All characters after the first character in each word must be lowercase. Remove all characters that are not an English letter, except the first '#'. Truncate the result to a maximum of 100 characters. Return the tag after performing the actions on caption.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"Leetcode daily streak achieved"

Example 2

"can I Go There"

Example 3

"hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh"

Step 02

Core Insight

What unlocks the optimal approach

Simulate as described

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 #3582: Generate Tag for Video Caption
class Solution {
    public String generateTag(String caption) {
        String[] words = caption.trim().split("\\s+");
        StringBuilder sb = new StringBuilder("#");

        for (int i = 0; i < words.length; i++) {
            String word = words[i];
            if (word.isEmpty()) {
                continue;
            }

            word = word.toLowerCase();
            if (i == 0) {
                sb.append(word);
            } else {
                sb.append(Character.toUpperCase(word.charAt(0)));
                if (word.length() > 1) {
                    sb.append(word.substring(1));
                }
            }

            if (sb.length() >= 100) {
                break;
            }
        }

        return sb.length() > 100 ? sb.substring(0, 100) : sb.toString();
    }
}

// Accepted solution for LeetCode #3582: Generate Tag for Video Caption
func generateTag(caption string) string {
	words := strings.Fields(caption)
	var builder strings.Builder
	builder.WriteString("#")

	for i, word := range words {
		word = strings.ToLower(word)
		if i == 0 {
			builder.WriteString(word)
		} else {
			runes := []rune(word)
			if len(runes) > 0 {
				runes[0] = unicode.ToUpper(runes[0])
			}
			builder.WriteString(string(runes))
		}
		if builder.Len() >= 100 {
			break
		}
	}

	ans := builder.String()
	if len(ans) > 100 {
		ans = ans[:100]
	}
	return ans
}

# Accepted solution for LeetCode #3582: Generate Tag for Video Caption
class Solution:
    def generateTag(self, caption: str) -> str:
        words = [s.capitalize() for s in caption.split()]
        if words:
            words[0] = words[0].lower()
        return "#" + "".join(words)[:99]

// Accepted solution for LeetCode #3582: Generate Tag for Video Caption
fn generate_tag(caption: String) -> String {
    let mut ret = String::new();
    ret.push('#');

    let mut is_first = true;
    let mut is_capital = true;
    for c in caption.chars() {
        if c == ' ' {
            is_capital = true;
            continue;
        }

        if ret.len() >= 100 {
            break;
        }

        if is_first {
            ret.push(c.to_ascii_lowercase());
            is_first = false;
            is_capital = false;
        } else if is_capital {
            is_capital = false;
            ret.push(c.to_ascii_uppercase());
        } else {
            ret.push(c.to_ascii_lowercase());
        }
    }

    ret
}

fn main() {
    let caption = "Leetcode daily streak archived".to_string();
    let ret = generate_tag(caption);
    println!("ret={ret}");
}

#[test]
fn test() {
    {
        let caption = "Leetcode daily streak achieved".to_string();
        let expected = "#leetcodeDailyStreakAchieved";
        let ret = generate_tag(caption);
        assert_eq!(ret, expected);
    }
    {
        let caption = "can I Go There".to_string();
        let expected = "#canIGoThere";
        let ret = generate_tag(caption);
        assert_eq!(ret, expected);
    }
    {
        let mut caption = String::new();
        for _ in 0..200 {
            caption.push('a');
        }
        let ret = generate_tag(caption);
        assert_eq!(ret.len(), 100);
    }
}

// Accepted solution for LeetCode #3582: Generate Tag for Video Caption
function generateTag(caption: string): string {
    const words = caption.trim().split(/\s+/);
    let ans = '#';
    for (let i = 0; i < words.length; i++) {
        const word = words[i].toLowerCase();
        if (i === 0) {
            ans += word;
        } else {
            ans += word.charAt(0).toUpperCase() + word.slice(1);
        }
        if (ans.length >= 100) {
            ans = ans.slice(0, 100);
            break;
        }
    }
    return ans;
}

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(n)

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.