What is the time complexity of Palindrome Partitioning IV?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Palindrome Partitioning IV?

This problem uses String, Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Palindrome Partitioning IV?

Insight 1: Pre-compute all possible palindromic substrings using dynamic programming to avoid redundant palindrome checks. Insight 2: Iterate through all possible split points to check if three palindromic substrings can be formed.

Which companies ask LeetCode 1745. Palindrome Partitioning IV in interviews?

This problem is frequently asked at tcs.

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Palindrome Partitioning IV - LeetCode 1745 Solution

Q: How to solve Palindrome Partitioning IV?

The solution uses dynamic programming to pre-compute all palindromic substrings of the input string `s`. It then iterates through all possible split points (i and j) to check if the substrings s[0:i], s[i:j], and s[j:n] are all palindromes, where n is the length of the string `s`. If such split points are found, it returns true; otherwise, it returns false.

Q: What is the space complexity of Palindrome Partitioning IV?

The space complexity is O(n^2).

Palindrome Partitioning IV - Complete Solution Guide

Palindrome Partitioning IV is LeetCode problem 1745, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 on

Detailed Explanation

The problem asks us to determine if a given string `s` can be split into three non-empty palindromic substrings. The input is the string `s`, and the output is a boolean value: `true` if the split is possible, and `false` otherwise. A palindrome is a string that reads the same forwards and backward. The string `s` consists only of lowercase English letters, and its length is between 3 and 2000 characters.

Solution Approach

The solution uses dynamic programming to pre-compute all palindromic substrings of the input string `s`. It then iterates through all possible split points (i and j) to check if the substrings s[0:i], s[i:j], and s[j:n] are all palindromes, where n is the length of the string `s`. If such split points are found, it returns true; otherwise, it returns false.

Step-by-Step Algorithm

Step 1: Create a 2D boolean array `is_pal` of size n x n, where n is the length of the string `s`. `is_pal[i][j]` will be true if the substring s[i:j+1] is a palindrome, and false otherwise.
Step 2: Populate the `is_pal` array using dynamic programming. Iterate from the bottom right to the top left of the array. For each cell `is_pal[i][j]`, check if s[i] == s[j]. If they are equal, and either the substring length is less than 2 (i.e., j - i < 2) or the inner substring s[i+1:j] is a palindrome (i.e., `is_pal[i+1][j-1]` is true), then `is_pal[i][j]` is true. Otherwise, `is_pal[i][j]` is false.
Step 3: Iterate through all possible split points `i` and `j` such that 1 <= i < n - 1 and i + 1 <= j < n. Check if the substrings s[0:i], s[i:j], and s[j:n] are all palindromes using the pre-computed `is_pal` array. Specifically, check if `is_pal[0][i-1]`, `is_pal[i][j-1]`, and `is_pal[j][n-1]` are all true.
Step 4: If a combination of `i` and `j` is found such that all three substrings are palindromes, return true.
Step 5: If no such combination is found after iterating through all possible split points, return false.

Key Insights

Insight 1: Pre-compute all possible palindromic substrings using dynamic programming to avoid redundant palindrome checks.
Insight 2: Iterate through all possible split points to check if three palindromic substrings can be formed.
Insight 3: Consider all possible combinations of split points to ensure that all valid combinations of palindromic substrings are considered.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

This problem involves: String, Dynamic Programming.

Companies

Asked at: tcs.

Palindrome Partitioning IV | LeetCode 1745

Hard

String Dynamic Programming

Time: O(n^2)Space: O(n^2)

Solution Code

Available in:

class Solution {
    public boolean checkPartitioning(String s) {
        int n = s.length();
        boolean[][] is_pal = new boolean[n][n];

        for (int i = n - 1; i >= 0; i--) {
            for (int j = i; j < n; j++) {
                if (s.charAt(i) == s.charAt(j) && (j - i < 2 || is_pal[i+1][j-1])) {
                    is_pal[i][j] = true;
                }
            }
        }

        for (int i = 1; i < n - 1; i++) {
            if (is_pal[0][i-1]) {
                for (int j = i + 1; j < n; j++) {
                    if (is_pal[i][j-1] && is_pal[j][n-1]) {
                        return true;
                    }
                }
            }
        }

        return false;
    }
}

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Problem Info

Description

Problem 1745: Palindrome Partitioning IV - Detailed explanation and solution approaches.

Topics

String Dynamic Programming

Companies

tcs

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1745

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Create a 2D boolean array `is_pal` of size n x n, where n is the length of the string `s`. `is_pal[i][j]` will be true if the substring s[i:j+1] is a palindrome, and false otherwise.
2
Step 2: Populate the `is_pal` array using dynamic programming. Iterate from the bottom right to the top left of the array. For each cell `is_pal[i][j]`, check if s[i] == s[j]. If they are equal, and either the substring length is less than 2 (i.e., j - i < 2) or the inner substring s[i+1:j] is a palindrome (i.e., `is_pal[i+1][j-1]` is true), then `is_pal[i][j]` is true. Otherwise, `is_pal[i][j]` is false.
3
Step 3: Iterate through all possible split points `i` and `j` such that 1 <= i < n - 1 and i + 1 <= j < n. Check if the substrings s[0:i], s[i:j], and s[j:n] are all palindromes using the pre-computed `is_pal` array. Specifically, check if `is_pal[0][i-1]`, `is_pal[i][j-1]`, and `is_pal[j][n-1]` are all true.
4
Step 4: If a combination of `i` and `j` is found such that all three substrings are palindromes, return true.
5
Step 5: If no such combination is found after iterating through all possible split points, return false.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity is O(n^2) because we iterate through all possible substrings to build the `is_pal` array, which takes O(n^2) time. Then we iterate through all possible pairs of split points (i, j), which also takes O(n^2) time. The space complexity is O(n^2) because we store the `is_pal` array, which is of size n x n.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Iterate through all possible split points and check if each substring is a palindrome. This would involve redundant palindrome checks.

Time: O(n^3)Space: O(1)

Trade-offs: This approach is simpler to understand but has a higher time complexity, making it less efficient for larger strings. The dynamic programming solution trades space complexity for time complexity.

✨

Key Insights

1Insight 1: Pre-compute all possible palindromic substrings using dynamic programming to avoid redundant palindrome checks.
2Insight 2: Iterate through all possible split points to check if three palindromic substrings can be formed.
3Insight 3: Consider all possible combinations of split points to ensure that all valid combinations of palindromic substrings are considered.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty string. The problem statement says the length of string s is between 3 and 2000, so this is not a valid edge case for this specific problem.
Edge case 2: String of length 3 consisting of the same character (e.g., "aaa"). The algorithm correctly identifies that this string can be split into three palindromes, namely "a", "a", and "a".
Edge case 3: String that is itself a palindrome (e.g., "aba"). The algorithm will correctly return `false` because it cannot be split into *three* non-empty palindromic substrings.
Edge case 4: String where only one split can be made (e.g., "aabba"). Even though "aa" and "bba" are palindromes, we need three palindromes to return `true`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check the length constraint of the palindromic substrings. The problem requires three *non-empty* palindromes.

✗Common mistake 2: Not correctly implementing the dynamic programming approach for palindrome detection, leading to incorrect results.

✗Common mistake 3: Incorrectly iterating through the possible split points, resulting in some valid combinations being missed.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Text processing and analysis, where identifying palindromic sequences can be useful for tasks like bioinformatics (e.g., analyzing DNA sequences) or natural language processing (e.g., sentiment analysis).

🏦Real-world scenario 2: Data compression algorithms, where identifying palindromic patterns can help reduce the amount of data needed to represent a string. Though, this specific 3-palindrome problem isn't directly used, the principle of palindrome detection is.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 4, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Palindrome Partitioning IV"?

For this hard-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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