Palindrome Partitioning I and II 🎯

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Palindrome Partitioning I and II 🎯

Welcome to CodeYourCraft! Today, we're going to dive into an exciting problem known as Palindrome Partitioning. This problem is a great way to understand the power of dynamic programming and its applications in real-world projects.

Let's first understand what a palindrome is: A palindrome is a word, phrase, number, or other sequence of characters that reads the same forward and backward. For example, "racecar", "madam", and "121" are palindromes.

Palindrome Partitioning Problem 📝

The Palindrome Partitioning problem asks us to partition a given string into substrings such that each substring is a palindrome. Let's take the word "babad", for example:

"bab", "a", "d" is one valid partition since "bab" and "d" are palindromes, and "a" is not a palindrome but it's the only option left.
"ba", "b", "d" is another valid partition.
"bab", "a", "b" is not a valid partition because "bab" is not a palindrome.

Palindrome Partitioning I 💡

Palindrome Partitioning I (also known as Partitioning into Substrings) asks for the number of ways to partition a given string into substrings such that each substring is a palindrome.

Example

Let's take the word "abca":

"a", "b", "c", "a" is one valid partition (4 ways).
"a", "b", "a", "c" is another valid partition (3 ways).
"a", "b", "c", "a" is not a valid partition because "abca" is not a palindrome.

The total number of ways to partition "abca" into palindromes is 7 (4 + 3).

Code Example

python

def count_palindromes(s):
    n = len(s)
    table = [[0 for _ in range(n)] for _ in range(n)]

    # Base case: single character palindrome
    for i in range(n):
        table[i][i] = 1

    # For each length greater than 1, check all possible starting indices
    for k in range(1, n):
        for i in range(n - k):
            j = i + k
            if s[i] == s[j] and k > 1 or k == 1:
                table[i][j] = table[i + 1][j - 1] + (1 if s[i] == s[j] else 0)
            else:
                table[i][j] = table[i][j - 1] + table[i + 1][j]

    return table[0][n - 1]

# Test the function
print(count_palindromes("abca"))  # Output: 7

Palindrome Partitioning II 💡

Palindrome Partitioning II asks us to find all possible partitions of a given string into substrings such that each substring is a palindrome.

Example

Let's take the word "babad":

"bab", "a", "d" is one valid partition.
"ba", "b", "d" is another valid partition.
"bab", "a", "b" is not a valid partition because "bab" is not a palindrome.

The total number of valid partitions for "babad" is 2.

Code Example

python

def partition(s):
    n = len(s)
    dp = [[False] * n for _ in range(n)]

    # Base case: single character palindrome
    for i in range(n):
        dp[i][i] = True

    # For each length greater than 1, check all possible starting indices
    for k in range(1, n):
        for i in range(n - k):
            j = i + k
            if s[i] == s[j] and k > 1 or k == 1:
                if dp[i + 1][j - 1]:
                    dp[i][j] = True
            elif s[i] == s[j - 1]:
                if dp[i + 1][j]:
                    dp[i][j] = True

    # Find all partitions
    partitions = []
    def find_partitions(i, j):
        if i == j:
            partitions.append([s[i:j + 1]])
        elif dp[i][j]:
            partitions.append(find_partitions(i + 1, j - 1))
            partitions.append(find_partitions(i, j - 1))

    find_partitions(0, n - 1)

    # Print the partitions
    for partition in partitions:
        print(partition)

# Test the function
partition("babad")

Quiz

Quick Quiz

Question 1 of 1

What is the main difference between Palindrome Partitioning I and Palindrome Partitioning II?