What is the time complexity of Count Palindromic Subsequences?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Count Palindromic Subsequences?

This problem uses String, Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Count Palindromic Subsequences?

Insight 1: Since we are looking for palindromes of length 5, they must have the form ABCBA, where A, B, and C can be any digit. Insight 2: Dynamic programming is a good approach since we want to count the number of ways to build a palindrome, and can break the problem down into prefix and suffix counts.

Which companies ask LeetCode 2484. Count Palindromic Subsequences in interviews?

This problem is frequently asked at Citadel, Goldman Sachs, Salesforce.

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Count Palindromic Subsequences - LeetCode 2484 Solution

Count Palindromic Subsequences - Complete Solution Guide

Count Palindromic Subsequences is LeetCode problem 2484, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 10 9 + 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",

Detailed Explanation

The problem asks us to find the number of palindromic subsequences of length 5 in a given string 's' consisting of digits. A subsequence is a string formed by deleting some characters from 's' without changing the order of the remaining characters. A palindrome is a string that reads the same forwards and backward. The answer must be returned modulo 10^9 + 7 because the count of palindromic subsequences can be very large. For example, in the string "103301", the subsequence "10301" is a palindrome, and we need to count how many such palindromes of length 5 exist.

Solution Approach

The solution uses dynamic programming to efficiently count the palindromic subsequences of length 5. It precomputes the counts of digit pairs in the suffix of the string and stores them in `suffix_pairs`. It also maintains counts of digits in the prefix of the string in `prefix_counts` and the counts of digit pairs in the prefix `prefix_pairs`. Then the code iterates through the string and, at each index `k`, considers that the digit at `k` could be the middle digit 'C' of the palindrome. It uses the `prefix_pairs` to find pairs of digits 'AB' before index `k`, and uses `suffix_pairs` to find matching pairs of digits 'BA' after index `k`. It then accumulates the counts of these combinations.

Step-by-Step Algorithm

Step 1: Initialize variables: `n` (length of string), `MOD` (10^9 + 7), `digits` (array of integer representation of digits in string), `suffix_pairs` (3D array to store the counts of digit pairs in the suffixes), `suffix_counts` (array to store the counts of digits in the suffixes).
Step 2: Calculate `suffix_pairs` and `suffix_counts`. Iterate from the end of the string to the beginning. For each digit `c` at index `k`, increment the count of `suffix_pairs[k][c][d2]` by `suffix_counts[d2]` for all digits `d2`. This counts the number of times a digit `c` is followed by a digit `d2` in the suffix. Update suffix_counts[c].
Step 3: Initialize variables: `total` (to store the total count of palindromes), `prefix_counts` (array to store the counts of digits in the prefixes), and `prefix_pairs` (2D array to store the counts of digit pairs in the prefixes).
Step 4: Iterate through the string from the beginning to the end. At each index `k`, the digit `c` is considered as the middle digit of the palindrome. Calculate `current_suffix_pairs` which stores suffix pairs after index k. Iterate through all possible digit pairs `x` and `y` (0-9). Update `total` by adding `prefix_pairs[x][y] * current_suffix_pairs[y][x]` to the total. This finds the number of palindromes ABCBA where C is the digit at index k.
Step 5: Update `prefix_pairs` and `prefix_counts`. Increment `prefix_pairs[d1][c]` by `prefix_counts[d1]` for all digits `d1`, where `c` is the digit at index `k`. Then increment `prefix_counts[c]` to count the occurrence of `c` in prefix.
Step 6: Return `total` modulo `MOD`.

Key Insights

Insight 1: Since we are looking for palindromes of length 5, they must have the form ABCBA, where A, B, and C can be any digit.
Insight 2: Dynamic programming is a good approach since we want to count the number of ways to build a palindrome, and can break the problem down into prefix and suffix counts.
Insight 3: We need to precompute the counts of digit pairs in the suffix to efficiently find matching digit pairs to form the palindromes. We also need to calculate prefix counts.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Dynamic Programming.

Companies

Asked at: Citadel, Goldman Sachs, Salesforce.

Count Palindromic Subsequences | LeetCode 2484

Hard

String Dynamic Programming

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int countPalindromes(String s) {
        int n = s.length();
        if (n < 5) {
            return 0;
        }

        int MOD = (int)1e9 + 7;
        int[] digits = new int[n];
        for (int i = 0; i < n; i++) {
            digits[i] = s.charAt(i) - '0';
        }

        int[][][] suffix_pairs = new int[n + 1][10][10];
        int[] suffix_counts = new int[10];

        for (int k = n - 1; k >= 0; k--) {
            int c = digits[k];
            if (k < n - 1) {
                for (int d1 = 0; d1 < 10; d1++) {
                    for (int d2 = 0; d2 < 10; d2++) {
                        suffix_pairs[k][d1][d2] = suffix_pairs[k+1][d1][d2];
                    }
                }
            }

            for (int d2 = 0; d2 < 10; d2++) {
                suffix_pairs[k][c][d2] = (suffix_pairs[k][c][d2] + suffix_counts[d2]);
            }
            suffix_counts[c]++;
        }

        int total = 0;
        int[] prefix_counts = new int[10];
        int[][] prefix_pairs = new int[10][10];

        for (int k = 0; k < n; k++) {
            int[][] current_suffix_pairs = suffix_pairs[k+1];

            for (int x = 0; x < 10; x++) {
                for (int y = 0; y < 10; y++) {
                    total = (int)(((long)total + (long)prefix_pairs[x][y] * current_suffix_pairs[y][x]) % MOD);
                }
            }

            int c = digits[k];
            for (int d1 = 0; d1 < 10; d1++) {
                prefix_pairs[d1][c] = (prefix_pairs[d1][c] + prefix_counts[d1]);
            }
            prefix_counts[c]++;
        }

        return total;
    }
}

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Problem Info

Description

Problem 2484: Count Palindromic Subsequences - Detailed explanation and solution approaches.

Topics

String Dynamic Programming

Companies

CitadelGoldman SachsSalesforce

Stats

Difficulty:Hard

Languages:4

Companies:3

Problem #:2484

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize variables: `n` (length of string), `MOD` (10^9 + 7), `digits` (array of integer representation of digits in string), `suffix_pairs` (3D array to store the counts of digit pairs in the suffixes), `suffix_counts` (array to store the counts of digits in the suffixes).
2
Step 2: Calculate `suffix_pairs` and `suffix_counts`. Iterate from the end of the string to the beginning. For each digit `c` at index `k`, increment the count of `suffix_pairs[k][c][d2]` by `suffix_counts[d2]` for all digits `d2`. This counts the number of times a digit `c` is followed by a digit `d2` in the suffix. Update suffix_counts[c].
3
Step 3: Initialize variables: `total` (to store the total count of palindromes), `prefix_counts` (array to store the counts of digits in the prefixes), and `prefix_pairs` (2D array to store the counts of digit pairs in the prefixes).
4
Step 4: Iterate through the string from the beginning to the end. At each index `k`, the digit `c` is considered as the middle digit of the palindrome. Calculate `current_suffix_pairs` which stores suffix pairs after index k. Iterate through all possible digit pairs `x` and `y` (0-9). Update `total` by adding `prefix_pairs[x][y] * current_suffix_pairs[y][x]` to the total. This finds the number of palindromes ABCBA where C is the digit at index k.
5
Step 5: Update `prefix_pairs` and `prefix_counts`. Increment `prefix_pairs[d1][c]` by `prefix_counts[d1]` for all digits `d1`, where `c` is the digit at index `k`. Then increment `prefix_counts[c]` to count the occurrence of `c` in prefix.
6
Step 6: Return `total` modulo `MOD`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is linear because the dominant operations are iterating through the string once for suffix calculations and again to count the palindromes, with constant-time operations within the loops. The space complexity arises primarily from storing the suffix counts, which requires O(n) space.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Generate all possible subsequences of length 5 and check if they are palindromes. This approach is highly inefficient.

Time: O(n^5), where n is the length of the string because it involves generating all combinations of selecting 5 characters and comparing them.Space: O(1), as it doesn't use any significant extra space.

Trade-offs: The brute force approach is simple to understand but extremely inefficient and not feasible for large input strings. The DP solution is much faster.

✨

Key Insights

1Insight 1: Since we are looking for palindromes of length 5, they must have the form ABCBA, where A, B, and C can be any digit.
2Insight 2: Dynamic programming is a good approach since we want to count the number of ways to build a palindrome, and can break the problem down into prefix and suffix counts.
3Insight 3: We need to precompute the counts of digit pairs in the suffix to efficiently find matching digit pairs to form the palindromes. We also need to calculate prefix counts.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: String length is less than 5. The code handles this by returning 0 at the beginning of the function.
Edge case 2: String contains repeating digits. The code handles this correctly by accumulating counts of all possible palindromes.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not taking the modulo at each step. The intermediate results can become very large and overflow the integer data type, leading to incorrect results. Need to apply modulo operation after each multiplication and addition.

✗Common mistake 2: Not handling the case where the string length is less than 5.

✗Common mistake 3: Not understanding the definition of subsequence. The characters do not have to be contiguous.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Bioinformatics: Finding palindromic sequences in DNA or RNA strings. Palindromic sequences often have biological significance.

🏦Real-world scenario 2: Data compression: Identifying and exploiting palindromic patterns in data for compression algorithms. Palindromic patterns in data can be represented more compactly.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 23, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Count Palindromic Subsequences"?

For this hard-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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