🥏Beginner-Friendly Guide to Solving "Lexicographically Largest String From the Box I" | LeetCode 3403(C++ | JavaScript | Python)

When you first read the problem “Find the Lexicographically Largest String From the Box”, it might sound complicated 🤔. However, with the right insight, the solution becomes both elegant and efficient. ✨

In this article, we’ll walk through the problem, break down the core concept, and implement the optimal solution in C++, JavaScript, and Python. 💻

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📝 Problem Recap

You are given:

• A string word 🧵
• An integer numFriends 👥

The game rules:

• Split word into exactly numFriends non-empty substrings.
• Every possible way to split word counts as a “round.” 🎲
• For each round, put all split substrings into a box 📦.
• After considering all possible rounds, find the lexicographically largest substring in the box. 🔠

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💡 Key Insight

At first glance, it looks like we need to consider all ways to split the string, which could be exponential in complexity ⚡.

But here’s the trick:

• If you split a string of length n into numFriends parts, the longest substring in any split must have length at least n - numFriends + 1. 📏

Why?

• Because at minimum, the other numFriends - 1 substrings each have length 1. 🧩
• So the longest substring length is constrained and can be calculated. ✔️

Therefore, the lexicographically largest substring that appears in any valid split must be a substring of word of length exactly n - numFriends + 1. 🔍

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🔧 How to Solve It?

• Find the lexicographically largest substring of length n - numFriends + 1.
• Return that substring. 🎯

This approach avoids any heavy recursion or DP 🛠️.

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💻 Implementations

C++

class Solution {
public:
    string answerString(string word, int numFriends) {
        if (numFriends == 1) return word;
        string res = "";
        int length = word.length() - numFriends + 1;
        for (int i = 0; i <= (int)word.size() - length; i++) {
            string temp = word.substr(i, length);
            if (temp > res) {
                res = temp;
            }
        }
        return res;
    }
};

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JavaScript

function answerString(word, numFriends) {
    if (numFriends === 1) return word;
    let length = word.length - numFriends + 1;
    let res = "";
    for (let i = 0; i <= word.length - length; i++) {
        let temp = word.substring(i, i + length);
        if (temp > res) {
            res = temp;
        }
    }
    return res;
}

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Python

class Solution:
    def answerString(self, word: str, numFriends: int) -> str:
        if numFriends == 1:
            return word
        length = len(word) - numFriends + 1
        res = ""
        for i in range(len(word) - length + 1):
            temp = word[i:i+length]
            if temp > res:
                res = temp
        return res

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🎯 Conclusion

This problem beautifully demonstrates how understanding problem constraints and properties can simplify seemingly complex problems. 🌟

Instead of brute forcing every split, we reduce the problem to a straightforward substring search, making our solution both simple and efficient. 🧠💡

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Feel free to try these implementations, and let me know if you have any questions or want to explore advanced optimizations! 🙌

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Happy coding! 👩‍💻👨‍💻🚀

Comments

[Nevo David]

Pretty cool seeing how one small trick makes the whole thing way easier. Always makes me want to look for more shortcuts like that.

[Om Shree]

Thank you so much! 😊
You can always find daily LeetCode problem solutions and walkthroughs on my blog, feel free to check it out anytime! 🚀🧠

[Dotallio]

Love how you broke down the trick with the longest substring - it makes it all click. Did you run into any weird edge cases while testing different word lengths?

[Om Shree]

Thanks a lot, @dotallio ! 🙌 I'm really glad the explanation helped make it click for you.

Regarding edge cases: yes, I did test a variety of scenarios, especially ones where the word length was exactly equal to numFriends, or where all characters were the same. In such cases, the logic ensures that we still extract the longest valid substring (len(word) - numFriends + 1) without duplication of splits.

Let me know if you ran into any interesting variations, always excited to discuss edge cases! 🚀

[Anna kowoski]

Clean and efficient solution—thanks for sharing!

[Om Shree]

Thank you