How do you solve Palindrome Partitioning on LeetCode?

Palindrome Partitioning (LeetCode #131) 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 how to efficiently check for palindromes is crucial for optimizing the solution.

What is the brute force solution for Palindrome Partitioning?

The brute force approach for Palindrome Partitioning is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach generates all possible substrings of the input string and checks each substring to see if it is a palindrome. If it is, we can recursively partition the rest of the string. This method is straightforward but inefficient due to the exponential number of possible partitions. The optimal Optimal Solution approach improves this to O(n²).

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

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

What are the interview tips for Palindrome Partitioning?

Always consider edge cases, such as single character strings or strings with all identical characters. Explain your thought process clearly while coding, as it shows your understanding of the problem. Practice writing both brute force and optimized solutions to solidify your understanding.

What are common mistakes when solving Palindrome Partitioning?

Not using a helper function to check for palindromes, leading to repeated code. Failing to backtrack correctly, which can lead to incorrect results.

#131

Palindrome Partitioning

Medium

StringDynamic ProgrammingBacktrackingBacktrackingDynamic Programming

LeetCode ↗

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 precompute a table that tells us if any substring is a palindrome. This allows us to efficiently build partitions without checking each substring multiple times.

⚙️

Algorithm

3 steps

1Step 1: Create a 2D array to store palindrome information for substrings.
2Step 2: Fill the array using dynamic programming to mark palindromic substrings.
3Step 3: Use backtracking to generate partitions based on the precomputed palindrome information.

solution.py25 lines

1def partition(s):
2    n = len(s)
3    dp = [[False] * n for _ in range(n)]
4    for i in range(n):
5        dp[i][i] = True
6    for length in range(2, n + 1):
7        for start in range(n - length + 1):
8            end = start + length - 1
9            dp[start][end] = (s[start] == s[end]) and (length == 2 or dp[start + 1][end - 1])
10    result = []
11    backtrack(0, [], result, s, dp)
12    return result
13
14def backtrack(start, path, result, s, dp):
15    if start == len(s):
16        result.append(path[:])
17        return
18    for end in range(start + 1, len(s) + 1):
19        if dp[start][end - 1]:
20            path.append(s[start:end])
21            backtrack(end, path, result, s, dp)
22            path.pop()
23
24# Example usage:
25print(partition('aab'))

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Complexity note: The time complexity is O(n²) due to filling the palindrome table and the backtracking process. The space complexity is O(n²) because of the 2D array used to store palindrome information.

1Understanding how to efficiently check for palindromes is crucial for optimizing the solution.
2Dynamic programming can significantly reduce the time complexity by avoiding redundant calculations.

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