Longest Palindromic Subsequence
Detailed guide and Python implementation for the 'Longest Palindromic Subsequence' problem.
1. Concept Overview
The 'Longest Palindromic Subsequence' problem is a key challenge in the Dynamic Programming section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Longest Palindromic Subsequence.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Longest Palindromic Subsequence carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Write a function lps(s) that finds the length of the longest palindromic subsequence in a string 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.
- •1 <= len(s) <= 1000
- •s consists of uppercase or lowercase English letters.
Examples
lps('BBABCBCAB')7
The longest palindromic subsequence has length 7, e.g., 'BABCBAB'.
lps('GEEKSFORGEEKS')5
The longest palindromic subsequence has length 5, e.g., 'EEKEE'.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
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Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def lps_opt(s):
n = len(s); dp = [[0] * n for _ in range(n)]
for i in range(n - 1, -1, -1):
dp[i][i] = 1
for j in range(i + 1, n):
if s[i] == s[j]: dp[i][j] = dp[i + 1][j - 1] + 2
else: dp[i][j] = max(dp[i + 1][j], dp[i][j - 1])
return dp[0][n - 1]Brute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def lps_brute(s):
def solve(i, j):
if i == j: return 1
if i > j: return 0
if s[i] == s[j]: return 2 + solve(i + 1, j - 1)
return max(solve(i + 1, j), solve(i, j - 1))
return solve(0, len(s) - 1)Algorithm Pattern Checklist
When dealing with Dynamic Programming data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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