Maximize Number of Subsequences in a String

Medium

Amazon

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Topics:

StringsDynamic Programming

You are given a 0-indexed string text and another 0-indexed string pattern of length 2, both of which consist of only lowercase English letters.

You can add either pattern[0] or pattern[1] anywhere in text exactly once. Note that the character can be added even at the beginning or at the end of text.

Return the maximum number of times pattern can occur as a subsequence of the modified text.

A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

For example:

text = "abdcdbc", pattern = "ac" If we add pattern[0] = 'a' in between text[1] and text[2], we get "abadcdbc". Now, the number of times "ac" occurs as a subsequence is 4.
text = "aabb", pattern = "ab" Some of the strings which can be obtained from text and have 6 subsequences "ab" are "aaabb", "aaabb", and "aabbb".

How would you efficiently solve this problem, considering the constraints on the input string lengths (up to 10^5)? What is the time and space complexity of your solution? Can you provide the code?

Maximize Number of Subsequences in a String

Medium

Amazon

6 views

Topics:

StringsDynamic Programming

You are given a 0-indexed string text and another 0-indexed string pattern of length 2, both of which consist of only lowercase English letters.

You can add either pattern[0] or pattern[1] anywhere in text exactly once. Note that the character can be added even at the beginning or at the end of text.

Return the maximum number of times pattern can occur as a subsequence of the modified text.

A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

For example:

text = "abdcdbc", pattern = "ac" If we add pattern[0] = 'a' in between text[1] and text[2], we get "abadcdbc". Now, the number of times "ac" occurs as a subsequence is 4.
text = "aabb", pattern = "ab" Some of the strings which can be obtained from text and have 6 subsequences "ab" are "aaabb", "aaabb", and "aabbb".

How would you efficiently solve this problem, considering the constraints on the input string lengths (up to 10^5)? What is the time and space complexity of your solution? Can you provide the code?

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 are the possible characters that can appear in the input string 'text' and the subsequence 'pattern'? Are they limited to lowercase English letters, or could they include uppercase letters, numbers, or special characters?
Can the input string 'text' or the subsequence 'pattern' be empty or null?
If the 'pattern' appears zero times in the 'text', what value should I return?
Does the order of characters in the 'pattern' matter? For example, if text = 'ababa' and pattern = 'ab', should 'a' from index 0 be matched with 'b' from index 1, or can the 'a' and 'b' come from anywhere as long as 'a' precedes 'b'?
If there are multiple ways to maximize the number of subsequences, is there a preferred solution (e.g., by minimizing the index where I insert the character)? Otherwise, any valid maximized count is acceptable?

Brute Force Solution

Approach

The brute force approach tries every single combination of letters from the given string to see if it matches the desired subsequence. Think of trying all possible ways to pick letters, one at a time, to form the subsequence. We check each possible combination to see how many times we can form the entire subsequence.

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

Consider every possible set of letters you can pick from the original string.
For each set of letters, check if the chosen letters form the desired subsequence exactly.
If they do, increase a counter. If not, move on to the next set of letters.
Repeat steps 1-3 for all possible sets of letters you can create from the string.
After checking all possibilities, the counter will hold the maximum number of non-overlapping times the subsequence appears.

Code Implementation

def maximize_subsequence_brute_force(text, pattern):
    number_of_subsequences = 0
    text_length = len(text)
    pattern_length = len(pattern)

    # Iterate through all possible subsets of the text
    for i in range(2 ** text_length):
        subset = ""
        for j in range(text_length):

            # Check if the j-th bit is set in i
            if (i >> j) & 1:
                subset += text[j]

        # Check if the subset matches the pattern
        if subset == pattern:
            number_of_subsequences += 1

    return number_of_subsequences

Big(O) Analysis

Time Complexity

O(2^n * m) – The brute force approach considers every possible subsequence of the given string. For a string of length n, there are 2^n possible subsequences. For each of these subsequences, we need to check if it exactly matches the target subsequence, which has a length of m. Therefore, for each of the 2^n subsequences, we perform at most m comparisons to verify if it is indeed the target subsequence. This results in a time complexity of O(2^n * m).

Space Complexity

O(1) – The brute force approach, as described, does not utilize any significant auxiliary data structures. Although it iterates through combinations, it checks each combination individually without storing a large number of intermediate combinations or sets. The dominant space usage comes from a counter variable, which remains constant regardless of the input string's length (N). Therefore, the space complexity is O(1).

Optimal Solution

Approach

The best way to solve this problem is to count the number of times the first character of our pattern appears before each instance of the second character. We can improve the count by strategically placing more of the first character at the beginning and more of the second character at the end of the input string.

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

First, go through the string and count how many times you see your pattern as a subsequence. To do this, keep track of the count of the first character. Whenever you see the second character, add the current count of the first character to your total subsequence count.
Next, think about adding the first character of your pattern to the very beginning of the string. How much would that increase the subsequence count? The increase is equal to the number of times the second character appears in the string.
Then, think about adding the second character of your pattern to the very end of the string. How much would that increase the subsequence count? The increase is equal to the number of times the first character appears in the string (including the one you might have added at the beginning).
Finally, compare the original count with the increased counts from adding the characters at the beginning and the end, and return the largest of the three counts. That's the most subsequences you can create!

Code Implementation

def maximize_number_of_subsequences(input_string, pattern):
    first_char_count = 0
    subsequence_count = 0

    for char in input_string:
        if char == pattern[0]:
            first_char_count += 1
        elif char == pattern[1]:
            subsequence_count += first_char_count

    original_count = subsequence_count

    # Calculate count if first char is added at the beginning
    second_char_count = input_string.count(pattern[1])
    subsequence_count_add_first =
    original_count + second_char_count

    # Calculate count if second char is added at the end
    # Need to add the initial first_char_count
    first_char_count = input_string.count(pattern[0])
    subsequence_count_add_second =
    original_count + first_char_count

    # If the pattern consists of same characters
    if pattern[0] == pattern[1]:
        first_char_count = input_string.count(pattern[0])
        subsequence_count_add_first = (first_char_count + 1) *
        (first_char_count) // 2
        subsequence_count_add_second = (first_char_count) *
        (first_char_count + 1) // 2
        original_count = (first_char_count) * (first_char_count - 1) // 2
        subsequence_count_add_first = original_count + first_char_count
        subsequence_count_add_second = original_count + first_char_count

    # Return the maximum subsequence count.
    return max(original_count,
               subsequence_count_add_first,
               subsequence_count_add_second)

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input string once to count the initial number of subsequences, which takes O(n) time where n is the length of the string. Counting the occurrences of the first and second characters also iterates through the string once, taking O(n) time. The remaining operations (comparing counts and returning the maximum) take constant time, O(1). 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 fixed number of variables to store counts (e.g., the count of the first character, the subsequence count, counts for comparison). The number of these variables does not depend on the length of the input string, denoted as N. Therefore, the auxiliary space used is constant, resulting in a space complexity of O(1).

Edge Cases

Case	How to Handle
Empty text or pattern string	Return 0 if either text or pattern is empty, as no subsequence can be formed.
Pattern with repeating characters, e.g., 'aa'	Count occurrences of 'a' in the modified text, where we add 'a' at the beginning and end.
Text with long sequences of one character in the pattern	Ensure the counts do not overflow (use long if necessary) when counting such subsequences.
Text contains all the characters of pattern but not in correct order.	The algorithm still counts valid subsequences formed by matching characters regardless of their initial order.
Pattern contains characters not present in the text.	The count for the missing characters remains zero which leads to zero subsequence count.
Text or pattern strings with maximum allowed length.	Check for potential integer overflows during the counting process and use long data types if needed.
Null text or pattern string	Throw IllegalArgumentException or return -1 to signify invalid input.
Text equals pattern.	The algorithm will correctly handle it and count one valid subsequence.