DSA Tracker

Medium

Longest Palindromic Subsequence

A medium Dynamic Programming problem included in Love Babbar 450, Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.

Topic
Dynamic Programming
Sheets
2
Core for
9 roles
Platform
LeetCode
Practice in the compilerWatch this solved step by step

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 2026-09-09

Compare 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

  1. 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.

  2. 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.

  3. 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.

What each operation costs

OperationTime
fill dynamic programming table of n statesO(n)
solve two-dimensional grid of m by n statesO(m * n)
space-optimized state transition keeping one rowO(n)
What usually goes wrong with Dynamic Programming
  • Filling a bottom-up table in an order where the current cell needs values that have not been computed yet, reading uninitialized zeros.
  • Failing to initialize base cases properly, such as filling a minimization table with zeros instead of infinity, which traps the answer at zero.
  • Overwriting values in a 1D space-optimized knapsack array by scanning in the wrong direction, allowing the same item to be chosen multiple times.

Which roles need this problem

Dynamic Programming is a core topic for these 9 roles — if you're targeting one of them, this problem is early in your path, not optional.

SDE / Backend EngineerML EngineerInformation Retrieval EngineerDatabase EngineerComputer Vision EngineerNLP EngineerCompiler / PL EngineerQuant DeveloperBioinformatics Engineer

Secondary for 5 more roles, including Game Developer, Cryptography Engineer, Performance Engineer.

Track this in your role's order

Pick your target role and all 370 problems — including this one — resequence to what that interview actually asks. Free.

Start free

More Dynamic Programming problems

Problem set and role mapping as of .