516. Longest Palindromic Subsequence
mediumFind the length of the longest subsequence of s that is also a palindrome. Classic interval DP: dp[i][j] = length of the longest palindromic subsequence inside s[i..j], filled in increasing-length order.
By Sam K., Founder, InterviewChamp.AI · Last verified
Problem
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.
Constraints
1 <= s.length <= 1000s consists only of lowercase English letters.
Examples
Example 1
s = "bbbab"4Explanation: One possible longest palindromic subsequence is "bbbb".
Example 2
s = "cbbd"2Explanation: One possible longest palindromic subsequence is "bb".
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Hints
Progressive — try the first before opening the next.
Hint 1
Interval DP. dp[i][j] = length of the longest palindromic subsequence in s[i..j].
Hint 2
If s[i] == s[j]: dp[i][j] = dp[i+1][j-1] + 2.
Hint 3
Else: dp[i][j] = max(dp[i+1][j], dp[i][j-1]).
Hint 4
Fill in increasing interval length so dp[i+1][j-1] is ready when needed. Base: dp[i][i] = 1.
Solution approach
Reveal approach
Bottom-up interval DP. dp[i][j] is the length of the longest palindromic subsequence in s[i..j]. Initialize the diagonal dp[i][i] = 1 (a single char is its own length-1 palindrome). Iterate length = 2..n and i = 0..n-length, set j = i + length - 1. If s[i] == s[j], dp[i][j] = dp[i+1][j-1] + 2 (with the convention dp[i+1][i] = 0 for length 2). Otherwise dp[i][j] = max(dp[i+1][j], dp[i][j-1]). Return dp[0][n-1]. Alternative framing: this equals LCS(s, reverse(s)) — same answer, same complexity, often easier to reason about for students who already know LCS.
Complexity
- Time
- O(n^2)
- Space
- O(n^2)
Related patterns
- dynamic-programming
- interval-dp
- string-dp
Related problems
Asked at
Companies reported asking this problem (sourced from public Glassdoor, Blind, and Levels.fyi interview posts).
- Amazon
- Uber
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