How do you solve Palindrome Partitioning IV on LeetCode?

Palindrome Partitioning IV (LeetCode #1745) 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: Preprocessing palindromes can save time in checking splits.

What is the brute force solution for Palindrome Partitioning IV?

The brute force approach for Palindrome Partitioning IV is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try every possible way to split the string into three parts and check if each part is a palindrome. This is straightforward but inefficient as it checks all combinations. The optimal Optimal Solution approach improves this to O(n²).

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

Palindrome Partitioning IV 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 IV?

Explain your thought process clearly while coding. Consider edge cases before finalizing your solution. Practice writing clean and efficient code.

What are common mistakes when solving Palindrome Partitioning IV?

Not checking for valid substring lengths (must be non-empty). Overlooking edge cases where the string length is minimal.

#1745

Palindrome Partitioning IV

Hard

StringDynamic ProgrammingDynamic ProgrammingTwo Pointers

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)

We can preprocess the string to find all palindromic substrings and then use this information to efficiently check for valid partitions. This reduces redundant checks.

⚙️

Algorithm

3 steps

1Step 1: Create a 2D array to store whether substrings are palindromes.
2Step 2: Fill this array by checking all substrings in a single pass.
3Step 3: Use two nested loops to find valid split points and check the preprocessed palindrome array.

solution.py15 lines

1def checkPartition(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            if s[start] == s[end]:
10                dp[start][end] = (length == 2) or dp[start + 1][end - 1]
11    for i in range(1, n - 1):
12        for j in range(i + 1, n):
13            if dp[0][i - 1] and dp[i][j - 1] and dp[j][n - 1]:
14                return True
15    return False

ℹ

Complexity note: The complexity is O(n²) due to the preprocessing step where we fill the palindrome table, but this is more efficient than the brute force approach because we avoid redundant palindrome checks.

1Preprocessing palindromes can save time in checking splits.
2Using a 2D array to store palindrome checks avoids redundant calculations.

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