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Optimal Partition of String

Medium

Bloomberg LP

0 views

Topics:

ArraysGreedy AlgorithmsStrings

Given a string s, partition the string into one or more substrings such that the characters in each substring are unique. That is, no letter appears in a single substring more than once.

Return the minimum number of substrings in such a partition.

Note that each character should belong to exactly one substring in a partition.

Example 1:

Input: s = "abacaba"
Output: 4
Explanation:
Two possible partitions are ("a","ba","cab","a") and ("ab","a","ca","ba").
It can be shown that 4 is the minimum number of substrings needed.

Example 2:

Input: s = "ssssss"
Output: 6
Explanation:
The only valid partition is ("s","s","s","s","s","s").

Constraints:

1 <= s.length <= 10⁵
s consists of only English lowercase letters.

Optimal Partition of String

Medium

Bloomberg LP

0 views

Topics:

ArraysGreedy AlgorithmsStrings

Return the minimum number of substrings in such a partition.

Note that each character should belong to exactly one substring in a partition.

Example 1:

Input: s = "abacaba"
Output: 4
Explanation:
Two possible partitions are ("a","ba","cab","a") and ("ab","a","ca","ba").
It can be shown that 4 is the minimum number of substrings needed.

Example 2:

Input: s = "ssssss"
Output: 6
Explanation:
The only valid partition is ("s","s","s","s","s","s").

Constraints:

1 <= s.length <= 10⁵
s consists of only English lowercase letters.

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 is the range of possible characters in the string, and is the string case-sensitive?
Can the input string be empty or null?
If the string cannot be partitioned into valid substrings according to the problem description, what should the function return?
Are we looking for the minimum number of substrings, or just *a* valid partitioning?
Could you provide an example of a slightly larger input to illustrate expected behavior more clearly?

Brute Force Solution

Approach

The brute force approach to partitioning a string involves exploring every possible way to cut the string into smaller pieces. We'll try every possible combination of cuts and evaluate each combination to see if it's a valid solution. Finally, we pick the best valid solution according to the problem's specific criteria.

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

Consider the string. We could make a cut after the first character, or after the second character, or the third, and so on, all the way to cutting it right before the last character.
For each place we could cut the string, imagine that's our first piece. Take that first piece and see if it meets the criteria of being a valid partition (as defined by the problem, such as containing unique characters or not exceeding a length limit).
If the first piece *is* valid, then focus on the remaining part of the string (the stuff that's left after the cut).
Repeat the cutting and checking process on this remaining part, considering all the possible ways to divide it into valid pieces.
Keep going, cutting and checking, until the whole string is divided into pieces, and all pieces are valid according to the problem rules.
Each complete way of cutting the string into valid pieces is a potential solution. Remember each of these solutions.
After exploring every single way to cut the string, look at all the valid solutions we found.
Choose the 'best' solution according to whatever the problem statement specifies. For example, the solution with the fewest number of pieces, or the solution that maximizes some other metric.

Code Implementation

def optimal_partition_brute_force(input_string):
    all_possible_partitions = []

    def is_valid_partition(substring):
        #  Check if the substring contains only unique characters
        return len(set(substring)) == len(substring)

    def generate_partitions(current_string, current_partition):
        if not current_string:
            all_possible_partitions.append(current_partition)
            return

        for index in range(1, len(current_string) + 1):
            first_piece = current_string[:index]

            # Ensure the first piece meets the validity criteria.
            if is_valid_partition(first_piece):

                # If valid, recursively partition the remaining string.
                generate_partitions(current_string[index:], current_partition + [first_piece])

    generate_partitions(input_string, [])

    # Finding the partition with the minimal number of substrings
    if not all_possible_partitions:
        return []

    best_partition = min(all_possible_partitions, key=len)
    return best_partition

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores all possible partitions of the string. For a string of length n, there are n-1 possible cut points. At each cut point, we can either make a cut or not. This leads to 2^(n-1) possible combinations of cuts, effectively exploring all subsets of cut points. Therefore, the time complexity is O(2^n) because in the worst case we check every possible substring to see if it is valid.

Space Complexity

O(N) – The brute force approach uses recursion to explore all possible partitions of the string. In the worst-case scenario, the recursion depth can reach N, where N is the length of the string. Each recursive call adds a new frame to the call stack, consuming memory. Thus, the auxiliary space used by the recursion stack grows linearly with the input size N, leading to a space complexity of O(N).

Optimal Solution

Approach

The goal is to cut the string into the fewest possible chunks where each chunk contains unique letters. Instead of exploring every single possible split, we focus on finding the furthest reach of each letter to guide our cuts. This lets us quickly decide where to divide the string without wasting time on bad options.

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

First, for each letter in the string, figure out the last place it appears.
Then, walk through the string from the beginning.
As you go, keep track of the furthest ending position of any letter you've seen so far in the current chunk.
If you reach the furthest ending position that you're tracking, you've found the end of a chunk, so make a cut.
Start a new chunk after the cut, and repeat the process.

Code Implementation

def optimal_partition_of_string(input_string):
    last_occurrence = {}
    for index, char in enumerate(input_string):
        last_occurrence[char] = index

    chunk_start = 0
    chunk_end = 0
    partitions = 0

    for index, char in enumerate(input_string):
        # Extend the current chunk to the furthest reach.
        chunk_end = max(chunk_end, last_occurrence[char])

        if index == chunk_end:
            # Found the end of a partition
            partitions += 1

            # Start the next partition
            chunk_start = index + 1
            chunk_end = index + 1

    return partitions

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input string once to find the last occurrence of each character, which takes O(n) time, where n is the length of the string. Then, it iterates through the string again to determine the optimal partitions. Within this second loop, each character is examined a constant number of times to update the current partition's end and check if a cut should be made. Therefore, the overall time complexity is dominated by the linear traversals of the string, resulting in O(n).

Space Complexity

O(1) – The algorithm uses a dictionary (or hash map or array) to store the last occurrence of each character. Since the input string consists of lowercase English letters, the dictionary will hold at most 26 entries. After that, it uses a few integer variables to track the current position and the end of the current chunk. Thus, the space used is constant and independent of the input string's length N.

Edge Cases

Case	How to Handle
Null or empty input string	Return an empty list immediately, as there are no partitions possible.
String with a single character	Return a list containing only '1', as the string can only be partitioned as one segment.
String with all identical characters (e.g., 'aaaa')	Return a list containing only '1', as the entire string forms a single partition.
String with maximum possible length (consider memory limits)	The algorithm should have linear time complexity to avoid timeouts, and the space complexity should also be linear.
String where each character only appears once	Return a list of all '1's with length equal to the string length, as each character forms its own partition.
String where the last character repeats a character from the beginning	The solution correctly extends the partition to include the first occurance and the current last character.
String with very long stretches of same characters followed by a single distinct character.	The last occurrence logic correctly identifies the partition boundary at the last character of the initial stretch.
Integer overflow if calculating length in a language with size limits	Use data structures and calculations that avoid integer overflow if the string length is large, such as `long` or avoid explicit length calculation.

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 is the range of possible characters in the string, and is the string case-sensitive?
Can the input string be empty or null?
If the string cannot be partitioned into valid substrings according to the problem description, what should the function return?
Are we looking for the minimum number of substrings, or just *a* valid partitioning?
Could you provide an example of a slightly larger input to illustrate expected behavior more clearly?

Brute Force Solution

Approach

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

Consider the string. We could make a cut after the first character, or after the second character, or the third, and so on, all the way to cutting it right before the last character.
For each place we could cut the string, imagine that's our first piece. Take that first piece and see if it meets the criteria of being a valid partition (as defined by the problem, such as containing unique characters or not exceeding a length limit).
If the first piece *is* valid, then focus on the remaining part of the string (the stuff that's left after the cut).
Repeat the cutting and checking process on this remaining part, considering all the possible ways to divide it into valid pieces.
Keep going, cutting and checking, until the whole string is divided into pieces, and all pieces are valid according to the problem rules.
Each complete way of cutting the string into valid pieces is a potential solution. Remember each of these solutions.
After exploring every single way to cut the string, look at all the valid solutions we found.
Choose the 'best' solution according to whatever the problem statement specifies. For example, the solution with the fewest number of pieces, or the solution that maximizes some other metric.

Code Implementation

def optimal_partition_brute_force(input_string):
    all_possible_partitions = []

    def is_valid_partition(substring):
        #  Check if the substring contains only unique characters
        return len(set(substring)) == len(substring)

    def generate_partitions(current_string, current_partition):
        if not current_string:
            all_possible_partitions.append(current_partition)
            return

        for index in range(1, len(current_string) + 1):
            first_piece = current_string[:index]

            # Ensure the first piece meets the validity criteria.
            if is_valid_partition(first_piece):

                # If valid, recursively partition the remaining string.
                generate_partitions(current_string[index:], current_partition + [first_piece])

    generate_partitions(input_string, [])

    # Finding the partition with the minimal number of substrings
    if not all_possible_partitions:
        return []

    best_partition = min(all_possible_partitions, key=len)
    return best_partition

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, for each letter in the string, figure out the last place it appears.
Then, walk through the string from the beginning.
As you go, keep track of the furthest ending position of any letter you've seen so far in the current chunk.
If you reach the furthest ending position that you're tracking, you've found the end of a chunk, so make a cut.
Start a new chunk after the cut, and repeat the process.

Code Implementation

def optimal_partition_of_string(input_string):
    last_occurrence = {}
    for index, char in enumerate(input_string):
        last_occurrence[char] = index

    chunk_start = 0
    chunk_end = 0
    partitions = 0

    for index, char in enumerate(input_string):
        # Extend the current chunk to the furthest reach.
        chunk_end = max(chunk_end, last_occurrence[char])

        if index == chunk_end:
            # Found the end of a partition
            partitions += 1

            # Start the next partition
            chunk_start = index + 1
            chunk_end = index + 1

    return partitions

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input string	Return an empty list immediately, as there are no partitions possible.
String with a single character	Return a list containing only '1', as the string can only be partitioned as one segment.
String with all identical characters (e.g., 'aaaa')	Return a list containing only '1', as the entire string forms a single partition.
String with maximum possible length (consider memory limits)	The algorithm should have linear time complexity to avoid timeouts, and the space complexity should also be linear.
String where each character only appears once	Return a list of all '1's with length equal to the string length, as each character forms its own partition.
String where the last character repeats a character from the beginning	The solution correctly extends the partition to include the first occurance and the current last character.
String with very long stretches of same characters followed by a single distinct character.	The last occurrence logic correctly identifies the partition boundary at the last character of the initial stretch.
Integer overflow if calculating length in a language with size limits	Use data structures and calculations that avoid integer overflow if the string length is large, such as `long` or avoid explicit length calculation.