What is the time complexity of Palindrome Partitioning?

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

What is the space complexity of Palindrome Partitioning?

The space complexity is O(N^2).

What data structures are used to solve Palindrome Partitioning?

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

What are the key insights for solving Palindrome Partitioning?

Insight 1: Dynamic programming can be used to precompute and store whether a substring is a palindrome efficiently. This avoids redundant palindrome checks during the backtracking process. Insight 2: Backtracking is suitable for exploring all possible partitions of the string. Each recursive call explores a potential substring, and if it's a palindrome, it's added to the current path.

Which companies ask LeetCode 131. Palindrome Partitioning in interviews?

This problem is frequently asked at Infosys, Uber, Walmart Labs, Yahoo.

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Palindrome Partitioning - LeetCode 131 Solution

Q: How to solve Palindrome Partitioning?

The provided solution uses a combination of dynamic programming and backtracking. First, it builds a 2D boolean array `dp` to store whether each substring of `s` is a palindrome. Then, it uses a recursive backtracking function to explore all possible partitions. The `dp` array is used to quickly check if a substring is a palindrome during the backtracking process, significantly improving performance.

Palindrome Partitioning - Complete Solution Guide

Palindrome Partitioning is LeetCode problem 131, a Medium 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 , partition s such that every substring of the partition is a palindrome . Return all possible palindrome partitioning of s . Example 1: Input: s = "aab" Output: [["a","a","b"],["aa","b"]] Example 2: Input: s = "a" Output: [["a"]] Constraints: 1 <= s.length <= 16 s contains only lowercase English letters.

Detailed Explanation

The Palindrome Partitioning problem requires finding all possible ways to split a given string `s` into substrings, such that each substring is a palindrome. A palindrome is a string that reads the same forwards and backward (e.g., "aba", "racecar"). The problem aims to return a list of lists, where each inner list represents a valid partition of the original string into palindromic substrings.

Solution Approach

The provided solution uses a combination of dynamic programming and backtracking. First, it builds a 2D boolean array `dp` to store whether each substring of `s` is a palindrome. Then, it uses a recursive backtracking function to explore all possible partitions. The `dp` array is used to quickly check if a substring is a palindrome during the backtracking process, significantly improving performance.

Step-by-Step Algorithm

Step 1: Initialize a 2D boolean array `dp` of size n x n, where n is the length of the string `s`. `dp[i][j]` will store whether the substring s[i...j] is a palindrome.
Step 2: Populate the `dp` array using dynamic programming. Iterate through the string `s` from the end to the beginning. For each substring s[i...j], check if `s[i] == s[j]` and if the inner substring s[i+1...j-1] is a palindrome (or if the substring has length 1 or 2). If both conditions are true, then `dp[i][j]` is set to true.
Step 3: Initialize an empty list `result` to store all valid palindrome partitions and an empty list `path` to store the current partition being explored.
Step 4: Define a recursive backtracking function that takes a `start` index as input. If `start` reaches the end of the string `s`, it means a complete partition has been found, so add a copy of the current `path` to the `result` list.
Step 5: In the backtracking function, iterate through all possible ending positions `end` from `start` to the end of the string. If the substring s[start...end] is a palindrome (i.e., `dp[start][end]` is true), add it to the `path` and recursively call the backtracking function with `start` set to `end + 1`.
Step 6: After the recursive call returns, remove the last element from the `path` to backtrack and explore other possible partitions.
Step 7: Call the backtracking function with `start` set to 0 to begin the search for palindrome partitions.
Step 8: Return the `result` list containing all valid palindrome partitions.

Key Insights

Insight 1: Dynamic programming can be used to precompute and store whether a substring is a palindrome efficiently. This avoids redundant palindrome checks during the backtracking process.
Insight 2: Backtracking is suitable for exploring all possible partitions of the string. Each recursive call explores a potential substring, and if it's a palindrome, it's added to the current path.
Insight 3: Base case for backtracking is when the 'start' index reaches the end of the string. At this point, the current path represents a complete and valid palindrome partitioning.

Complexity Analysis

Time Complexity: O(N*2^N)

Space Complexity: O(N^2)

Topics

This problem involves: String, Dynamic Programming, Backtracking.

Companies

Asked at: Infosys, Uber, Walmart Labs, Yahoo.

Palindrome Partitioning | LeetCode 131

Medium

String Dynamic Programming Backtracking

Time: O(N*2^N)Space: O(N^2)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<List<String>> partition(String s) {
        int n = s.length();
        boolean[][] dp = 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 || dp[i + 1][j - 1])) {
                    dp[i][j] = true;
                }
            }
        }

        List<List<String>> result = new ArrayList<>();
        List<String> path = new ArrayList<>();

        backtrack(s, 0, dp, result, path);
        return result;
    }

    private void backtrack(String s, int start, boolean[][] dp, List<List<String>> result, List<String> path) {
        int n = s.length();
        if (start == n) {
            result.add(new ArrayList<>(path));
            return;
        }

        for (int end = start; end < n; end++) {
            if (dp[start][end]) {
                path.add(s.substring(start, end + 1));
                backtrack(s, end + 1, dp, result, path);
                path.remove(path.size() - 1);
            }
        }
    }
}

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Problem Info

Description

Problem 131: Palindrome Partitioning - Detailed explanation and solution approaches.

Topics

String Dynamic Programming Backtracking

Companies

InfosysUberWalmart LabsYahoo

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:131

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

8 steps

1
Step 1: Initialize a 2D boolean array `dp` of size n x n, where n is the length of the string `s`. `dp[i][j]` will store whether the substring s[i...j] is a palindrome.
2
Step 2: Populate the `dp` array using dynamic programming. Iterate through the string `s` from the end to the beginning. For each substring s[i...j], check if `s[i] == s[j]` and if the inner substring s[i+1...j-1] is a palindrome (or if the substring has length 1 or 2). If both conditions are true, then `dp[i][j]` is set to true.
3
Step 3: Initialize an empty list `result` to store all valid palindrome partitions and an empty list `path` to store the current partition being explored.
4
Step 4: Define a recursive backtracking function that takes a `start` index as input. If `start` reaches the end of the string `s`, it means a complete partition has been found, so add a copy of the current `path` to the `result` list.
5
Step 5: In the backtracking function, iterate through all possible ending positions `end` from `start` to the end of the string. If the substring s[start...end] is a palindrome (i.e., `dp[start][end]` is true), add it to the `path` and recursively call the backtracking function with `start` set to `end + 1`.
6
Step 6: After the recursive call returns, remove the last element from the `path` to backtrack and explore other possible partitions.
7
Step 7: Call the backtracking function with `start` set to 0 to begin the search for palindrome partitions.
8
Step 8: Return the `result` list containing all valid palindrome partitions.

💡

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^N)

💾

Space Complexity

O(N^2)

📝

Why?

The time complexity arises from the exponential nature of exploring all possible partitions using backtracking. The dynamic programming precomputation helps reduce the complexity of each palindrome check to O(1), but the overall number of partitions to explore remains exponential. The space complexity is dominated by the `dp` array used for storing palindrome information. The recursion stack and the path list contribute a smaller O(N) component to the overall space complexity.

📖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

Without Dynamic Programming

Instead of precomputing palindromes with DP, check if a substring is a palindrome each time it's encountered during backtracking. This can be achieved by reversing the substring and comparing it to the original.

Time: O(N * N * 2^N)Space: O(N)

Trade-offs: Avoids the O(N^2) space of DP, but increases time complexity due to repeated palindrome checks (O(N) for each check). Suitable for very small input strings where the DP overhead might be more significant.

✨

Key Insights

1Insight 1: Dynamic programming can be used to precompute and store whether a substring is a palindrome efficiently. This avoids redundant palindrome checks during the backtracking process.
2Insight 2: Backtracking is suitable for exploring all possible partitions of the string. Each recursive call explores a potential substring, and if it's a palindrome, it's added to the current path.
3Insight 3: Base case for backtracking is when the 'start' index reaches the end of the string. At this point, the current path represents a complete and valid palindrome partitioning.

📚

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 algorithm implicitly handles this case, returning an empty list of partitions. No explicit check is needed.
Edge case 2: Single-character string - The algorithm correctly identifies the single character as a palindrome and returns a list containing a single partition with the single character.
Edge case 3: String with no palindromic partitions (e.g., 'abc') - The algorithm correctly explores all possibilities and returns an empty list.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to create a copy of the path before adding it to the result list. The `path` list is modified during backtracking, so directly adding it to the result would lead to all partitions being the same.

✗Common mistake 2: Incorrectly implementing the dynamic programming logic for palindrome detection. A common error is not considering the base cases (single-character or two-character palindromes) correctly.

✗Common mistake 3: Not properly handling the backtracking step, leading to incorrect or incomplete partitions. It's crucial to remove the last added substring from the path before exploring other possibilities.

✅

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 - Identifying palindromic phrases or words within a text for linguistic analysis or pattern recognition.

🏦Real-world scenario 2: Bioinformatics - Identifying palindromic sequences in DNA or RNA, which can have biological significance.

🎮Real-world scenario 3: Data compression - Palindromic patterns can be exploited for data compression by storing only one half of the palindrome.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

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

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 10, 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

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

@AlgoMasterJul 14

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

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

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.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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"?

For this medium-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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