LeetCode #3403 — MEDIUM

Find the Lexicographically Largest String From the Box I

Move from brute-force thinking to an efficient approach using two pointers strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given a string word, and an integer numFriends.

Alice is organizing a game for her numFriends friends. There are multiple rounds in the game, where in each round:

word is split into numFriends non-empty strings, such that no previous round has had the exact same split.
All the split words are put into a box.

Find the lexicographically largest string from the box after all the rounds are finished.

Example 1:

Input: word = "dbca", numFriends = 2

Output: "dbc"

Explanation:

All possible splits are:

"d" and "bca".
"db" and "ca".
"dbc" and "a".

Example 2:

Input: word = "gggg", numFriends = 4

Output: "g"

Explanation:

The only possible split is: "g", "g", "g", and "g".

Constraints:

1 <= word.length <= 5 * 10³
word consists only of lowercase English letters.
1 <= numFriends <= word.length

Step 01

Brute Force Baseline

Problem summary: You are given a string word, and an integer numFriends. Alice is organizing a game for her numFriends friends. There are multiple rounds in the game, where in each round: word is split into numFriends non-empty strings, such that no previous round has had the exact same split. All the split words are put into a box. Find the lexicographically largest string from the box after all the rounds are finished.

Baseline thinking

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

Pattern signal: Two Pointers

Example 1

"dbca"
2

Example 2

"gggg"
4

Core Insight

What unlocks the optimal approach

Find lexicographically largest substring of size <code>n - numFriends + 1</code> or less starting at every index.

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 #3403: Find the Lexicographically Largest String From the Box I
class Solution {
    public String answerString(String word, int numFriends) {
        if (numFriends == 1) {
            return word;
        }
        int n = word.length();
        String ans = "";
        for (int i = 0; i < n; ++i) {
            String t = word.substring(i, Math.min(n, i + n - (numFriends - 1)));
            if (ans.compareTo(t) < 0) {
                ans = t;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3403: Find the Lexicographically Largest String From the Box I
func answerString(word string, numFriends int) (ans string) {
	if numFriends == 1 {
		return word
	}
	n := len(word)
	for i := 0; i < n; i++ {
		t := word[i:min(n, i+n-(numFriends-1))]
		ans = max(ans, t)
	}
	return
}

# Accepted solution for LeetCode #3403: Find the Lexicographically Largest String From the Box I
class Solution:
    def answerString(self, word: str, numFriends: int) -> str:
        if numFriends == 1:
            return word
        n = len(word)
        return max(word[i : i + n - (numFriends - 1)] for i in range(n))

// Accepted solution for LeetCode #3403: Find the Lexicographically Largest String From the Box I
/**
 * [3403] Find the Lexicographically Largest String From the Box I
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn answer_string(word: String, num_friends: i32) -> String {
        // When there is only one person, the word can not be split.
        if num_friends == 1 {
            return word;
        }

        let word: Vec<u8> = word.bytes().collect();
        let length = word.len() - num_friends as usize + 1;
        let mut result: Option<&[u8]> = None;

        for i in 0..word.len() {
            let end = (i + length).min(word.len());
            let w = &word[i..(i + length).min(word.len())];

            if let Some(r) = result {
                if w > r {
                    result = Some(w)
                }
            } else {
                result = Some(w);
            }
        }

        String::from_utf8(result.unwrap().into_iter().map(|x| *x).collect()).unwrap()
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3403() {
        assert_eq!("gh", Solution::answer_string("gh".to_string(), 1));
        assert_eq!("dbc", Solution::answer_string("dbca".to_string(), 2));
        assert_eq!("g", Solution::answer_string("gggg".to_string(), 4));
    }
}

// Accepted solution for LeetCode #3403: Find the Lexicographically Largest String From the Box I
function answerString(word: string, numFriends: number): string {
    if (numFriends === 1) {
        return word;
    }
    const n = word.length;
    let ans = '';
    for (let i = 0; i < n; i++) {
        const t = word.slice(i, Math.min(n, i + n - (numFriends - 1)));
        ans = t > ans ? t : ans;
    }
    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^2)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.