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Remove Palindromic Subsequences

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StringsTwo Pointers

You are given a string s consisting only of letters 'a' and 'b'. In a single step you can remove one palindromic subsequence from s.

Return the minimum number of steps to make the given string empty.

A string is a subsequence of a given string if it is generated by deleting some characters of a given string without changing its order. Note that a subsequence does not necessarily need to be contiguous.

A string is called palindrome if is one that reads the same backward as well as forward.

Example 1:

Input: s = "ababa"
Output: 1
Explanation: s is already a palindrome, so its entirety can be removed in a single step.

Example 2:

Input: s = "abb"
Output: 2
Explanation: "abb" -> "bb" -> "". 
Remove palindromic subsequence "a" then "bb".

Example 3:

Input: s = "baabb"
Output: 2
Explanation: "baabb" -> "b" -> "". 
Remove palindromic subsequence "baab" then "b".

Constraints:

1 <= s.length <= 1000
s[i] is either 'a' or 'b'.

Try coding it in LeetCode

Remove Palindromic Subsequences

Easy

Asked by:

14 views

Topics:

StringsTwo Pointers

You are given a string s consisting only of letters 'a' and 'b'. In a single step you can remove one palindromic subsequence from s.

Return the minimum number of steps to make the given string empty.

A string is called palindrome if is one that reads the same backward as well as forward.

Example 1:

Input: s = "ababa"
Output: 1
Explanation: s is already a palindrome, so its entirety can be removed in a single step.

Example 2:

Input: s = "abb"
Output: 2
Explanation: "abb" -> "bb" -> "". 
Remove palindromic subsequence "a" then "bb".

Example 3:

Input: s = "baabb"
Output: 2
Explanation: "baabb" -> "b" -> "". 
Remove palindromic subsequence "baab" then "b".

Constraints:

1 <= s.length <= 1000
s[i] is either 'a' or 'b'.

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:

Can the input string `s` be empty or null?
Is the input string `s` guaranteed to consist only of the characters 'a' and 'b'?
By 'subsequence,' do you mean non-contiguous characters are allowed, or does it have to be a substring (contiguous)?
If the string is already a palindrome, should I return 1 or is there a specific return value for that case?
Could you provide a few examples of how the function should behave with different inputs?

Brute Force Solution

Approach

The problem asks for the fewest steps to make a string empty by removing palindromic subsequences. A brute force approach means we consider every possible subsequence removal, which boils down to exploring all combinations of characters to remove at each step.

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

First, see if the entire string is already a palindrome. If so, you can remove it in one step, and you're done.
If not, try removing every possible single-character subsequence that is a palindrome. See what's left. Think of trying every single letter removal on the first step.
Then, try removing every possible two-character subsequence that is a palindrome. Again, see what's left.
Continue this process, considering longer and longer palindromic subsequences for removal at each step. In other words, try every possible combination of characters you can remove from the string in one go, making sure that the removed characters form a palindrome.
After each removal, check if what's left is a palindrome. If so, you can remove the rest in one more step. If not, repeat the process on the remaining string.
Keep track of the fewest number of steps it takes to empty the string for each removal combination you've tried.
Finally, compare all the different removal routes you explored and pick the one that took the fewest steps overall.

Code Implementation

def remove_palindromic_subsequences_brute_force(input_string):
    def is_palindrome(substring):
        return substring == substring[::-1]

    def solve(current_string, steps):
        if not current_string:
            return steps

        if is_palindrome(current_string):
            return steps + 1

        min_steps = float('inf')

        # Iterate through all possible subsequences
        for i in range(1 << len(current_string)):
            subsequence = ""
            remaining_string = ""
            indices_to_remove = []

            for j in range(len(current_string)):
                if (i >> j) & 1:
                    subsequence += current_string[j]
                    indices_to_remove.append(j)
                else:
                    remaining_string += current_string[j]

            if is_palindrome(subsequence):
                # Explore removing the palindromic subsequence
                min_steps = min(min_steps, solve(remaining_string, steps + 1))

        return min_steps

    # Start with an initial step count of 0
    return solve(input_string, 0)

Big(O) Analysis

Time Complexity

O(2^n) – The described brute force approach involves exploring all possible subsequences. In the worst case, we might have to consider every possible combination of characters to remove as a palindromic subsequence at each step. Since there are 2^n possible subsequences in a string of length n, and for each subsequence, we might need to verify if it is a palindrome (which takes O(n) time in the worst case), this leads to a time complexity of O(2^n * n). However, the dominant factor here is the exploration of all 2^n subsequences, so we simplify this to O(2^n).

Space Complexity

O(N) – The described brute force approach recursively explores all possible subsequences. In the worst-case scenario, the depth of the recursion can be proportional to the length of the input string, N. Each recursive call creates a new stack frame, potentially storing intermediate string results and variables for tracking removed subsequences. Therefore, the auxiliary space used by the recursion stack can grow up to O(N), where N is the length of the input string. The algorithm also implicitly gathers and stores these intermediate results across all branches.

Optimal Solution

Approach

The key to solving this problem quickly lies in recognizing a pattern about palindromes. The string only contains 'a' and 'b'. Therefore, you don't need to analyze every possible subsequence. Instead, you can use the palindrome properties and the limited characters to simplify the problem.

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

First, check if the entire input string is a palindrome. If it is, you can remove it in one step.
If the string is empty, you don't need any steps, so the answer is zero.
If the string is not a palindrome, since it only contains 'a' and 'b', you can always remove all 'a's in one step, and then remove all 'b's in the next step.
Therefore, if the string is not a palindrome, the answer is always two.

Code Implementation

def remove_palindromic_subsequences(input_string): 
    if not input_string:
        return 0

    # Check if the entire string is already a palindrome.
    if input_string == input_string[::-1]:
        return 1

    # If not a palindrome, it takes 2 steps.
    # One to remove all 'a's and another for all 'b's.
    return 2

Big(O) Analysis

Time Complexity

O(n) – The provided solution's primary operation is checking if the input string 's' is a palindrome. This check involves iterating through the string once, comparing characters from the beginning and end towards the middle. This iteration examines at most n/2 characters, where n is the length of the string. The other steps, such as checking for an empty string or returning 2 if it's not a palindrome, take constant time. Therefore, the dominant operation determines the overall time complexity, which is proportional to the length of the string, resulting in O(n).

Space Complexity

O(1) – The algorithm checks if the input string is a palindrome, but this check can be performed in-place using two pointers without allocating any additional data structures that scale with the input size N (length of the string). No auxiliary arrays, hash maps, or other data structures are created to store intermediate results. Thus, the space used is constant, independent of the input string's length.

Edge Cases

Case	How to Handle
Empty string input	Return 0 as an empty string is already a palindrome subsequence of length 0.
Null string input	Throw an IllegalArgumentException or return 0 indicating invalid input based on problem constraints.
Single character string	Return 1 since a single character string is a palindrome subsequence.
String is already a palindrome	Return 1 since the entire string can be removed in one operation.
String consists of only one distinct character (e.g., 'aaaa')	Return 1 since this is already a palindrome subsequence.
String consists of two distinct characters repeated alternatively (e.g., 'ababab')	Return 2 as we can remove all 'a's as one palindrome subsequence and then all 'b's.
Very long strings	The solution should handle long strings without stack overflow issues, avoiding recursion if possible and aiming for O(n) time.
String with unicode characters	Ensure the palindrome check handles unicode characters correctly (e.g., consider case sensitivity).

Empty string input

How to Handle:

Return 0 as an empty string is already a palindrome subsequence of length 0.

Null string input

How to Handle:

Throw an IllegalArgumentException or return 0 indicating invalid input based on problem constraints.

Single character string

How to Handle:

Return 1 since a single character string is a palindrome subsequence.

String is already a palindrome

How to Handle:

Return 1 since the entire string can be removed in one operation.

String consists of only one distinct character (e.g., 'aaaa')

How to Handle:

Return 1 since this is already a palindrome subsequence.

String consists of two distinct characters repeated alternatively (e.g., 'ababab')

How to Handle:

Return 2 as we can remove all 'a's as one palindrome subsequence and then all 'b's.

Very long strings

How to Handle:

The solution should handle long strings without stack overflow issues, avoiding recursion if possible and aiming for O(n) time.

String with unicode characters

How to Handle:

Ensure the palindrome check handles unicode characters correctly (e.g., consider case sensitivity).

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:

Can the input string `s` be empty or null?
Is the input string `s` guaranteed to consist only of the characters 'a' and 'b'?
By 'subsequence,' do you mean non-contiguous characters are allowed, or does it have to be a substring (contiguous)?
If the string is already a palindrome, should I return 1 or is there a specific return value for that case?
Could you provide a few examples of how the function should behave with different inputs?

Brute Force Solution

Approach

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

First, see if the entire string is already a palindrome. If so, you can remove it in one step, and you're done.
If not, try removing every possible single-character subsequence that is a palindrome. See what's left. Think of trying every single letter removal on the first step.
Then, try removing every possible two-character subsequence that is a palindrome. Again, see what's left.
Continue this process, considering longer and longer palindromic subsequences for removal at each step. In other words, try every possible combination of characters you can remove from the string in one go, making sure that the removed characters form a palindrome.
After each removal, check if what's left is a palindrome. If so, you can remove the rest in one more step. If not, repeat the process on the remaining string.
Keep track of the fewest number of steps it takes to empty the string for each removal combination you've tried.
Finally, compare all the different removal routes you explored and pick the one that took the fewest steps overall.

Code Implementation

def remove_palindromic_subsequences_brute_force(input_string):
    def is_palindrome(substring):
        return substring == substring[::-1]

    def solve(current_string, steps):
        if not current_string:
            return steps

        if is_palindrome(current_string):
            return steps + 1

        min_steps = float('inf')

        # Iterate through all possible subsequences
        for i in range(1 << len(current_string)):
            subsequence = ""
            remaining_string = ""
            indices_to_remove = []

            for j in range(len(current_string)):
                if (i >> j) & 1:
                    subsequence += current_string[j]
                    indices_to_remove.append(j)
                else:
                    remaining_string += current_string[j]

            if is_palindrome(subsequence):
                # Explore removing the palindromic subsequence
                min_steps = min(min_steps, solve(remaining_string, steps + 1))

        return min_steps

    # Start with an initial step count of 0
    return solve(input_string, 0)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, check if the entire input string is a palindrome. If it is, you can remove it in one step.
If the string is empty, you don't need any steps, so the answer is zero.
If the string is not a palindrome, since it only contains 'a' and 'b', you can always remove all 'a's in one step, and then remove all 'b's in the next step.
Therefore, if the string is not a palindrome, the answer is always two.

Code Implementation

def remove_palindromic_subsequences(input_string): 
    if not input_string:
        return 0

    # Check if the entire string is already a palindrome.
    if input_string == input_string[::-1]:
        return 1

    # If not a palindrome, it takes 2 steps.
    # One to remove all 'a's and another for all 'b's.
    return 2

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty string input	Return 0 as an empty string is already a palindrome subsequence of length 0.
Null string input	Throw an IllegalArgumentException or return 0 indicating invalid input based on problem constraints.
Single character string	Return 1 since a single character string is a palindrome subsequence.
String is already a palindrome	Return 1 since the entire string can be removed in one operation.
String consists of only one distinct character (e.g., 'aaaa')	Return 1 since this is already a palindrome subsequence.
String consists of two distinct characters repeated alternatively (e.g., 'ababab')	Return 2 as we can remove all 'a's as one palindrome subsequence and then all 'b's.
Very long strings	The solution should handle long strings without stack overflow issues, avoiding recursion if possible and aiming for O(n) time.
String with unicode characters	Ensure the palindrome check handles unicode characters correctly (e.g., consider case sensitivity).

Empty string input

How to Handle:

Return 0 as an empty string is already a palindrome subsequence of length 0.

Null string input

How to Handle:

Throw an IllegalArgumentException or return 0 indicating invalid input based on problem constraints.

Single character string

How to Handle:

Return 1 since a single character string is a palindrome subsequence.

String is already a palindrome

How to Handle:

Return 1 since the entire string can be removed in one operation.

String consists of only one distinct character (e.g., 'aaaa')

How to Handle:

Return 1 since this is already a palindrome subsequence.

String consists of two distinct characters repeated alternatively (e.g., 'ababab')

How to Handle:

Return 2 as we can remove all 'a's as one palindrome subsequence and then all 'b's.

Very long strings

How to Handle:

The solution should handle long strings without stack overflow issues, avoiding recursion if possible and aiming for O(n) time.

String with unicode characters

How to Handle:

Ensure the palindrome check handles unicode characters correctly (e.g., consider case sensitivity).