LeetCode Problem 3403

Difficulty: 🟡 Medium
Topics: Two Pointers, String, Enumeration

Solution

LeetCode 3403 - Find the Lexicographically Largest String From the Box I

Problem Understanding

We are given a string word of length n and an integer numFriends.

In each round, the string is split into exactly numFriends non-empty contiguous pieces. Every distinct way to place the numFriends - 1 cuts is considered a different round. After each split, all resulting pieces are placed into a box.

After all possible valid splits have been considered, the box contains every string that can appear as one of the pieces in at least one valid split.

The task is to find the lexicographically largest string that ever appears in the box.

A key observation is that every piece produced by a split is a contiguous substring of word. Therefore, the problem is equivalent to determining which substrings can appear as pieces in some valid split, then finding the lexicographically largest among them.

The constraints are:

• 1 <= word.length <= 5000
• 1 <= numFriends <= word.length

Since the length is only 5000, an O(n²) solution is acceptable, but anything cubic would be too slow.

Several edge cases are important:

• numFriends = 1, where the only possible split is the entire string itself.
• numFriends = n, where every piece must have length 1.
• Strings containing many equal characters, where lexicographic comparisons depend on length.
• The lexicographically largest answer may not be a suffix of the string, but it will always be a maximal allowed prefix of some suffix.

Approaches

Brute Force

The most direct approach is to enumerate every valid split of the string into numFriends non-empty parts.

For each split:

1. Generate all pieces.
2. Insert them into a collection.
3. Track the lexicographically largest piece encountered.

This is correct because it explicitly examines every possible round and every string placed into the box.

However, the number of splits is:

$$\binom{n-1}{\text{numFriends}-1}$$

which can be enormous. Even for moderate values of n, this becomes completely infeasible.

Key Insight

Suppose a piece has length L.

For that piece to appear in a split, the remaining characters must be distributed among the other numFriends - 1 non-empty pieces.

The remaining length is:

$$n - L$$

Since each of the other pieces must contain at least one character, we need:

$$n - L \ge \text{numFriends} - 1$$

or equivalently

$$L \le n - \text{numFriends} + 1$$

Define:

$$m = n - \text{numFriends} + 1$$

Then every substring whose length is at most m can appear in some valid split, and no longer substring can.

Now consider all substrings starting at a fixed index i.

Among all lengths ≤ m, the lexicographically largest choice is always the longest allowed one, because a string is lexicographically larger than its own proper prefix.

Therefore, for each starting position i, the best candidate is:

$$\text{word}[i : \min(n, i+m)]$$

The answer is simply the lexicographically largest candidate among all starting positions.

Approach Comparison

Approach	Time Complexity	Space Complexity	Notes
Brute Force	Exponential	Exponential	Enumerates all valid splits
Optimal	O(n²)	O(1) extra	Checks one maximal candidate per starting position

Algorithm Walkthrough

1. Let n be the length of word.
2. If numFriends == 1, return word.

In this case there is only one valid split, namely the entire string. 3. Compute

$$m = n - \text{numFriends} + 1$$

This is the maximum length any piece can have. 4. Initialize an empty answer string. 5. For every starting position i from 0 to n-1:

• Construct the longest substring beginning at i whose length does not exceed m.
• This substring is

word[i : min(n, i + m)]

6. Compare this candidate with the current answer.

If the candidate is lexicographically larger, replace the answer. 7. After all starting positions have been processed, return the answer.

Why it works

A substring can appear in some valid split if and only if its length is at most n - numFriends + 1.

For a fixed starting position, every feasible substring is a prefix of the longest feasible substring starting there. A string is always lexicographically smaller than any of its proper extensions, so the longest feasible substring is the best candidate from that starting position.

Therefore, examining only the longest feasible substring from each starting position considers every possible optimal answer. Taking the maximum among them yields the lexicographically largest string that can appear in the box.

Problem Understanding

This problem asks us to simulate a game where a string word is split into numFriends non-empty substrings in all possible ways, and then we need to determine the lexicographically largest string that appears in any of these splits. In other words, for each way of partitioning the string into numFriends contiguous parts, all the resulting substrings are placed into a box, and the goal is to find the substring that would come last if the box were sorted in dictionary order.

The input word is a lowercase string of length up to 5000, and numFriends is an integer between 1 and word.length. The output is a single string, the lexicographically largest substring from all splits. Edge cases include when numFriends equals 1 (the entire word is the only substring) or when numFriends equals word.length (each character becomes its own substring). The problem guarantees that numFriends is always less than or equal to word.length, so we will never need to handle impossible splits.

Approaches

The brute-force approach would be to generate all possible partitions of word into numFriends non-empty substrings, collect all substrings in a list, and then find the largest lexicographically. While this is conceptually simple and guaranteed to give the correct answer, the number of partitions grows combinatorially with word.length and numFriends, making it completely infeasible for word.length up to 5000.

The key observation for an optimal solution is that when we look for the lexicographically largest substring from all splits, the first split part will dominate the result because it can potentially contain the initial characters of word in different lengths. This means we only need to consider the first segment of length ranging from 1 to len(word) - numFriends + 1 because the remaining characters must accommodate the other numFriends - 1 non-empty substrings. Then, we can take the maximum character sequence from these candidate first substrings.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(C(n-1, k-1) * n)	O(n^2)	Generate all splits and collect all substrings
Optimal	O(n)	O(1)	Consider only prefixes of length 1 to n - numFriends + 1

Algorithm Walkthrough

1. Compute the maximum length allowed for the first substring. This is max_len = len(word) - numFriends + 1. Any longer, and there would not be enough characters left for the other friends.
2. Initialize an empty string best to keep track of the lexicographically largest substring seen so far.
3. Iterate over all possible prefix lengths i from 1 to max_len.
4. For each prefix word[:i], compare it with best using standard string comparison.
5. If the current prefix is larger than best, update best with this prefix.
6. After iterating through all prefix lengths, return best.

Why it works: By only considering the first substring in each split, we guarantee that we do not miss the largest possible lexicographical substring, since subsequent substrings cannot influence the ordering of the first substring. The bounds on prefix lengths ensure that all splits remain valid.

Python Solution

class Solution:
    def answerString(self, word: str, numFriends: int) -> str:
        n = len(word)

        if numFriends == 1:
            return word

        max_len = n - numFriends + 1
        answer = ""

        for start in range(n):
            candidate = word[start:start + max_len]
            if candidate > answer:
                answer = candidate

        return answer

The implementation follows the proof directly.

The variable max_len stores the largest possible length of any piece. For every starting position, the code extracts the longest feasible substring beginning there. Since this substring dominates all shorter feasible substrings with the same starting position, only one candidate per position must be checked.

The variable answer maintains the largest candidate seen so far, and the final value is returned. max_len = len(word) - numFriends + 1 best = "" for i in range(1, max_len + 1): prefix = word[:i] if prefix > best: best = prefix return best

In this Python implementation, we calculate the maximum length of the first substring and iterate through all valid prefix lengths. The comparison `prefix > best` uses Python’s built-in lexicographical string comparison, which is exactly what the problem requires. The solution efficiently returns the largest substring without generating all splits.

## Go Solution

```go
func answerString(word string, numFriends int) string {
	n := len(word)

	if numFriends == 1 {
		return word
	}

	maxLen := n - numFriends + 1
	ans := ""

	for start := 0; start < n; start++ {
		end := start + maxLen
		if end > n {
			end = n
		}

		candidate := word[start:end]
		if candidate > ans {
			ans = candidate
		}
	}

	return ans
}

The Go implementation is nearly identical to the Python version.

The only notable difference is that Go requires explicit bounds handling when computing the end index. String comparison with > is lexicographic, so it naturally performs the required ordering comparison.

Worked Examples

Example 1

Input

word = "dbca"
numFriends = 2

We compute:

n = 4
max_len = 4 - 2 + 1 = 3

Candidates:

Start	Candidate
0	"dbc"
1	"bca"
2	"ca"
3	"a"

Comparison process:

Candidate	Current Best
"dbc"	"dbc"
"bca"	"dbc"
"ca"	"dbc"
"a"	"dbc"

Final answer:

"dbc"

Example 2

Input

word = "gggg"
numFriends = 4

We compute:

n = 4
max_len = 1

Candidates:

Start	Candidate
0	"g"
1	"g"
2	"g"
3	"g"

All candidates are identical.

Final answer:

"g"

maxLen := len(word) - numFriends + 1
best := ""
for i := 1; i <= maxLen; i++ {
    prefix := word[:i]
    if prefix > best {
        best = prefix
    }
}
return best

}

The Go implementation follows the same logic. Slicing strings in Go works similarly to Python, but we need to handle indices explicitly. Lexicographical comparison of strings is also built-in using `>`.

## Worked Examples

**Example 1:** word = "dbca", numFriends = 2

| Prefix Length | Prefix | Best So Far |
| --- | --- | --- |
| 1 | "d" | "d" |
| 2 | "db" | "db" |
| 3 | "dbc" | "dbc" |

Output: `"dbc"`

**Example 2:** word = "gggg", numFriends = 4

| Prefix Length | Prefix | Best So Far |
| --- | --- | --- |
| 1 | "g" | "g" |

Output: `"g"`

## Complexity Analysis

| Measure | Complexity | Explanation |
| --- | --- | --- |
| Time | O(n²) | Up to `n` candidates, each comparison may examine `O(n)` characters |
| Space | O(1) extra | Only a few variables are maintained |

The algorithm processes exactly one candidate per starting position. Since string comparisons can require scanning many characters, the worst case occurs when many candidates share long common prefixes, yielding `O(n²)` total time. With `n ≤ 5000`, this is easily acceptable.
| Time | O(n) | Only iterate through the first `len(word) - numFriends + 1` characters, a linear scan |
| Space | O(1) | Only store the `best` substring; no additional data structures |

The reasoning is that we are only scanning prefixes and updating a single variable, so the operations scale linearly with the input length.

## Test Cases

s = Solution()

assert s.answerString("dbca", 2) == "dbc" # example 1 assert s.answerString("gggg", 4) == "g" # example 2

assert s.answerString("a", 1) == "a" # minimum size assert s.answerString("abcde", 1) == "abcde" # only one friend

assert s.answerString("abcde", 5) == "e" # every piece length 1 assert s.answerString("zzzzz", 3) == "zzz" # repeated characters

assert s.answerString("azzz", 2) == "zzz" # best candidate not at start assert s.answerString("abab", 2) == "bab" # overlapping candidates

assert s.answerString("cba", 2) == "cb" # descending characters assert s.answerString("leetcode", 3) == "tcode" # general case

assert s.answerString("aaaaa", 2) == "aaaa" # equal-prefix comparisons assert s.answerString("bacab", 3) == "cab" # middle position gives answer

### Test Summary

| Test | Why |
| --- | --- |
| `("dbca", 2)` | Official example |
| `("gggg", 4)` | Official example |
| `("a", 1)` | Smallest input |
| `("abcde", 1)` | Entire string must be returned |
| `("abcde", 5)` | Maximum number of friends |
| `("zzzzz", 3)` | Repeated characters |
| `("azzz", 2)` | Best answer occurs later in string |
| `("abab", 2)` | Overlapping candidate substrings |
| `("cba", 2)` | Descending lexicographic order |
| `("leetcode", 3)` | Typical mixed input |
| `("aaaaa", 2)` | Proper-prefix comparison behavior |
| `("bacab", 3)` | Optimal substring starts in the middle |

## Edge Cases

### `numFriends = 1`

When there is only one friend, no cuts are made. The only valid split consists of the entire string. A common bug is to continue applying the general logic and accidentally choose a suffix instead of the whole string. The implementation handles this explicitly by returning `word` immediately.

### `numFriends = n`

In this situation every piece must contain exactly one character. Therefore the answer is simply the largest character appearing in the string. The formula

$$n - \text{numFriends} + 1 = 1$$

naturally restricts every candidate to length one, so the implementation works without any special handling.

### Many Equal Characters

Strings such as `"aaaaa"` can expose mistakes in lexicographic reasoning. The substring `"aaaa"` is lexicographically larger than `"aaa"` because the shorter string is a proper prefix. By always selecting the longest feasible substring from each starting position, the algorithm correctly handles this situation and never misses the optimal answer.
# Basic examples
assert Solution().answerString("dbca", 2) == "dbc"  # Example 1
assert Solution().answerString("gggg", 4) == "g"    # Example 2

# Edge cases
assert Solution().answerString("a", 1) == "a"       # Single character
assert Solution().answerString("abc", 1) == "abc"  # Only one friend
assert Solution().answerString("abcd", 4) == "a"   # Each character is a substring
assert Solution().answerString("zxy", 2) == "zx"   # Lexicographical order matters
assert Solution().answerString("aaaab", 3) == "aaa" # Repeated characters

Test	Why
"dbca", 2	Checks general functionality and lexicographical comparison
"gggg", 4	All characters identical, single prefix possible
"a", 1	Minimum input size
"abc", 1	Single friend, whole string is output
"abcd", 4	Maximum number of friends, smallest substrings
"zxy", 2	Lexicographical order impacts selection
"aaaab", 3	Repeated characters, ensures correct prefix choice

Edge Cases

One important edge case is when numFriends equals the length of the string. Each friend will get exactly one character, so the largest substring is always the first character. Another edge case is when numFriends is 1, in which case the entire word is the only substring. Strings with repeated characters or sequences where lexicographical ordering changes mid-string can also be tricky; for example, "aaaab" with 3 friends must select "aaa" as the best prefix. Our implementation handles all these cases by correctly iterating only over valid prefix lengths and using string comparison.