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Rearrange String k Distance Apart

#115 Most AskedHard

Rearrange String k Distance Apart #115 Most Asked

Hard

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StringsGreedy AlgorithmsArrays

Given a string s and an integer k, rearrange s such that the same characters are at least distance k from each other. If it is not possible to rearrange the string, return an empty string "".

Example 1:

Input: s = "aabbcc", k = 3
Output: "abcabc"
Explanation: The same letters are at least distance 3 from each other.

Example 2:

Input: s = "aaabc", k = 2
Output: "abcab"
Explanation: The same letters are at least distance 2 from each other.

Example 3:

Input: s = "aaadbbcc", k = 2
Output: "abacabcd"
Explanation: The same letters are at least distance 2 from each other.

Constraints:

1 <= s.length <= 3 * 10⁵
0 <= k <= s.length
s consists of only lowercase English letters.

Try coding it in LeetCode

Rearrange String k Distance Apart

#115 Most AskedHard

Rearrange String k Distance Apart #115 Most Asked

Hard

TikTok

4 views

Topics:

StringsGreedy AlgorithmsArrays

Example 1:

Input: s = "aabbcc", k = 3
Output: "abcabc"
Explanation: The same letters are at least distance 3 from each other.

Example 2:

Input: s = "aaabc", k = 2
Output: "abcab"
Explanation: The same letters are at least distance 2 from each other.

Example 3:

Input: s = "aaadbbcc", k = 2
Output: "abacabcd"
Explanation: The same letters are at least distance 2 from each other.

Constraints:

1 <= s.length <= 3 * 10⁵
0 <= k <= s.length
s consists of only lowercase English letters.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What should I return if it's not possible to rearrange the string such that identical characters are k distance apart? Should I return an empty string, null, or throw an exception?
What are the possible characters in the input string? Is it limited to lowercase English letters, or can it include uppercase letters, numbers, or other characters?
Is k always a non-negative integer? Can k be zero? What happens if k is larger than the length of the input string?
If there are multiple valid rearrangements, is any valid rearrangement acceptable, or is there a specific order I should aim for?
Can the input string be empty or null? If so, what should I return?

Brute Force Solution

Approach

The basic idea is to try all possible arrangements of the letters in the string and check if any of them satisfy the distance requirement. We generate each possible arrangement and then verify if the distance between the same letters is at least k.

Here's how the algorithm would work step-by-step:

Generate all possible orderings of the letters in the string.
For each ordering, check if the distance between every pair of the same letter is at least k.
If an ordering satisfies the distance requirement, we have found a solution.
If we have checked all possible orderings and none of them satisfy the requirement, then no solution exists.

Code Implementation

def rearrange_string_k_distance_apart_brute_force(input_string, distance):
    import itertools

    def is_valid_arrangement(arrangement, distance):
        # Check if the arrangement satisfies the distance requirement
        for first_index in range(len(arrangement)):
            for second_index in range(first_index + 1, len(arrangement)):
                if arrangement[first_index] == arrangement[second_index]:
                    if abs(first_index - second_index) < distance:
                        return False
        return True

    # Generate all possible permutations of the string
    all_permutations = itertools.permutations(input_string)

    for permutation in all_permutations:
        arrangement = ''.join(permutation)

        # Check if the current arrangement is valid
        if is_valid_arrangement(arrangement, distance):
            return arrangement

    # No valid arrangement found
    return ""

Big(O) Analysis

Time Complexity

O(n!) – The algorithm generates all possible permutations of the input string of length n. Generating all permutations has a time complexity of O(n!). For each permutation, it checks if the distance between every pair of identical characters is at least k. Checking the distance takes O(n^2) in the worst case where you need to compare all pairs. Since the permutation generation dominates the cost, the overall time complexity is approximately O(n! * n^2). However, since O(n!) grows faster than O(n! * n^2), we can simplify the total complexity to O(n!).

Space Complexity

O(N!) – The provided solution generates all possible orderings of the letters in the string, which involves storing these permutations. In the worst case, where all characters are distinct, there are N! permutations, where N is the length of the input string. Storing these permutations requires O(N!) space. Additionally, each permutation has a length of N, but since the problem specifies generating all permutations, the space occupied by them dominates, making the auxiliary space O(N!).

Optimal Solution

Approach

The best approach to rearrange a string with a distance constraint is to prioritize the most frequent characters first. This strategy uses a system to track how many of each character we have, and a waiting list to ensure we don't place the same character too close together. By strategically placing the most available character each time, we efficiently build the rearranged string.

Here's how the algorithm would work step-by-step:

First, count how many times each character appears in the string.
Create a list of characters, ordering them from most frequent to least frequent.
Also create a 'waiting list'. Think of this as characters that have been used recently and can't be used again yet due to the distance constraint.
Start building the rearranged string. At each position, select the most frequent character that isn't in the waiting list.
Place that character in the string. Reduce its count by one, and add it to the waiting list.
After adding to the waiting list, if the distance constraint is met for the earliest character in the list, remove that character from the waiting list. That character can now be used again.
Repeat the process of selecting the most frequent available character until you've used all characters. If at any point you can't find a character to place, that means it's impossible to rearrange the string according to the distance constraint.

Code Implementation

from collections import Counter, deque

def rearrange_string(input_string, distance):
    character_counts = Counter(input_string)

    # Prioritize characters by frequency.
    character_heap = [(-count, char) for char, count in character_counts.items()]
    import heapq
    heapq.heapify(character_heap)

    result = []
    waiting_queue = deque()

    while character_heap:
        count, character = heapq.heappop(character_heap)
        result.append(character)

        # Add the character to the waiting queue.
        waiting_queue.append((character, len(result) + distance))

        # Free up characters in the waiting queue.
        while waiting_queue and waiting_queue[0][1] == len(result):
            freed_character, _ = waiting_queue.popleft()
            freed_character_count = character_counts[freed_character] - 1

            # Re-add to the heap if count > 0
            if freed_character_count > 0:
                heapq.heappush(character_heap, (-freed_character_count, freed_character))
                character_counts[freed_character] = freed_character_count

    # Check to make sure it was actually possible.
    if len(result) != len(input_string):
        return ""

    return ''.join(result)

Big(O) Analysis

Time Complexity

O(n log n) – Counting character frequencies takes O(n) time, where n is the length of the string. Sorting the characters based on frequency takes O(m log m) time, where m is the number of distinct characters (m <= n). In the main loop, we iterate up to n times. Inside the loop, finding the most frequent available character from the heap takes O(log m) time in each iteration. The waiting list operations (adding and removing) also take O(1) time on average, but could theoretically be O(m) depending on implementation. The heap operations dominate within the main loop, leading to O(n log m). Since m <= n, the overall time complexity is dominated by O(n log n) due to the initial sorting step and the heap operations within the loop.

Space Complexity

O(1) – The space complexity is dominated by the character counts and the waiting list. The character counts require space proportional to the number of distinct characters, which is at most 26 (English alphabet) regardless of the input string length N. The waiting list stores characters temporarily, and its maximum size is dictated by the distance k, not N. Therefore, the space used by both character counts and the waiting list is independent of the input string length N, resulting in constant extra space.

Edge Cases

Case	How to Handle
Empty string	Return an empty string immediately as there's nothing to rearrange.
k is zero	Return the original string, as any arrangement fulfills the 0-distance requirement.
k is greater than string length	Return the original string since k-distance apart is impossible so just return original.
String with all identical characters	Return an empty string if k > 1, signifying no valid rearrangement; otherwise, return the original string.
No valid rearrangement possible	Return an empty string to signal that a valid rearrangement does not exist (e.g., AAAABB with k=2).
Very large string exceeding memory limits	The algorithm needs to be optimized for memory efficiency potentially using generators or streaming approaches if the entire string can't be held in memory.
String with Unicode characters	Ensure character counting and processing handle Unicode characters correctly, considering UTF-8 or UTF-16 encoding.
String length is equal to k	If all characters are different, simply return the string, otherwise return empty string if same characters occur more than once, because k distance between them isn't possible.

Empty string

How to Handle:

Return an empty string immediately as there's nothing to rearrange.

k is zero

How to Handle:

Return the original string, as any arrangement fulfills the 0-distance requirement.

k is greater than string length

How to Handle:

Return the original string since k-distance apart is impossible so just return original.

String with all identical characters

How to Handle:

Return an empty string if k > 1, signifying no valid rearrangement; otherwise, return the original string.

No valid rearrangement possible

How to Handle:

Return an empty string to signal that a valid rearrangement does not exist (e.g., AAAABB with k=2).

Very large string exceeding memory limits

How to Handle:

The algorithm needs to be optimized for memory efficiency potentially using generators or streaming approaches if the entire string can't be held in memory.

String with Unicode characters

How to Handle:

Ensure character counting and processing handle Unicode characters correctly, considering UTF-8 or UTF-16 encoding.

String length is equal to k

How to Handle:

If all characters are different, simply return the string, otherwise return empty string if same characters occur more than once, because k distance between them isn't possible.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What should I return if it's not possible to rearrange the string such that identical characters are k distance apart? Should I return an empty string, null, or throw an exception?
What are the possible characters in the input string? Is it limited to lowercase English letters, or can it include uppercase letters, numbers, or other characters?
Is k always a non-negative integer? Can k be zero? What happens if k is larger than the length of the input string?
If there are multiple valid rearrangements, is any valid rearrangement acceptable, or is there a specific order I should aim for?
Can the input string be empty or null? If so, what should I return?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Generate all possible orderings of the letters in the string.
For each ordering, check if the distance between every pair of the same letter is at least k.
If an ordering satisfies the distance requirement, we have found a solution.
If we have checked all possible orderings and none of them satisfy the requirement, then no solution exists.

Code Implementation

def rearrange_string_k_distance_apart_brute_force(input_string, distance):
    import itertools

    def is_valid_arrangement(arrangement, distance):
        # Check if the arrangement satisfies the distance requirement
        for first_index in range(len(arrangement)):
            for second_index in range(first_index + 1, len(arrangement)):
                if arrangement[first_index] == arrangement[second_index]:
                    if abs(first_index - second_index) < distance:
                        return False
        return True

    # Generate all possible permutations of the string
    all_permutations = itertools.permutations(input_string)

    for permutation in all_permutations:
        arrangement = ''.join(permutation)

        # Check if the current arrangement is valid
        if is_valid_arrangement(arrangement, distance):
            return arrangement

    # No valid arrangement found
    return ""

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, count how many times each character appears in the string.
Create a list of characters, ordering them from most frequent to least frequent.
Also create a 'waiting list'. Think of this as characters that have been used recently and can't be used again yet due to the distance constraint.
Start building the rearranged string. At each position, select the most frequent character that isn't in the waiting list.
Place that character in the string. Reduce its count by one, and add it to the waiting list.
After adding to the waiting list, if the distance constraint is met for the earliest character in the list, remove that character from the waiting list. That character can now be used again.
Repeat the process of selecting the most frequent available character until you've used all characters. If at any point you can't find a character to place, that means it's impossible to rearrange the string according to the distance constraint.

Code Implementation

from collections import Counter, deque

def rearrange_string(input_string, distance):
    character_counts = Counter(input_string)

    # Prioritize characters by frequency.
    character_heap = [(-count, char) for char, count in character_counts.items()]
    import heapq
    heapq.heapify(character_heap)

    result = []
    waiting_queue = deque()

    while character_heap:
        count, character = heapq.heappop(character_heap)
        result.append(character)

        # Add the character to the waiting queue.
        waiting_queue.append((character, len(result) + distance))

        # Free up characters in the waiting queue.
        while waiting_queue and waiting_queue[0][1] == len(result):
            freed_character, _ = waiting_queue.popleft()
            freed_character_count = character_counts[freed_character] - 1

            # Re-add to the heap if count > 0
            if freed_character_count > 0:
                heapq.heappush(character_heap, (-freed_character_count, freed_character))
                character_counts[freed_character] = freed_character_count

    # Check to make sure it was actually possible.
    if len(result) != len(input_string):
        return ""

    return ''.join(result)

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty string	Return an empty string immediately as there's nothing to rearrange.
k is zero	Return the original string, as any arrangement fulfills the 0-distance requirement.
k is greater than string length	Return the original string since k-distance apart is impossible so just return original.
String with all identical characters	Return an empty string if k > 1, signifying no valid rearrangement; otherwise, return the original string.
No valid rearrangement possible	Return an empty string to signal that a valid rearrangement does not exist (e.g., AAAABB with k=2).
Very large string exceeding memory limits	The algorithm needs to be optimized for memory efficiency potentially using generators or streaming approaches if the entire string can't be held in memory.
String with Unicode characters	Ensure character counting and processing handle Unicode characters correctly, considering UTF-8 or UTF-16 encoding.
String length is equal to k	If all characters are different, simply return the string, otherwise return empty string if same characters occur more than once, because k distance between them isn't possible.

Empty string

How to Handle:

Return an empty string immediately as there's nothing to rearrange.

k is zero

How to Handle:

Return the original string, as any arrangement fulfills the 0-distance requirement.

k is greater than string length

How to Handle:

Return the original string since k-distance apart is impossible so just return original.

String with all identical characters

How to Handle:

Return an empty string if k > 1, signifying no valid rearrangement; otherwise, return the original string.

No valid rearrangement possible

How to Handle:

Return an empty string to signal that a valid rearrangement does not exist (e.g., AAAABB with k=2).

Very large string exceeding memory limits

How to Handle:

The algorithm needs to be optimized for memory efficiency potentially using generators or streaming approaches if the entire string can't be held in memory.

String with Unicode characters

How to Handle:

Ensure character counting and processing handle Unicode characters correctly, considering UTF-8 or UTF-16 encoding.

String length is equal to k

How to Handle:

If all characters are different, simply return the string, otherwise return empty string if same characters occur more than once, because k distance between them isn't possible.

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