What is the time complexity of Longest Palindromic Subsequence?

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

What data structures are used to solve Longest Palindromic Subsequence?

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

What are the key insights for solving Longest Palindromic Subsequence?

Insight 1: The core idea is to use dynamic programming to build up solutions for smaller subproblems and combine them to solve the larger problem. Insight 2: We can define a 2D DP table where dp[i][j] stores the length of the longest palindromic subsequence of the substring s[i...j].

Which companies ask LeetCode 516. Longest Palindromic Subsequence in interviews?

This problem is frequently asked at Cisco, LinkedIn, Nutanix.

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Longest Palindromic Subsequence - LeetCode 516 Solution

Q: How to solve Longest Palindromic Subsequence?

The provided solutions use dynamic programming to solve this problem. A 2D array 'dp' is used to store the lengths of the longest palindromic subsequences for all possible substrings of the input string 's'. The 'dp' array is filled in a bottom-up manner, starting with substrings of length 1 and gradually increasing the substring length. The algorithm iterates through the string from the end to the beginning (outer loop), and for each index i, it iterates from i+1 to the end of the string (inner loop). The value of dp[i][j] is determined based on whether the characters s[i] and s[j] are equal. If they are equal, it means we can extend the palindromic subsequence found in s[i+1...j-1] by including s[i] and s[j]. If they are not equal, we take the maximum length between the longest palindromic subsequence found by excluding either s[i] or s[j]. Finally, dp[0][n-1] will hold the length of the longest palindromic subsequence of the entire string.

Q: What is the space complexity of Longest Palindromic Subsequence?

The space complexity is O(n^2).

Longest Palindromic Subsequence - Complete Solution Guide

Longest Palindromic Subsequence is LeetCode problem 516, 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 , find the longest palindromic subsequence 's length in s . A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements. Example 1: Input: s = "bbbab" Output: 4 Explanation: One possible longest palindromic subsequence is "bbbb". Example 2: Input: s = "cbbd" Output: 2 Explanation: One possible longest palindromic subsequence is "bb". Constraints: 1 <= s.length <= 1000 s consists only o

Detailed Explanation

The problem asks us to find the length of the longest palindromic subsequence within a given string 's'. A subsequence is formed by deleting zero or more characters from the original string without changing the order of the remaining characters. A palindrome is a sequence that reads the same forwards and backward. The goal is to find the longest subsequence of 's' that is also a palindrome.

Solution Approach

The provided solutions use dynamic programming to solve this problem. A 2D array 'dp' is used to store the lengths of the longest palindromic subsequences for all possible substrings of the input string 's'. The 'dp' array is filled in a bottom-up manner, starting with substrings of length 1 and gradually increasing the substring length. The algorithm iterates through the string from the end to the beginning (outer loop), and for each index i, it iterates from i+1 to the end of the string (inner loop). The value of dp[i][j] is determined based on whether the characters s[i] and s[j] are equal. If they are equal, it means we can extend the palindromic subsequence found in s[i+1...j-1] by including s[i] and s[j]. If they are not equal, we take the maximum length between the longest palindromic subsequence found by excluding either s[i] or s[j]. Finally, dp[0][n-1] will hold the length of the longest palindromic subsequence of the entire string.

Step-by-Step Algorithm

Step 1: Initialize a 2D array 'dp' of size n x n with all elements set to 0, where n is the length of the input string 's'.
Step 2: Iterate through the string 's' from the last character to the first (i from n-1 down to 0).
Step 3: For each character s[i], set dp[i][i] = 1, because a single character is a palindrome of length 1.
Step 4: Iterate through the string from i+1 to the end (j from i+1 to n-1).
Step 5: If s[i] == s[j], then dp[i][j] = 2 + dp[i+1][j-1]. This means the longest palindromic subsequence of s[i...j] includes both s[i] and s[j], adding 2 to the length of the longest palindromic subsequence of s[i+1...j-1].
Step 6: If s[i] != s[j], then dp[i][j] = max(dp[i+1][j], dp[i][j-1]). This means the longest palindromic subsequence of s[i...j] is either the longest palindromic subsequence of s[i+1...j] or s[i...j-1], whichever is longer.
Step 7: After the iterations are complete, dp[0][n-1] will contain the length of the longest palindromic subsequence of the entire string 's'.
Step 8: Return dp[0][n-1].

Key Insights

Insight 1: The core idea is to use dynamic programming to build up solutions for smaller subproblems and combine them to solve the larger problem.
Insight 2: We can define a 2D DP table where dp[i][j] stores the length of the longest palindromic subsequence of the substring s[i...j].
Insight 3: The relationship between dp[i][j] and its subproblems is crucial: if s[i] == s[j], then dp[i][j] = 2 + dp[i+1][j-1]; otherwise, dp[i][j] = max(dp[i+1][j], dp[i][j-1]).

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

This problem involves: String, Dynamic Programming.

Companies

Asked at: Cisco, LinkedIn, Nutanix.

Longest Palindromic Subsequence | LeetCode 516

Medium

String Dynamic Programming

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

Solution Code

Available in:

class Solution {
    public int longestPalindromeSubseq(String s) {
        int n = s.length();
        int[][] dp = new int[n][n];

        for (int i = n - 1; i >= 0; i--) {
            dp[i][i] = 1;
            for (int j = i + 1; j < n; j++) {
                if (s.charAt(i) == s.charAt(j)) {
                    dp[i][j] = 2 + (i + 1 <= j - 1 ? dp[i + 1][j - 1] : 0);
                } else {
                    dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
                }
            }
        }
        
        return dp[0][n - 1];
    }
}

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Problem Info

Description

Problem 516: Longest Palindromic Subsequence - Detailed explanation and solution approaches.

Topics

String Dynamic Programming

Companies

CiscoLinkedInNutanix

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:516

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 array 'dp' of size n x n with all elements set to 0, where n is the length of the input string 's'.
2
Step 2: Iterate through the string 's' from the last character to the first (i from n-1 down to 0).
3
Step 3: For each character s[i], set dp[i][i] = 1, because a single character is a palindrome of length 1.
4
Step 4: Iterate through the string from i+1 to the end (j from i+1 to n-1).
5
Step 5: If s[i] == s[j], then dp[i][j] = 2 + dp[i+1][j-1]. This means the longest palindromic subsequence of s[i...j] includes both s[i] and s[j], adding 2 to the length of the longest palindromic subsequence of s[i+1...j-1].
6
Step 6: If s[i] != s[j], then dp[i][j] = max(dp[i+1][j], dp[i][j-1]). This means the longest palindromic subsequence of s[i...j] is either the longest palindromic subsequence of s[i+1...j] or s[i...j-1], whichever is longer.
7
Step 7: After the iterations are complete, dp[0][n-1] will contain the length of the longest palindromic subsequence of the entire string 's'.
8
Step 8: Return dp[0][n-1].

💡

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 algorithm uses nested loops to fill the dp table. The outer loop iterates 'n' times, and the inner loop also iterates at most 'n' times. This results in a time complexity of O(n*n) = O(n^2). The algorithm uses a 2D array 'dp' of size n x n to store the lengths of the longest palindromic subsequences. Therefore, the space complexity is O(n^2).

📖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

Recursion with Memoization

We can solve this problem using a recursive approach with memoization. The idea is to define a recursive function that computes the length of the longest palindromic subsequence for a given substring. We can memoize the results to avoid recomputing the same subproblems.

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

Trade-offs: Recursion with memoization is conceptually similar to dynamic programming but can be slightly slower due to the overhead of recursive calls. It might be more intuitive for some developers.

Space Optimized DP

The DP solution can be optimized for space from O(n^2) to O(n). Since each row of the DP table only depends on the previous row and elements within the current row, we can maintain just the current and previous rows. This can be achieved using two 1D arrays.

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

Trade-offs: This alternative approach reduces the memory requirement significantly. It's a good idea when memory usage is a constraint.

✨

Key Insights

1Insight 1: The core idea is to use dynamic programming to build up solutions for smaller subproblems and combine them to solve the larger problem.
2Insight 2: We can define a 2D DP table where dp[i][j] stores the length of the longest palindromic subsequence of the substring s[i...j].
3Insight 3: The relationship between dp[i][j] and its subproblems is crucial: if s[i] == s[j], then dp[i][j] = 2 + dp[i+1][j-1]; otherwise, dp[i][j] = max(dp[i+1][j], dp[i][j-1]).

📚

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 code doesn't explicitly handle an empty string as input. However, it will implicitly work, as the loops will not execute, and the initial dp array will remain all zeros, resulting in 0 being returned. If the problem definition specified a different expected behavior for empty strings it would need explicit handling.
Edge case 2: Single character string: The algorithm correctly handles single-character strings as the initialization step dp[i][i] = 1 covers this and will result in the correct output of 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to initialize the dp array properly. The diagonal elements (dp[i][i]) should be initialized to 1 because a single character is a palindrome of length 1.

✗Common mistake 2: Incorrectly defining the recurrence relation. The conditions for s[i] == s[j] and s[i] != s[j] must be handled correctly to ensure the correct result.

✗Common mistake 3: Iterating the dp array in the wrong order. The algorithm must fill the dp array in a bottom-up manner to ensure that all necessary subproblems are solved before they are needed.

✅

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: Bioinformatics: Finding similarities between DNA sequences. Palindromic subsequences can be used to identify regions of similarity or conserved regions in DNA or RNA sequences.

🏦Real-world scenario 2: Text editing: Spell checkers and autocorrect tools can use longest palindromic subsequence to find the closest correct word when a user makes a typo.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: November 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

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

@AlgoMasterJul 13

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

@InterviewProJul 6

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Longest Palindromic Subsequence"?

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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