LeetCode #2484 — HARD

Count Palindromic Subsequences

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

Given a string of digits s, return the number of palindromic subsequences of s having length 5. Since the answer may be very large, return it modulo 109 + 7.

Note:

  • A string is palindromic if it reads the same forward and backward.
  • A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

Example 1:

Input: s = "103301"
Output: 2
Explanation: 
There are 6 possible subsequences of length 5: "10330","10331","10301","10301","13301","03301". 
Two of them (both equal to "10301") are palindromic.

Example 2:

Input: s = "0000000"
Output: 21
Explanation: All 21 subsequences are "00000", which is palindromic.

Example 3:

Input: s = "9999900000"
Output: 2
Explanation: The only two palindromic subsequences are "99999" and "00000".

Constraints:

  • 1 <= s.length <= 104
  • s consists of digits.
Patterns Used
📊Dynamic Programming→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Given a string of digits s, return the number of palindromic subsequences of s having length 5. Since the answer may be very large, return it modulo 109 + 7. Note: A string is palindromic if it reads the same forward and backward. A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Dynamic Programming

Example 1

"103301"

Example 2

"0000000"

Example 3

"9999900000"

Related Problems

  • Arithmetic Slices II - Subsequence (arithmetic-slices-ii-subsequence)
  • Count Different Palindromic Subsequences (count-different-palindromic-subsequences)
  • Unique Length-3 Palindromic Subsequences (unique-length-3-palindromic-subsequences)
Step 02

Core Insight

What unlocks the optimal approach

  • There are 100 possibilities for the first two characters of the palindrome.
  • Iterate over all characters, letting the current character be the center of the palindrome.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2484: Count Palindromic Subsequences
class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int countPalindromes(String s) {
        int n = s.length();
        int[][][] pre = new int[n + 2][10][10];
        int[][][] suf = new int[n + 2][10][10];
        int[] t = new int[n];
        for (int i = 0; i < n; ++i) {
            t[i] = s.charAt(i) - '0';
        }
        int[] c = new int[10];
        for (int i = 1; i <= n; ++i) {
            int v = t[i - 1];
            for (int j = 0; j < 10; ++j) {
                for (int k = 0; k < 10; ++k) {
                    pre[i][j][k] = pre[i - 1][j][k];
                }
            }
            for (int j = 0; j < 10; ++j) {
                pre[i][j][v] += c[j];
            }
            c[v]++;
        }
        c = new int[10];
        for (int i = n; i > 0; --i) {
            int v = t[i - 1];
            for (int j = 0; j < 10; ++j) {
                for (int k = 0; k < 10; ++k) {
                    suf[i][j][k] = suf[i + 1][j][k];
                }
            }
            for (int j = 0; j < 10; ++j) {
                suf[i][j][v] += c[j];
            }
            c[v]++;
        }
        long ans = 0;
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j < 10; ++j) {
                for (int k = 0; k < 10; ++k) {
                    ans += (long) pre[i - 1][j][k] * suf[i + 1][j][k];
                    ans %= MOD;
                }
            }
        }
        return (int) ans;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(100 × n)
Space
O(100 × n)

Approach Breakdown

RECURSIVE
O(2ⁿ) time
O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING
O(n × m) time
O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Related Problems
#5Longest Palindromic Substringmedium→#10Regular Expression Matchinghard→#32Longest Valid Parentheseshard→#44Wildcard Matchinghard→#45Jump Game IImedium→