Is Palindrome Partitioning IV easy or hard?

Palindrome Partitioning IV is labeled Hard because it combines substring DP with partition enumeration. The challenge is recognizing that repeated palindrome checks must be precomputed to avoid O(n^3) brute-force solutions.

Palindrome Partitioning IV Python/Java solution

Python and Java implementations typically build a boolean dp[n][n] table for palindrome checks, then iterate over two partition indices to verify three palindromic segments. The logic is identical across languages with O(n^2) preprocessing and constant-time substring validation.

How to solve Palindrome Partitioning IV in O(n^2)?

Precompute whether every substring is a palindrome using dynamic programming or center expansion. After preprocessing, iterate over two split indices i and j and verify if s[0..i], s[i+1..j], and s[j+1..n-1] are all palindromes. Each check becomes O(1), keeping the total runtime O(n^2).

What is the best approach for Palindrome Partitioning IV?

Dynamic programming with palindrome precomputation is the most common approach. Build a DP table where dp[i][j] indicates whether substring s[i..j] is a palindrome, then iterate over two partition points to check if three segments are palindromes. This runs in O(n^2) time and O(n^2) space and avoids repeated substring checks.

Is Palindrome Partitioning IV asked at Google/Amazon/Meta?

Palindrome partitioning and substring DP problems frequently appear in interviews at companies like Google, Amazon, and Meta. Variants involving palindrome checks, partition counts, or minimum cuts are common dynamic programming interview questions.

What data structure is used in Palindrome Partitioning IV?

Most solutions use a 2D boolean DP table to store whether substring ranges are palindromes. Some implementations use center expansion with a matrix or set of palindrome ranges. The core techniques involve string processing and dynamic programming.

What is the time complexity of Palindrome Partitioning IV?

The optimal solutions run in O(n^2) time. Both the dynamic programming approach and the center expansion technique spend O(n^2) time identifying palindromic substrings and testing possible partition points. Space complexity is typically O(n^2) for storing palindrome ranges.

Description Hints 2 Solution Video 10:09

1745. Palindrome Partitioning IV

Hard45.8% acceptance1 company

Given a string s, return true if it is possible to split the string s into three non-empty palindromic substrings. Otherwise, return false.

A string is said to be palindrome if it the same string when reversed.

Example 1:

Input: s = "abcbdd"
Output: true
Explanation: "abcbdd" = "a" + "bcb" + "dd", and all three substrings are palindromes.

Example 2:

Input: s = "bcbddxy"
Output: false
Explanation: s cannot be split into 3 palindromes.

Constraints:

3 <= s.length <= 2000
s consists only of lowercase English letters.

Asked at 1 company

TCS

Topics

Dynamic Programming String

Hints open one at a time

Hint 1

Preprocess checking palindromes in O(1)

Hint 2

Note that one string is a prefix and another one is a suffix you can try brute forcing the rest

10:09

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