Longest Palindromic Subsequence
A medium Dynamic Programming interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Dynamic Programming
- Sheets
- 2
- Core for
- 9 roles
The problem
Given a string, find the length of the longest palindromic subsequence. A subsequence is derived by deleting some characters without changing the order of the remaining characters.
Example 1
- Input
- s = "bbbab"
- Output
- 4
- Why
- One longest palindromic subsequence is "bbbb", length 4.
Example 2
- Input
- s = "cbbd"
- Output
- 2
- Why
- The longest palindromic subsequence is "bb", length 2.
Example 3
- Input
- s = "a"
- Output
- 1
- Why
- A single character is a palindrome of length 1.
Constraints
- 1 <= s.length <= 1000
- s consists of lowercase English letters
How to think about it
Updated 2026-09-29Compare the boundary characters s[i] and s[j]. If they match, both belong to the outer shell of the palindrome, adding 2 to the longest palindromic subsequence of the interior s[i+1..j-1]. If they differ, the best palindrome must lie entirely within s[i+1..j] or s[i..j-1].
Approaches, worst first
Recursive boundary contraction
time O(2^n) · space O(n)
Recurse on bounds (i, j). If s[i] == s[j], return 2 + solve(i+1, j-1); else return max(solve(i+1, j), solve(i, j-1)). Explores exponential redundant intervals.
Reduction to LCS with reversed string
time O(n^2) · space O(n)
Reverse string s and compute the Longest Common Subsequence between s and its reverse. The resulting LCS length is identical to the longest palindromic subsequence.
Direct interval dynamic programmingWrite this one
time O(n^2) · space O(n^2)
Fill table dp[i][j] iterating substring lengths from 1 to n. For length 1, dp[i][i] = 1. For matching characters, dp[i][j] = dp[i+1][j-1] + 2; otherwise dp[i][j] = max(dp[i+1][j], dp[i][j-1]).
Where people lose marks · 3
- Length 2 base case: when j == i + 1 and characters match, dp[i+1][j-1] refers to an empty range (length 0); taking 2 + 0 = 2 avoids negative-length lookups.
- Incorrect iteration order: calculating interval (i, j) before interior (i+1, j-1) is solved.
- Assuming longest palindromic substring equals longest palindromic subsequence. Subsequences allow deleting intermediate characters, making them longer or equal.
The theory behind it
Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems
What Dynamic Programming is
Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.
When to reach for it
Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.
How the pattern works
Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.
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