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Palindrome Partitioning II

Hard

Asked by:

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Topics:

StringsDynamic Programming

Given a string s, partition s such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of s.

Example 1:

Input: s = "aab"
Output: 1
Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.

Example 2:

Input: s = "a"
Output: 0

Example 3:

Input: s = "ab"
Output: 1

Constraints:

1 <= s.length <= 2000
s consists of lowercase English letters only.

Palindrome Partitioning II

Hard

Asked by:

16 views

Topics:

StringsDynamic Programming

Given a string s, partition s such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of s.

Example 1:

Input: s = "aab"
Output: 1
Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.

Example 2:

Input: s = "a"
Output: 0

Example 3:

Input: s = "ab"
Output: 1

Constraints:

1 <= s.length <= 2000
s consists of lowercase English letters only.

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 maximum length of the input string `s`? Are there any specific performance considerations given the string length?
Can the input string `s` be empty or null? If so, what should be the expected return value?
Are we looking for the minimum number of cuts required to partition the string into palindromic substrings, or is there another objective?
Does case matter when determining if a substring is a palindrome? For example, should "Racecar" be considered a palindrome?
If the string `s` is already a palindrome, should the return value be 0?

Brute Force Solution

Approach

The brute force approach to this problem involves checking every possible way to break the input string into smaller parts. For each possible split, we verify if all the parts are palindromes. Finally, we determine the minimum number of cuts needed to achieve this.

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

Start by considering the whole string without any cuts.
Then, try making one cut in every possible position.
For each of these splits, check if the two resulting pieces are palindromes. If they are, you've found a potential solution.
Next, consider making two cuts in every possible combination of positions.
Again, check if all three resulting pieces are palindromes.
Continue this process, adding more cuts each time, until you've explored all possible cut combinations.
For each valid combination where every part is a palindrome, count the number of cuts.
Finally, pick the arrangement that requires the fewest cuts among all valid combinations.

Code Implementation

def palindrome_partitioning_brute_force(input_string):
    string_length = len(input_string)

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

    def min_cuts_recursive(start_index):
        if start_index == string_length:
            return -1  # No more cuts needed

        minimum_cuts_needed = float('inf')
        for end_index in range(start_index, string_length):
            substring_to_check = input_string[start_index:end_index + 1]

            # Check if the current substring is a palindrome
            if is_palindrome(substring_to_check):

                # Recursively find min cuts for remaining string
                remaining_cuts = min_cuts_recursive(end_index + 1)
                minimum_cuts_needed = min(minimum_cuts_needed, 1 + remaining_cuts)

        return minimum_cuts_needed

    # Handle the case of an empty string
    if not input_string:
        return 0

    return min_cuts_recursive(0)

Try it yourself on LeetCode

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 2^(n-1) possible ways to insert cuts. For each of these partitions, we must verify if each substring created by the cuts is a palindrome, which takes O(n) time in the worst case for each partition. Therefore, the time complexity is O(n * 2^(n-1)), which simplifies to O(2^n) since the exponential term dominates. The palindrome check within each partition contributes a linear factor that doesn't affect the overall exponential complexity.

Space Complexity

O(N^2) – The described brute force approach implicitly explores all possible substrings to check for palindromes. In the worst-case scenario, determining if all parts are palindromes requires storing boolean results for all possible substrings to avoid redundant calculations. This means we'd need to cache palindrome information for all substrings, leading to a table or similar structure of size N * N, where N is the length of the input string. Therefore, the auxiliary space complexity is O(N^2).

Optimal Solution

Approach

The best approach is to use a technique called dynamic programming. We first figure out all the smaller chunks of the original input that are palindromes and save that information. Then, we use this knowledge to build up to the overall solution efficiently.

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

First, determine which substrings are palindromes. A palindrome is a word or phrase that reads the same forwards and backward.
Next, for each position in the original input, find the minimum number of cuts needed to split the string into palindromes up to that position.
To do this, check if the substring from the start to the current position is already a palindrome. If it is, then no cuts are needed.
Otherwise, check all possible cut positions before the current one. Look up if the substring after the cut is a palindrome (you've already pre-calculated this).
If you find such a cut, see if using that cut gives you the minimum number of cuts from prior calculated results.
Keep track of the smallest number of cuts needed to get a final answer.
By leveraging information about smaller palindromic substrings and previously calculated minimum cuts, we can determine the overall minimum cuts without recomputing the palindrome checks or the cut counts at each step.

Code Implementation

def palindrome_partitioning_ii(input_string):
    string_length = len(input_string)

    # is_palindrome[i][j] will be True if the substring
    # input_string[i:j+1] is a palindrome.
    is_palindrome = [[False] * string_length for _ in range(string_length)]

    # min_cuts[i] will store the minimum number of cuts needed
    # for the substring input_string[0:i+1].
    min_cuts = [0] * string_length

    for end_index in range(string_length):
        minimum_cuts_needed = end_index
        for start_index in range(end_index + 1):
            if input_string[start_index:end_index + 1] == input_string[start_index:end_index + 1][::-1]:
                is_palindrome[start_index][end_index] = True

                # If the substring from start_index to end_index is a palindrome,
                # we update the minimum cuts needed.
                if start_index == 0:
                    minimum_cuts_needed = 0
                else:

                    #We leverage previously calculated min_cuts for optimization.
                    minimum_cuts_needed = min(minimum_cuts_needed, min_cuts[start_index - 1] + 1)

        min_cuts[end_index] = minimum_cuts_needed

    # The result is the minimum number of cuts needed for the entire string.
    return min_cuts[string_length - 1]

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n³) – The algorithm first identifies all palindromic substrings, which takes O(n²) time because we iterate through all possible start and end positions within the string. Then, for each index i from 0 to n-1, we iterate from 0 to i to find the minimum cuts needed. Inside this nested loop, we access a pre-calculated palindrome table in O(1) time, but the nested loops themselves contribute O(n²) operations. Therefore, the dominant operations are the initial palindrome identification O(n²) and calculating the minimum cuts which has a time complexity of O(n²). Taking these factors into account along with the string slicing during palindrome verification, the overall time complexity is approximately O(n³).

Space Complexity

O(N^2) – The algorithm utilizes dynamic programming, which involves creating a two-dimensional boolean array, isPalindrome, of size N x N to store whether each substring is a palindrome. Additionally, an array cuts of size N is used to store the minimum number of cuts needed for each position. Therefore, the auxiliary space is dominated by the isPalindrome array, which takes O(N^2) space, and the cuts array which is O(N). Since N^2 grows faster than N, the overall space complexity is O(N^2), where N is the length of the input string.

Edge Cases

Case	How to Handle
Empty string input	Return 0 because an empty string requires no cuts to be partitioned into palindromes.
Null string input	Throw an IllegalArgumentException or return -1 indicating an invalid input.
String with a single character	Return 0 because a single-character string is already a palindrome and requires no cuts.
String that is already a palindrome	Return 0 as no cuts are needed.
String with all identical characters (e.g., 'aaaa')	Return 0 as the string is already a palindrome without any cuts.
String requiring maximum cuts (e.g., 'ababa')	The dynamic programming approach should correctly calculate the minimum cuts even when many partitions are considered.
Very long string (close to maximum string length)	Ensure the solution's time and space complexity are efficient enough to handle large inputs, and avoid potential stack overflow errors if using recursion.
String with a palindrome at the beginning or end	The algorithm should identify and leverage existing palindromic substrings to minimize cuts.

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 maximum length of the input string `s`? Are there any specific performance considerations given the string length?
Can the input string `s` be empty or null? If so, what should be the expected return value?
Are we looking for the minimum number of cuts required to partition the string into palindromic substrings, or is there another objective?
Does case matter when determining if a substring is a palindrome? For example, should "Racecar" be considered a palindrome?
If the string `s` is already a palindrome, should the return value be 0?

Brute Force Solution

Approach

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

Start by considering the whole string without any cuts.
Then, try making one cut in every possible position.
For each of these splits, check if the two resulting pieces are palindromes. If they are, you've found a potential solution.
Next, consider making two cuts in every possible combination of positions.
Again, check if all three resulting pieces are palindromes.
Continue this process, adding more cuts each time, until you've explored all possible cut combinations.
For each valid combination where every part is a palindrome, count the number of cuts.
Finally, pick the arrangement that requires the fewest cuts among all valid combinations.

Code Implementation

def palindrome_partitioning_brute_force(input_string):
    string_length = len(input_string)

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

    def min_cuts_recursive(start_index):
        if start_index == string_length:
            return -1  # No more cuts needed

        minimum_cuts_needed = float('inf')
        for end_index in range(start_index, string_length):
            substring_to_check = input_string[start_index:end_index + 1]

            # Check if the current substring is a palindrome
            if is_palindrome(substring_to_check):

                # Recursively find min cuts for remaining string
                remaining_cuts = min_cuts_recursive(end_index + 1)
                minimum_cuts_needed = min(minimum_cuts_needed, 1 + remaining_cuts)

        return minimum_cuts_needed

    # Handle the case of an empty string
    if not input_string:
        return 0

    return min_cuts_recursive(0)

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, determine which substrings are palindromes. A palindrome is a word or phrase that reads the same forwards and backward.
Next, for each position in the original input, find the minimum number of cuts needed to split the string into palindromes up to that position.
To do this, check if the substring from the start to the current position is already a palindrome. If it is, then no cuts are needed.
Otherwise, check all possible cut positions before the current one. Look up if the substring after the cut is a palindrome (you've already pre-calculated this).
If you find such a cut, see if using that cut gives you the minimum number of cuts from prior calculated results.
Keep track of the smallest number of cuts needed to get a final answer.
By leveraging information about smaller palindromic substrings and previously calculated minimum cuts, we can determine the overall minimum cuts without recomputing the palindrome checks or the cut counts at each step.

Code Implementation

def palindrome_partitioning_ii(input_string):
    string_length = len(input_string)

    # is_palindrome[i][j] will be True if the substring
    # input_string[i:j+1] is a palindrome.
    is_palindrome = [[False] * string_length for _ in range(string_length)]

    # min_cuts[i] will store the minimum number of cuts needed
    # for the substring input_string[0:i+1].
    min_cuts = [0] * string_length

    for end_index in range(string_length):
        minimum_cuts_needed = end_index
        for start_index in range(end_index + 1):
            if input_string[start_index:end_index + 1] == input_string[start_index:end_index + 1][::-1]:
                is_palindrome[start_index][end_index] = True

                # If the substring from start_index to end_index is a palindrome,
                # we update the minimum cuts needed.
                if start_index == 0:
                    minimum_cuts_needed = 0
                else:

                    #We leverage previously calculated min_cuts for optimization.
                    minimum_cuts_needed = min(minimum_cuts_needed, min_cuts[start_index - 1] + 1)

        min_cuts[end_index] = minimum_cuts_needed

    # The result is the minimum number of cuts needed for the entire string.
    return min_cuts[string_length - 1]

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty string input	Return 0 because an empty string requires no cuts to be partitioned into palindromes.
Null string input	Throw an IllegalArgumentException or return -1 indicating an invalid input.
String with a single character	Return 0 because a single-character string is already a palindrome and requires no cuts.
String that is already a palindrome	Return 0 as no cuts are needed.
String with all identical characters (e.g., 'aaaa')	Return 0 as the string is already a palindrome without any cuts.
String requiring maximum cuts (e.g., 'ababa')	The dynamic programming approach should correctly calculate the minimum cuts even when many partitions are considered.
Very long string (close to maximum string length)	Ensure the solution's time and space complexity are efficient enough to handle large inputs, and avoid potential stack overflow errors if using recursion.
String with a palindrome at the beginning or end	The algorithm should identify and leverage existing palindromic substrings to minimize cuts.