Palindrome Partitioning Leetcode Problem 131 [Python Solution]

Welcome to another coding session!

Today, we're going to tackle the Palindrome Partitioning problem from LeetCode.

This problem falls under the category of backtracking and can be quite intriguing.

The task at hand is to take a given string, s, and partition it in such a way that every substring within the partition is a palindrome.

We'll be exploring how to solve this problem step by step, providing a clear Python solution for you to understand.

If you want to check out the original problem on LeetCode, you can find it here.

Let's start by understanding the problem in detail.

Problem Overview

Given a string s, we need to partition it into substrings such that every substring in the partition is a palindrome.

Once we've accomplished this, we should return all possible palindrome partitions of the string s.

Let's look at a couple of examples to get a better grasp of what's expected:

Example 1:

Input:

s = "aab"


Output:

[["a", "a", "b"], ["aa", "b"]]


In this example, we partition the string "aab" into two possible sets of substrings, both of which meet the palindrome condition.

Example 2:

Input:

s = "a"


Output:

[["a"]]


For the single-character string "a," the only possible partition is itself, which is a valid palindrome.

Now that we have a good grasp of the problem, let's delve deeper into how we can approach it.

Understanding Palindrome Constraints

Before we dive into the solution, it's crucial to understand what makes a palindrome.

A palindrome is a string that reads the same forwards and backwards.

For example, "racecar" is a palindrome because it remains the same when reversed.

To create valid partitions, we'll need to ensure that each substring we select from the input string is a palindrome.

This means that if we reverse any of these substrings, it should remain the same.

Let's move on to the solution itself.

Palindrome Partitioning LeetCode Problem Solution

To solve this problem efficiently, we will use a backtracking approach.

The idea is to generate every possible partition and check if it meets the palindrome condition.

If it does, we add it to our result list.

Let's break down the Python solution step by step:

1. Bruteforce Approach

We will start with a brute force approach.

The basic idea is to use recursion to explore every possible partition of the input string s.

If we find a valid partition, we add it to our results.

def partition(self, s: str) -> List[List[str]]:
    res, part = [], []

    def dfs(i):
        if i >= len(s):
            res.append(part.copy())
            return
        for j in range(i, len(s)):
            if self.isPali(s, i, j):
                part.append(s[i : j + 1])
                dfs(j + 1)
                part.pop()

    dfs(0)
    return res

The partition function takes the input string s and returns a list of lists of strings, representing the valid partitions.

Here's how it works:

• We maintain two variables, res (results) and part (current partition). res will store all the valid partitions we find, and part is used to build and track the current partition under consideration.

• Inside the dfs (depth-first search) function, we take an index i as an argument.

This index represents the current position within the string s.

• The base case for the recursion is when i goes out of bounds, indicating that we've explored the entire string.

When this happens, we add the current partition (part) to the results list (res).

• We then iterate over all possible ending positions j for the substring, starting from the current position i.

If the substring from index i to index j is a palindrome, we add it to the current partition and make a recursive call to dfs with the next index j + 1.

• After the recursive call, we remove the last added substring from the current partition, which allows us to explore other possibilities.

The isPali function is used to check if a given substring is a palindrome.

We'll define this function shortly.

2. Efficient Approach

The efficient approach relies on the concept of backtracking, as discussed in the brute force solution.

However, it includes an additional helper function called isPali to determine if a substring is a palindrome.

This approach is more efficient and helps avoid unnecessary checks.

Now, let's define the isPali function.

def isPali(self, s, l, r):
    while l < r:
        if s[l] != s[r]:
            return False
        l, r = l + 1, r - 1
    return True

The isPali function checks if a given substring within the string s is a palindrome.

It takes three arguments: the input string s, the left boundary index l, and the right boundary index r.

The function works as follows:

• We use a while loop that continues as long as l is less than r.

This loop compares characters at positions l and r in the string.

• If the characters at these positions are not equal, it means the substring is not a palindrome, so we return False.

• If the characters at positions l and r match, we move l one position to the right and r one position to the left to continue checking the next pair of characters.

• If the loop completes without finding any non-matching characters, the function returns True, indicating that the substring is a palindrome.

With the isPali function in place, we can efficiently determine whether a substring is a palindrome, making our backtracking solution more optimized.

Time and Space Complexity

Let's discuss the time and space complexity of our solution:

Time Complexity:

The time complexity of our backtracking solution is exponential and can be at least O(2^n), where 'n' is the length of the input string s.

This is because we are exploring all possible partitions, and there can be a large number of valid partitions in the worst case.

Space Complexity:

The space complexity is influenced by the recursion stack and the space required to store the current partition.

In the worst case, we might have 2^n recursive calls on the stack, leading to a space complexity of O(2^n).

Additionally, the space needed for the part list is O(n) where 'n' is the length of the input string.

Reasoning Behind Our Approach

Our approach is based on backtracking, which is a powerful technique for exploring all possible combinations or permutations of a problem.

We efficiently generate partitions of the input string, checking if each partition is a palindrome.

If it is, we add it to our results.

The backtracking approach allows us to explore all possible

combinations without missing any valid partitions.

We keep track of the current partition (part) and add substrings to it while ensuring they are palindromes.

When we reach the end of the string or find a partition that doesn't form a palindrome, we backtrack to explore other possibilities.

By efficiently checking for palindromes and using a recursive depth-first search, we can find all valid partitions without unnecessary duplication.

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Conclusion

In this coding session, we tackled the Palindrome Partitioning problem from LeetCode.

We learned how to efficiently generate all possible partitions of a string while ensuring that each substring is a palindrome.

The backtracking approach we discussed provides a robust solution to the problem.

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