Problem Description

You are given a string word and an integer numFriends. Alice wants to play a game with her numFriends friends that consists of multiple rounds.

In each round of the game:

• The string word must be split into exactly numFriends non-empty substrings
• Each split must be different from all previous rounds (no two rounds can have the exact same split)
• All the substring pieces from the split are collected into a box

After playing all possible rounds with different splits, you need to find and return the lexicographically largest string that was put into the box across all rounds.

For example, if word = "abc" and numFriends = 2, you could split it as:

• Round 1: "a" and "bc" → box contains {"a", "bc"}
• Round 2: "ab" and "c" → box contains {"a", "bc", "ab", "c"}

From all the strings in the box after all rounds, you would return the lexicographically largest one.

The key insight is that when numFriends = 1, the entire word goes into the box. Otherwise, you want to maximize the lexicographical value of one substring while ensuring the remaining numFriends - 1 friends each get at least one character. This means for any starting position i, the maximum length substring you can take is n - (numFriends - 1) characters, where n is the length of the word.

Intuition

When thinking about this problem, we need to maximize the lexicographical value of strings that end up in the box. The key realization is that we have control over how we split the string, and we want to create the largest possible substring.

First, let's consider the special case where numFriends = 1. In this case, there's only one way to split - the entire word goes to one friend. So we simply return the whole word.

For cases where numFriends > 1, we need to be strategic. Since we want the lexicographically largest string possible, we should try to create the longest substring starting from each position. Why? Because given two strings starting at the same position, the longer one will either be lexicographically larger or equal to the shorter one.

The constraint is that we must split into exactly numFriends parts, and each part must be non-empty. This means if we take a very long substring for one friend, the remaining friends must still each get at least one character.

If we start a substring at position i and want to maximize its length, we need to reserve at least numFriends - 1 characters for the other friends (one character each). Therefore, the maximum number of characters we can take starting from position i is n - (numFriends - 1), where n is the total length of the word.

By trying every possible starting position i from 0 to n-1 and taking the maximum possible length from each position, we ensure we've considered all the potentially largest substrings. Among all these candidates, we pick the lexicographically largest one.

This approach works because any substring that appears in any valid split must start at some position and can be at most n - (numFriends - 1) characters long. By checking all such possibilities, we're guaranteed to find the optimal answer.

Learn more about Two Pointers patterns.

Solution Approach

The solution implements a straightforward approach based on the intuition of enumerating all possible substring candidates.

Step 1: Handle the special case

if numFriends == 1:
    return word

When there's only one friend, the entire word becomes a single substring, so we return it directly.

Step 2: Calculate the maximum possible substring length

n = len(word)

We first get the length of the word. The maximum length of any substring we can extract is n - (numFriends - 1). This ensures that after taking this substring, we still have at least numFriends - 1 characters left for the remaining friends (one character each).

Step 3: Enumerate all possible substrings

return max(word[i : i + n - (numFriends - 1)] for i in range(n))

This line does the heavy lifting:

• We iterate through each possible starting position i from 0 to n-1
• For each starting position i, we extract a substring from index i to index i + n - (numFriends - 1)
• The slicing word[i : i + n - (numFriends - 1)] gives us the substring starting at position i with maximum possible length
• Python's slicing automatically handles cases where the end index exceeds the string length, so we don't need explicit boundary checks

Step 4: Find the lexicographically largest substring The max() function with string arguments returns the lexicographically largest string among all the generated substrings. Python's string comparison naturally follows lexicographical ordering.

Time Complexity: O(n²) where n is the length of the word, as we generate n substrings and each substring operation takes O(n) time in the worst case.

Space Complexity: O(n) for storing the substrings during comparison.

Example Walkthrough

Let's walk through an example with word = "dbca" and numFriends = 2.

Step 1: Check special case

• Since numFriends = 2 (not 1), we proceed to the general solution.

Step 2: Calculate maximum substring length

• n = len("dbca") = 4
• Maximum substring length = n - (numFriends - 1) = 4 - (2 - 1) = 3
• This means we can take at most 3 characters in any substring (leaving at least 1 character for the other friend).

Step 3: Generate all possible substrings Let's enumerate all substrings starting from each position with maximum length 3:

• Starting at index 0: word[0:3] = "dbc"
• Starting at index 1: word[1:4] = "bca"
• Starting at index 2: word[2:5] = "ca" (slicing stops at string end)
• Starting at index 3: word[3:6] = "a" (slicing stops at string end)

Step 4: Find the lexicographically largest Comparing all candidates: {"dbc", "bca", "ca", "a"}

Lexicographical comparison:

• "dbc" starts with 'd'
• "bca" starts with 'b'
• "ca" starts with 'c'
• "a" starts with 'a'

Since 'd' > 'c' > 'b' > 'a', the lexicographically largest substring is "dbc".

Verification: This substring can indeed be obtained from a valid split:

• Split the word as "dbc" and "a" → both friends get non-empty substrings
• The substring "dbc" goes into the box and is our answer

The algorithm correctly identifies "dbc" as the lexicographically largest substring that can appear in the box across all possible valid splits.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n²)

The code iterates through all possible starting positions in the string (n iterations in the range). For each starting position i, it creates a substring word[i : i + n - (numFriends - 1)]. The maximum length of each substring is n - (numFriends - 1), which in the worst case (when numFriends = 2) is n - 1. Creating each substring takes O(n) time in Python. Additionally, the max() function compares these n substrings lexicographically, where each comparison can take up to O(n) time. Therefore, the overall time complexity is O(n) × O(n) = O(n²).

Space Complexity: O(n)

The generator expression creates substrings one at a time, but the max() function needs to keep track of the current maximum substring, which can be up to length n - (numFriends - 1). In the worst case, this is O(n) space. Although multiple substrings are generated during execution, they are not all stored simultaneously due to the generator expression, so the space complexity remains O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Misunderstanding the Problem Constraints

The Mistake: A common misinterpretation is thinking that we need to find ALL possible ways to split the string into numFriends parts and then return the lexicographically largest substring from all those splits. This would lead to an overly complex solution involving recursion or dynamic programming to generate all possible partitions.

Why It Happens: The problem statement mentions "all possible rounds with different splits," which can mislead developers into thinking they need to enumerate every valid partition of the string.

The Reality: The key insight is that to maximize the lexicographical value of one substring, we should give that substring as many characters as possible while ensuring the remaining friends get at least one character each. This means we only need to consider substrings of maximum possible length starting from each position.

Correct Approach:

# Instead of generating all possible splits like this (WRONG):
def wrong_approach(word, numFriends):
    all_substrings = set()
  
    def generate_splits(index, remaining_friends, current_split):
        if remaining_friends == 0 and index == len(word):
            all_substrings.update(current_split)
            return
        # ... complex recursion to generate all splits
  
    # This is unnecessarily complex!

# Simply find the largest substring of maximum possible length (CORRECT):
def correct_approach(word, numFriends):
    if numFriends == 1:
        return word
    max_length = len(word) - (numFriends - 1)
    return max(word[i:i + max_length] for i in range(len(word)))

Pitfall 2: Incorrect Maximum Length Calculation

The Mistake: Calculating the maximum substring length incorrectly, such as using len(word) // numFriends or len(word) - numFriends.

Why It Happens: Confusion about how to ensure all friends get at least one character while maximizing one friend's substring.

Example of the Error:

# WRONG: Dividing equally
max_length = len(word) // numFriends  

# WRONG: Off by one error
max_length = len(word) - numFriends

# CORRECT: Reserve one character for each of the other friends
max_length = len(word) - (numFriends - 1)

Illustration: If word = "abcde" (length 5) and numFriends = 3:

• Wrong calculation (5 // 3 = 1): Would only consider substrings of length 1
• Wrong calculation (5 - 3 = 2): Would only consider substrings of length 2
• Correct calculation (5 - (3 - 1) = 3): Correctly considers substrings of length 3

With the correct calculation, we can take 3 characters for one friend and leave 2 characters for the other 2 friends (one each).

Pitfall 3: Forgetting Edge Cases

The Mistake: Not handling the case where numFriends = 1 separately, or assuming the word length is always greater than numFriends.

The Solution: Always check for numFriends = 1 first, as it's a special case where the entire word should be returned:

def answerString(word, numFriends):
    # Essential edge case check
    if numFriends == 1:
        return word
  
    # Also consider: what if len(word) == numFriends?
    # In this case, max_length would be 1, and we'd be comparing single characters
    # The current solution handles this correctly, but it's worth being aware of
  
    n = len(word)
    max_length = n - (numFriends - 1)
    return max(word[i:i + max_length] for i in range(n))