[Dynamic Programming] Count All Palindromic Substrings - Medium

Dynamic programming is a strong method we use to solve tough problems. We can do this by breaking them into easier smaller problems. When we count all palindromic substrings, we build a solution step by step. We store results from earlier calculations. This way, we do not repeat work. In this article, we will look at different ways to count all palindromic substrings in a string. We will show examples in Java, Python, and C++.

In this article, we will talk about different ways to count palindromic substrings. We will cover dynamic programming, iterative methods, and the expand-around-center technique. We will give clear examples in several programming languages. We will also check the space complexity of each method. Plus, we will discuss common mistakes and good practices in dynamic programming. At the end, we will answer some frequently asked questions about palindromic substrings.

• Dynamically Counting All Palindromic Substrings Using Dynamic Programming in Java
• Implementing Palindromic Substring Count with Dynamic Programming in Python
• C++ Solution for Counting All Palindromic Substrings Using Dynamic Programming
• Optimized Space Complexity for Palindromic Substrings in Java
• Iterative Approach to Count Palindromic Substrings in Python
• Expand Around Center Technique for Palindromic Substrings in C++
• Comparative Analysis of Different Approaches to Count Palindromic Substrings
• Common Pitfalls and Best Practices in Dynamic Programming for Palindromic Substrings
• Frequently Asked Questions

For more information on dynamic programming ideas, we think you may like these articles: Count Ways to Climb Stairs and Longest Palindromic Subsequence.

Implementing Palindromic Substring Count with Dynamic Programming in Python

We can count all palindromic substrings in a string using dynamic programming in Python. We will use a 2D table to track these substrings. Here is how we can do it:

def countPalindromicSubstrings(s: str) -> int:
    n = len(s)
    if n == 0:
        return 0
    
    # Create a 2D array to store if substring is palindrome
    dp = [[False] * n for _ in range(n)]
    count = 0
    
    # Single letter palindromes
    for i in range(n):
        dp[i][i] = True
        count += 1
    
    # Check for two-letter palindromes
    for i in range(n - 1):
        if s[i] == s[i + 1]:
            dp[i][i + 1] = True
            count += 1
            
    # Check for palindromes longer than two letters
    for length in range(3, n + 1):  # length of the substring
        for i in range(n - length + 1):
            j = i + length - 1
            # A substring s[i:j+1] is a palindrome if the ends match and s[i+1:j] is also a palindrome
            if s[i] == s[j] and dp[i + 1][j - 1]:
                dp[i][j] = True
                count += 1

    return count

# Example usage
string = "abc"
print(countPalindromicSubstrings(string))  # Output: 3 (substrings: 'a', 'b', 'c')

Explanation:

• Initialization: We start with a 2D list dp to store if substrings are palindromic.
• Single Characters: Every character is a palindrome by itself.
• Two Characters: We check if two adjacent characters are the same.
• Longer Substrings: For substrings longer than two, we check if the first and last characters are the same. Also we check if the substring between them is a palindrome.

This dynamic programming method runs in O(n^2) time and uses O(n^2) space. It is good for counting all palindromic substrings in a string.

For more examples and uses of dynamic programming, we can refer to Dynamic Programming: Count All Palindromic Substrings.

C++ Solution for Counting All Palindromic Substrings Using Dynamic Programming

We will count all palindromic substrings in a string using dynamic programming in C++. We can use a 2D array to track if substrings are palindromic. The main idea is to fill this table based on what we know about palindromes:

1. A single character is always a palindrome.
2. Two same characters next to each other are palindromes.
3. For substrings longer than two characters, a substring s[i...j] is a palindrome if the characters at both ends are the same and the substring s[i+1...j-1] is also a palindrome.

Here is how we can do this in C++:

#include <iostream>
#include <vector>
#include <string>

using namespace std;

int countPalindromicSubstrings(const string& s) {
    int n = s.size();
    if (n == 0) return 0;

    vector<vector<bool>> dp(n, vector<bool>(n, false));
    int count = 0;

    // All substrings of length 1 are palindromic
    for (int i = 0; i < n; i++) {
        dp[i][i] = true;
        count++;
    }

    // Check for substrings of length 2
    for (int i = 0; i < n - 1; i++) {
        if (s[i] == s[i + 1]) {
            dp[i][i + 1] = true;
            count++;
        }
    }

    // Check for substrings of length 3 or more
    for (int length = 3; length <= n; length++) {
        for (int i = 0; i < n - length + 1; i++) {
            int j = i + length - 1;
            if (s[i] == s[j] && dp[i + 1][j - 1]) {
                dp[i][j] = true;
                count++;
            }
        }
    }

    return count;
}

int main() {
    string str = "abc";
    cout << "Total palindromic substrings: " << countPalindromicSubstrings(str) << endl;
    return 0;
}

Explanation of the Code

We have a function called countPalindromicSubstrings. It starts by making a 2D vector dp to track palindromic substrings.

First, we count all single-character palindromes and update the count.

Next, we look for two-character palindromes.

Finally, we check substrings of length 3 and more. We update the dp table and the count of palindromic substrings based on our rules.

Time Complexity

The time complexity for this algorithm is (O(n^2)). This happens because we have nested loops to fill the DP table.

Space Complexity

The space complexity is also (O(n^2)`. We need the 2D array to store the palindrome status of substrings.

This C++ solution counts all palindromic substrings using dynamic programming. It is efficient for this problem.

Optimized Space Complexity for Palindromic Substrings in Java

We want to make space use better when we count palindromic substrings in Java. We can do this by making the size of the dynamic programming table smaller. Instead of using a big 2D array, we can use a 1D array to remember the palindromes we find. This works because the state only needs the previous states.

Approach

1. Use a 1D Array: We use one array to track palindromic substrings instead of a 2D boolean array.
2. Count Palindromic Substrings: We start counting single characters and pairs.
3. Update Counts: We use two loops to update counts based on the values we calculated before.

Code Implementation

public class PalindromicSubstrings {

    public int countSubstrings(String s) {
        int n = s.length();
        int count = 0;

        // 1D array to keep track of palindromic substrings
        boolean[] dp = new boolean[n];

        // Check for substrings of length 1
        for (int i = 0; i < n; i++) {
            dp[i] = true; // Every single character is a palindrome
            count++;
        }

        // Check for substrings of length 2 and more
        for (int len = 2; len <= n; len++) {
            boolean[] newDp = new boolean[n];
            for (int start = 0; start <= n - len; start++) {
                int end = start + len - 1;
                if (s.charAt(start) == s.charAt(end)) {
                    if (len == 2) {
                        newDp[start] = true;
                    } else {
                        newDp[start] = dp[start + 1];
                    }
                    if (newDp[start]) {
                        count++;
                    }
                }
            }
            dp = newDp; // Move to the next length
        }

        return count;
    }

    public static void main(String[] args) {
        PalindromicSubstrings solution = new PalindromicSubstrings();
        String input = "abc";
        System.out.println("Count of palindromic substrings: " + solution.countSubstrings(input)); // Output: 3
    }
}

Explanation of Code

• We use a 1D boolean array dp to keep track if substrings are palindromic.
• The outer loop goes over different lengths of substrings.
• The inner loop checks if the first and last characters of the substring are the same.
• If they are the same and the length is 2 or the inner substring (from start + 1 to end - 1) is palindromic, we count it as a palindromic substring.

This way, we make space use go from (O(n^2)) to (O(n)). It is more efficient and still counts all palindromic substrings. For more learning about dynamic programming, we can check out related topics like Dynamic Programming - Palindromic Substrings Count.

Iterative Approach to Count Palindromic Substrings in Python

We can use an iterative way to count all palindromic substrings in a string. This method uses some ideas from dynamic programming. We will keep a 2D table to see which substrings are palindromic. The steps are:

1. Create a 2D list dp. Here, dp[i][j] is True if the substring s[i:j+1] is a palindrome.
2. Start by marking all substrings of length 1 as palindromes.
3. For substrings of length 2, set dp[i][i+1] to True if both characters are equal.
4. For longer substrings, we use this rule: dp[i][j] = (s[i] == s[j]) and dp[i+1][j-1].
5. We will count all palindromic substrings while we fill the dp table.

Here is the code in Python:

def countPalindromicSubstrings(s: str) -> int:
    n = len(s)
    if n == 0:
        return 0

    # Create a 2D DP list initialized to False
    dp = [[False] * n for _ in range(n)]
    count = 0

    # Single character substrings are palindromes
    for i in range(n):
        dp[i][i] = True
        count += 1

    # Check for substrings of length 2
    for i in range(n - 1):
        if s[i] == s[i + 1]:
            dp[i][i + 1] = True
            count += 1

    # Check for substrings of length 3 and more
    for length in range(3, n + 1):  # length of substring
        for i in range(n - length + 1):
            j = i + length - 1
            if s[i] == s[j] and dp[i + 1][j - 1]:
                dp[i][j] = True
                count += 1

    return count

# Example usage
s = "abcba"
result = countPalindromicSubstrings(s)
print(f"Total palindromic substrings in '{s}' is: {result}")

This code counts all palindromic substrings in the string s. The time it takes is (O(n^2)) because of the nested loops. The space needed is also (O(n^2)) for the DP table. This method works well for strings that are not too long and counts all unique palindromic substrings.

Expand Around Center Technique for Palindromic Substrings in C++

The Expand Around Center technique is a good way to count all palindromic substrings in a string. This method uses each character and each pair of characters as a center of a palindrome. Then, it expands outwards to check if it is a palindrome.

Key Properties:

• Time Complexity: O(n^2), where n is the length of the string.
• Space Complexity: O(1) because we use only a few variables to count.

Implementation Steps:

1. We loop through each character in the string.
2. For each character, we expand around it to find odd-length palindromes.
3. For each pair of characters, we expand around them to find even-length palindromes.
4. We count the total number of palindromic substrings that we find during expansion.

C++ Code Example:

#include <iostream>
#include <string>
using namespace std;

class Solution {
public:
    int countSubstrings(string s) {
        int count = 0;
        int n = s.length();

        for (int center = 0; center < n; center++) {
            // Count odd-length palindromes
            count += expandAroundCenter(s, center, center);
            // Count even-length palindromes
            count += expandAroundCenter(s, center, center + 1);
        }

        return count;
    }

private:
    int expandAroundCenter(string& s, int left, int right) {
        int count = 0;
        while (left >= 0 && right < s.length() && s[left] == s[right]) {
            count++;
            left--;
            right++;
        }
        return count;
    }
};

int main() {
    Solution sol;
    string s = "abc";
    cout << "Total palindromic substrings: " << sol.countSubstrings(s) << endl; // Output: 3
    return 0;
}

Explanation of the Code:

• The countSubstrings function looks at each character in the string and treats it as a center.
• The expandAroundCenter function checks for palindromes by expanding out from the center. It goes until the characters on the left and right do not match anymore.
• The total count of palindromic substrings is returned.

This technique works well for counting palindromic substrings without needing extra data structures. It is also a space-efficient solution. If we want to learn more advanced ways, we can look at the Dynamic Programming Count Palindromic Substrings.

Comparative Analysis of Different Approaches to Count Palindromic Substrings

We can count palindromic substrings in different ways. Each way has its own pros and cons. These include time complexity, space complexity, and how easy it is to implement. Here, we will look at three main methods: Dynamic Programming, Expand Around Center, and Iterative methods.

1. Dynamic Programming

Time Complexity: O(n^2)
Space Complexity: O(n^2)

In this method, we create a 2D table. We say dp[i][j] is true if the substring from index i to j is a palindrome. We fill this table in a bottom-up way.

public int countSubstrings(String s) {
    int n = s.length();
    boolean[][] dp = new boolean[n][n];
    int count = 0;

    for (int j = 0; j < n; j++) {
        for (int i = 0; i <= j; i++) {
            if (s.charAt(i) == s.charAt(j) && (j - i < 3 || dp[i + 1][j - 1])) {
                dp[i][j] = true;
                count++;
            }
        }
    }
    return count;
}

2. Expand Around Center

Time Complexity: O(n^2)
Space Complexity: O(1)

This method checks each possible center for a palindrome and expands out. Each character and each pair of characters can be a center.

def countSubstrings(s: str) -> int:
    count = 0
    n = len(s)

    for center in range(2 * n - 1):
        left = center // 2
        right = left + center % 2
        while left >= 0 and right < n and s[left] == s[right]:
            count += 1
            left -= 1
            right += 1

    return count

3. Iterative Approach

Time Complexity: O(n^2)
Space Complexity: O(1)

This approach uses one loop to check for palindromic substrings. It finds the start and end points while checking if they are equal.

int countSubstrings(string s) {
    int count = 0, n = s.size();

    for (int i = 0; i < n; ++i) {
        // Odd length palindromes
        for (int j = 0; i - j >= 0 && i + j < n && s[i - j] == s[i + j]; ++j) {
            count++;
        }
        // Even length palindromes
        for (int j = 1; i - j + 1 >= 0 && i + j < n && s[i - j + 1] == s[i + j]; ++j) {
            count++;
        }
    }
    return count;
}

Summary of Approaches

• Dynamic Programming helps us understand the links between substrings but it needs more space.
• Expand Around Center is best for space and is easy to use with low space needs.
• Iterative Approach is simple and clear. It counts palindromic substrings without needing more space.

If we want to learn more about dynamic programming, we can check out Dynamic Programming: Count Ways to Climb Stairs.

Common Pitfalls and Best Practices in Dynamic Programming for Palindromic Substrings

When we try to use dynamic programming for counting palindromic substrings, we can run into some common mistakes. If we understand these mistakes and follow some good practices, we can write code that works better and has fewer errors.

Common Pitfalls

1. Improper Initialization:
   • We need to make sure that the DP table starts correctly. For counting palindromic substrings, single-character substrings are always palindromes.

   for (int i = 0; i < n; i++) {
       dp[i][i] = true; // single character
   }

2. Not Considering Edge Cases:
   • We should handle cases with strings of length 0 or 1. These cases are easy because they are always palindromic.
3. Incorrect State Transition:
   • We must define the state transition correctly. For palindromes, if s[i] == s[j], then dp[i][j] = dp[i + 1][j - 1] if j - i > 1.
4. Forgetting to Count Substrings:
   • Sometimes, we forget to count palindromic substrings. We need to remember to add to our count every time we find a palindrome.
5. Space Complexity Issues:
   • We need to pay attention to space complexity when using a DP table. If we don’t handle it well, it can use too much memory.

Best Practices

1. Use Memoization:
   • If our approach creates overlapping subproblems, we should use memoization to save results of states we already calculated.
2. Iterative vs Recursive:
   • We should prefer iterative methods because they help us avoid stack overflow problems that can happen with deep recursion in longer strings.
3. Expand Around Center:
   • We can use the expand around center method for a solution that uses less space. This method counts palindromes by expanding around each character and each pair of characters.

   def countPalindromicSubstrings(s: str) -> int:
       count = 0
       for i in range(len(s)):
           count += expandAroundCenter(s, i, i)   # Odd length
           count += expandAroundCenter(s, i, i+1) # Even length
       return count
   
   def expandAroundCenter(s: str, left: int, right: int) -> int:
       count = 0
       while left >= 0 and right < len(s) and s[left] == s[right]:
           count += 1
           left -= 1
           right += 1
       return count

4. Use a 1D Array for Space Optimization:
   • If we only need results from the last row of the DP table, we can use a 1D array instead of a 2D array to save space.

   boolean[] dp = new boolean[n]; // for storing palindromic state for each character

5. Profile and Optimize:
   • We should profile our solution to find slow parts. We can optimize the time complexity by cutting down on unnecessary calculations or loops.
6. Testing on Edge Cases:
   • We must test our code with different string lengths. This includes edge cases like empty strings, strings with all the same characters, and strings with no palindromes.

By following these best practices and avoiding common mistakes, we can build a strong and efficient dynamic programming solution for counting palindromic substrings. If we want to learn more about dynamic programming techniques, we can check out Dynamic Programming: Count Ways to Climb Stairs for more understanding and examples.

Frequently Asked Questions

1. What is the Dynamic Programming approach for counting palindromic substrings?

We use the Dynamic Programming approach to count palindromic substrings by making a 2D table. Each entry dp[i][j] shows if the substring from index i to j is a palindrome. First, we treat all single-character substrings as palindromes. Then, we check larger substrings using results we already have to see if they are palindromes.

2. How does the Expand Around Center technique work for palindromic substrings?

The Expand Around Center technique finds palindromic substrings by looking at each character and each pair of neighboring characters as possible centers. For each center, we expand outwards while the characters on both sides are the same. We count all valid palindromic substrings while doing this. This method is fast, working in O(n^2) time and using O(1) space.

3. Why is space optimization important in counting palindromic substrings?

Space optimization is very important when counting palindromic substrings. It helps to lower memory use, especially for big strings. By using a rolling array or changing the space from O(n^2) to O(n), we can make our program run better without losing accuracy. Such optimizations matter a lot for real-world applications where we have limited resources.

4. Can Dynamic Programming be applied to find the longest palindromic substring?

Yes, we can use Dynamic Programming to find the longest palindromic substring. We keep a table to check if substrings are palindromes. This helps us track the longest palindrome we find. This method is similar to counting palindromic substrings but focuses on finding the maximum length.

5. What are common pitfalls when implementing Dynamic Programming for palindromic substrings?

Common mistakes in using Dynamic Programming for palindromic substrings include not setting up the DP table correctly, missing edge cases like empty strings, and not handling overlapping subproblems well. It is very important to set accurate base cases and know how the subproblems relate to each other to count palindromic substrings successfully.

For more reading on related dynamic programming ideas, we can check articles on Dynamic Programming - Fibonacci Numbers and Dynamic Programming - Unique Paths in a Grid. These topics will help us understand dynamic programming techniques that we can use in similar situations.