How do you solve Palindrome Partitioning II on LeetCode?

Palindrome Partitioning II (LeetCode #132) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n²) space complexity. Key insight: Understanding palindromes is crucial for this problem.

What is the brute force solution for Palindrome Partitioning II?

The brute force approach for Palindrome Partitioning II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible substring of the string to see if it is a palindrome. If it is, we can make a cut there and continue checking the rest of the string recursively. This method is simple but inefficient. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Palindrome Partitioning II?

Palindrome Partitioning II uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Two Pointers.

What are the interview tips for Palindrome Partitioning II?

Explain your thought process clearly. Discuss edge cases, like empty strings or single characters. Practice writing clean and efficient code.

What are common mistakes when solving Palindrome Partitioning II?

Not checking for palindromes efficiently. Forgetting to store intermediate results in dynamic programming.

#132

Palindrome Partitioning II

Hard

StringDynamic ProgrammingDynamic ProgrammingTwo Pointers

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n²)

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Intuition

Time O(n²)Space O(n²)

The optimal solution uses dynamic programming to store results of subproblems. We can create a table to keep track of whether substrings are palindromes and another array to store the minimum cuts needed for each position in the string.

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Algorithm

4 steps

1Step 1: Create a 2D array `isPalindrome` to track if substrings are palindromes.
2Step 2: Create a 1D array `cuts` to store the minimum cuts needed up to each index.
3Step 3: Fill the `isPalindrome` table by checking each substring.
4Step 4: For each character, calculate the minimum cuts needed using the `cuts` array.

solution.py19 lines

1# Full working Python code
2
3def min_cuts(s):
4    n = len(s)
5    if n == 0:
6        return 0
7    is_palindrome = [[False] * n for _ in range(n)]
8    cuts = [0] * n
9    for i in range(n):
10        min_cuts_needed = i
11        for j in range(i + 1):
12            if s[j] == s[i] and (i - j < 2 or is_palindrome[j + 1][i - 1]):
13                is_palindrome[j][i] = True
14                min_cuts_needed = 0 if j == 0 else min(min_cuts_needed, cuts[j - 1] + 1)
15        cuts[i] = min_cuts_needed
16    return cuts[-1]
17
18# Example usage
19print(min_cuts('aab'))

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Complexity note: The time complexity is O(n²) due to the nested loops checking each substring for palindrome status. The space complexity is O(n²) because we maintain a 2D array to store palindrome checks.

1Understanding palindromes is crucial for this problem.
2Dynamic programming can optimize recursive solutions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.