Problem Description

You are given a string word and an integer numFriends. Alice wants to organize a game for her friends where they play multiple rounds. In each round:

1. The string word must be split into exactly numFriends non-empty substrings
2. No two rounds can have the exact same split (each round must use a different way to split the string)
3. All the substring pieces from all rounds are collected into a box

Your task is to find the lexicographically largest string that ends up in the box after all possible rounds are completed.

Lexicographical Order: A string a is lexicographically smaller than string b if at the first position where they differ, string a has a character that comes earlier in the alphabet than the corresponding character in b. If all characters match up to the length of the shorter string, then the shorter string is considered lexicographically smaller.

Example Understanding:

• If word = "abc" and numFriends = 2, you could split it as ["a", "bc"] or ["ab", "c"]
• From these splits, you'd have strings: "a", "bc", "ab", "c" in the box
• The lexicographically largest would be "c"

The key insight is that when numFriends = 1, the entire word goes into the box as is. When numFriends > 1, you need to find the optimal way to create the largest possible substring while ensuring all friends get at least one character.

Intuition

When thinking about this problem, let's first consider the special case: if numFriends = 1, we don't split the string at all, so the entire word goes into the box and that's our answer.

For numFriends > 1, we need to be strategic about how we split the string to get the lexicographically largest possible substring. Here's the key observation: to maximize our result, we want to create the longest possible substring that starts with the lexicographically largest suffix of the original string.

Why focus on suffixes? Because any substring we create through splitting is essentially a suffix of some prefix of the original string. The lexicographically largest suffix gives us the best starting point.

Now, how long can we make this optimal substring? Since we need to split into numFriends pieces and each piece must be non-empty:

• The other numFriends - 1 friends need at least 1 character each
• This leaves us with at most len(word) - (numFriends - 1) characters for our optimal substring

The solution cleverly uses the "lexicographically largest suffix" algorithm (also known as finding the last substring). This algorithm efficiently finds the suffix that would appear last if we sorted all possible suffixes of the string. For example, in "dbca", the suffixes are "dbca", "bca", "ca", "a", and the lexicographically largest is "dbca".

Once we have this optimal suffix starting position, we take the first len(word) - numFriends + 1 characters from it. This ensures we can still distribute the remaining characters to the other friends (giving each at least one character), while maximizing the length of our target substring.

The two-pointer technique in lastSubstring efficiently finds this optimal starting position by comparing potential suffix candidates and eliminating inferior ones without checking all possibilities explicitly.

Learn more about Two Pointers patterns.

Solution Approach

The solution consists of two main components: the main function answerString and a helper function lastSubstring that finds the lexicographically largest suffix.

Main Function Logic

def answerString(self, word: str, numFriends: int) -> str:
    if numFriends == 1:
        return word
    s = self.lastSubstring(word)
    return s[: len(word) - numFriends + 1]

1. Base Case: If numFriends == 1, return the entire word since no splitting is needed.
2. Find Optimal Suffix: Call lastSubstring(word) to find the lexicographically largest suffix.
3. Truncate to Valid Length: Return the first len(word) - numFriends + 1 characters of this suffix to ensure we can still allocate at least one character to each of the other friends.

Finding the Lexicographically Largest Suffix

The lastSubstring function uses a two-pointer algorithm with three variables:

def lastSubstring(self, s: str) -> str:
    i, j, k = 0, 1, 0

• i: Index of the current best candidate for the starting position
• j: Index of the challenger being compared against position i
• k: Offset for character-by-character comparison

Algorithm Flow:

1. Comparison Loop: While j + k < len(s):

2. Equal Characters: If s[i + k] == s[j + k]:

   • Increment k to compare the next characters

3. Candidate is Smaller: If s[i + k] < s[j + k]:

   • The suffix starting at j is better than the one at i
   • Update i = i + k + 1 (skip past the compared portion)
   • Reset k = 0
   • If i >= j, set j = i + 1 to maintain j > i

4. Challenger is Smaller: If s[i + k] > s[j + k]:

   • The suffix starting at i remains better
   • Update j = j + k + 1 (skip past the compared portion)
   • Reset k = 0

5. Return Result: Return s[i:], the suffix starting at the best position found.

Why This Works:

The algorithm efficiently eliminates inferior candidates without checking all possible suffixes. When we find that one suffix is lexicographically smaller at position k, we can skip k + 1 positions because any suffix starting within that range would also be inferior (they would share the same losing prefix).

Time Complexity: O(n) where n is the length of the word, as each character is visited at most twice. Space Complexity: O(1) for the variables used in the algorithm (not counting the output string).

Example Walkthrough

Let's walk through the solution with word = "dbca" and numFriends = 2.

Step 1: Check if numFriends = 1

• Since numFriends = 2, we proceed to find the optimal suffix.

Step 2: Find the lexicographically largest suffix using lastSubstring("dbca")

Let's trace through the algorithm:

• Initialize: i = 0, j = 1, k = 0
• The suffixes of "dbca" are: "dbca" (index 0), "bca" (index 1), "ca" (index 2), "a" (index 3)

Iteration 1:

• Compare s[0+0] with s[1+0]: 'd' vs 'b'
• Since 'd' > 'b', suffix at index 0 is better
• Update: j = 1 + 0 + 1 = 2, k = 0

Iteration 2:

• Compare s[0+0] with s[2+0]: 'd' vs 'c'
• Since 'd' > 'c', suffix at index 0 is still better
• Update: j = 2 + 0 + 1 = 3, k = 0

Iteration 3:

• Compare s[0+0] with s[3+0]: 'd' vs 'a'
• Since 'd' > 'a', suffix at index 0 remains the best
• Update: j = 3 + 0 + 1 = 4, k = 0

Termination:

• j + k = 4 >= len(s) = 4, so we exit the loop
• Return s[0:] = "dbca"

Step 3: Truncate to valid length

• We have the suffix "dbca" starting at index 0
• Maximum length = len(word) - numFriends + 1 = 4 - 2 + 1 = 3
• Return the first 3 characters: "dbc"

Verification: Why is "dbc" the answer? Let's consider possible splits:

• Split 1: ["d", "bca"] → gives us "d" and "bca"
• Split 2: ["db", "ca"] → gives us "db" and "ca"
• Split 3: ["dbc", "a"] → gives us "dbc" and "a"

From all substrings {"d", "bca", "db", "ca", "dbc", "a"}, the lexicographically largest is "dbc".

The algorithm found this by identifying that the suffix starting at 'd' was optimal and taking the maximum possible length while ensuring at least one character remains for the other friend.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n) where n is the length of the input string word.

The analysis breaks down into two main parts:

• The lastSubstring function uses a two-pointer approach with variables i, j, and k to find the lexicographically largest substring. While it appears to have nested iterations due to the while loop and the variable k, each character in the string is visited at most twice. When either i or j advances, they skip over characters that have already been compared (by adding k + 1), ensuring linear traversal. This results in O(n) time complexity.
• The slicing operation s[: len(word) - numFriends + 1] takes O(len(word) - numFriends + 1) which is O(n) in the worst case.
• The overall time complexity remains O(n).

Space Complexity: O(n) where n is the length of the input string word.

The space analysis considers:

• The lastSubstring function uses only a constant amount of extra space (O(1)) for variables i, j, and k.
• The substring extraction s[i:] creates a new string that could be up to n characters long in the worst case, requiring O(n) space.
• The final slicing operation s[: len(word) - numFriends + 1] also creates a new string of up to n characters.
• The dominant space usage is O(n) for storing the substring results.

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

Common Pitfalls

1. Incorrect Maximum Length Calculation

Pitfall: A common mistake is incorrectly calculating the maximum length of the substring that can be taken. Developers might use len(word) - numFriends instead of len(word) - numFriends + 1.

Why it happens: The confusion arises from thinking about how many characters to "leave behind" rather than how many can be taken. If you have 5 characters and 3 friends, you need to ensure 2 other friends get at least 1 character each, so you can take up to 3 characters (5 - 3 + 1 = 3).

Incorrect Implementation:

def answerString(self, word: str, numFriends: int) -> str:
    if numFriends == 1:
        return word
    s = self.lastSubstring(word)
    return s[:len(word) - numFriends]  # Wrong! Off by one

Correct Implementation:

def answerString(self, word: str, numFriends: int) -> str:
    if numFriends == 1:
        return word
    s = self.lastSubstring(word)
    return s[:len(word) - numFriends + 1]  # Correct

2. Misunderstanding the Problem When numFriends = 1

Pitfall: Forgetting to handle the special case when numFriends = 1, or incorrectly assuming that the algorithm should still find the largest suffix even when there's only one friend.

Why it happens: The problem statement might lead to overthinking. When there's only one friend, there's only one way to "split" the string (giving the entire string to that friend), so the entire word goes into the box.

Incorrect Implementation:

def answerString(self, word: str, numFriends: int) -> str:
    # Missing the base case entirely
    s = self.lastSubstring(word)
    return s[:len(word) - numFriends + 1]

Correct Implementation:

def answerString(self, word: str, numFriends: int) -> str:
    if numFriends == 1:
        return word  # Must handle this special case
    s = self.lastSubstring(word)
    return s[:len(word) - numFriends + 1]

3. Infinite Loop in lastSubstring Function

Pitfall: Not properly updating the j pointer when i moves past or equals j, causing an infinite loop.

Why it happens: When comparing suffixes and finding that i needs to move forward, if i jumps to or past j, we need to ensure j is always ahead of i to maintain the algorithm's invariant.

Incorrect Implementation:

def lastSubstring(self, s: str) -> str:
    i, j, k = 0, 1, 0
    while j + k < len(s):
        if s[i + k] == s[j + k]:
            k += 1
        elif s[i + k] < s[j + k]:
            i += k + 1
            k = 0
            # Missing: if i >= j: j = i + 1
        else:
            j += k + 1
            k = 0
    return s[i:]

Correct Implementation:

def lastSubstring(self, s: str) -> str:
    i, j, k = 0, 1, 0
    while j + k < len(s):
        if s[i + k] == s[j + k]:
            k += 1
        elif s[i + k] < s[j + k]:
            i += k + 1
            k = 0
            if i >= j:  # Critical: maintain j > i
                j = i + 1
        else:
            j += k + 1
            k = 0
    return s[i:]

4. Returning Index Instead of Substring

Pitfall: Returning the index i instead of the actual substring s[i:] from the lastSubstring function.

Why it happens: During implementation, developers might focus on finding the correct starting position and forget to return the actual substring.

Incorrect Implementation:

def lastSubstring(self, s: str) -> str:
    i, j, k = 0, 1, 0
    # ... algorithm logic ...
    return i  # Wrong! Returns an integer

Correct Implementation:

def lastSubstring(self, s: str) -> str:
    i, j, k = 0, 1, 0
    # ... algorithm logic ...
    return s[i:]  # Correct: Returns the substring