Count Palindromic Substrings

🎯

Count Palindromic Substrings

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Imagine you are analyzing DNA sequences or text data and want to find how many palindromic patterns exist within a string. This helps in pattern recognition and data compression.

💡 This problem is about counting all substrings that read the same forwards and backwards. Beginners often struggle because it requires checking many substrings efficiently, and naive methods are too slow. Understanding how to break down the problem into smaller overlapping subproblems is key.

📋

Problem Statement

Given a string s, return the number of palindromic substrings in it. A substring is a contiguous sequence of characters within the string. Palindromic substrings read the same backward as forward.

1 ≤ s.length ≤ 10^5s consists of lowercase English letters.

💡

Example

Input"abc"

Output3

The palindromic substrings are 'a', 'b', 'c'.

Input"aaa"

Output6

The palindromic substrings are 'a', 'a', 'a', 'aa', 'aa', 'aaa'.

• Empty string → 0
• Single character string → 1
• String with all identical characters → n*(n+1)/2 palindromes
• String with no repeated characters → n palindromes

🔁

Why DP?

Why greedy fails:

A greedy approach might try to count palindromes by expanding from each character once, but it fails to count overlapping palindromes correctly. For example, in 'aaa', greedy might miss counting all substrings like 'aa' starting at different positions.

DP state:

dp[i][j] = True if the substring s[i..j] is a palindrome, otherwise False.

Recurrence:dp[i][j] = (s[i] == s[j]) && (j - i < 3 || dp[i+1][j-1])

A substring is palindrome if the characters at the ends match and the inner substring is palindrome, or if the substring length is less than 3 (which means it's either one or two characters).

⚠️

Common Mistakes

Not handling substrings of length 1 and 2 correctly in DP

Incorrect palindrome counts or index errors

✅ Explicitly mark substrings of length 1 as palindrome and handle length 2 separately

Using brute force palindrome check inside nested loops without memoization

Time limit exceeded on large inputs

✅ Use DP or expand around center to avoid repeated checks

Filling DP table in wrong order (e.g., start index before length)

Accessing uninitialized dp[i+1][j-1] causing wrong results

✅ Fill DP by increasing substring length to ensure dependencies are computed

Forgetting to count single character substrings as palindromes

Under-counting total palindromic substrings

✅ Initialize count with all single characters or mark dp[i][i] = True

🧠

Brute Force (Check All Substrings)

💡 This approach exists to build intuition by directly checking every possible substring. It is simple but inefficient, helping beginners understand the problem's nature before optimizing.

Intuition

Enumerate all substrings and check if each is a palindrome by comparing characters from both ends.

Algorithm

1. Generate all possible substrings of s.
2. For each substring, check if it is a palindrome by comparing characters from start and end.
3. If palindrome, increment count.
4. Return the total count.

💡 The nested loops and palindrome check are straightforward but lead to repeated work and high time complexity.

</>

Code

def countSubstrings(s: str) -> int:
    n = len(s)
    count = 0
    for start in range(n):
        for end in range(start, n):
            left, right = start, end
            is_palindrome = True
            while left < right:
                if s[left] != s[right]:
                    is_palindrome = False
                    break
                left += 1
                right -= 1
            if is_palindrome:
                count += 1
    return count

# Driver code
if __name__ == '__main__':
    print(countSubstrings('aaa'))  # Expected output: 6

Line Notes

for start in range(n):Outer loop picks the start index of substring

for end in range(start, n):Inner loop picks the end index of substring

while left < right:Check characters from both ends moving inward

if s[left] != s[right]:If mismatch, substring is not palindrome

public class Solution {
    public int countSubstrings(String s) {
        int n = s.length();
        int count = 0;
        for (int start = 0; start < n; start++) {
            for (int end = start; end < n; end++) {
                int left = start, right = end;
                boolean isPalindrome = true;
                while (left < right) {
                    if (s.charAt(left) != s.charAt(right)) {
                        isPalindrome = false;
                        break;
                    }
                    left++;
                    right--;
                }
                if (isPalindrome) count++;
            }
        }
        return count;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.countSubstrings("aaa")); // Expected output: 6
    }
}

Line Notes

for (int start = 0; start < n; start++) {Outer loop selects substring start

for (int end = start; end < n; end++) {Inner loop selects substring end

while (left < right) {Check palindrome by comparing ends

if (s.charAt(left) != s.charAt(right)) {Mismatch means not palindrome

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

int countSubstrings(string s) {
    int n = s.size();
    int count = 0;
    for (int start = 0; start < n; start++) {
        for (int end = start; end < n; end++) {
            int left = start, right = end;
            bool isPalindrome = true;
            while (left < right) {
                if (s[left] != s[right]) {
                    isPalindrome = false;
                    break;
                }
                left++;
                right--;
            }
            if (isPalindrome) count++;
        }
    }
    return count;
}

int main() {
    cout << countSubstrings("aaa") << endl; // Expected output: 6
    return 0;
}

Line Notes

for (int start = 0; start < n; start++) {Start index of substring

for (int end = start; end < n; end++) {End index of substring

while (left < right) {Check palindrome by inward comparison

if (s[left] != s[right]) {Mismatch breaks palindrome check

function countSubstrings(s) {
    const n = s.length;
    let count = 0;
    for (let start = 0; start < n; start++) {
        for (let end = start; end < n; end++) {
            let left = start, right = end;
            let isPalindrome = true;
            while (left < right) {
                if (s[left] !== s[right]) {
                    isPalindrome = false;
                    break;
                }
                left++;
                right--;
            }
            if (isPalindrome) count++;
        }
    }
    return count;
}

// Test
console.log(countSubstrings('aaa')); // Expected output: 6

Line Notes

for (let start = 0; start < n; start++) {Outer loop for substring start

for (let end = start; end < n; end++) {Inner loop for substring end

while (left < right) {Check palindrome by comparing ends

if (s[left] !== s[right]) {Mismatch means substring not palindrome

⏱

Complexity

TimeO(n^3)

SpaceO(1)

There are O(n^2) substrings and each palindrome check takes O(n) in worst case.

💡 For n=20, this means about 8000 checks, which is slow but still manageable for small inputs.

Interview Verdict: TLE

This approach is too slow for large inputs but is useful to understand the problem and baseline.

🧠

Dynamic Programming (Top-Down Memoization)

💡 This approach improves brute force by caching results of palindrome checks to avoid repeated work, introducing the core DP concept of overlapping subproblems.

Intuition

Use recursion with memoization to check if substrings are palindrome, storing results to avoid recomputation.

Algorithm

1. Define a recursive function that checks if s[i..j] is palindrome.
2. If already computed, return cached result.
3. Check ends and recurse on inner substring.
4. Count palindromic substrings by calling this function for all substrings.

💡 Memoization avoids repeated palindrome checks, drastically improving efficiency over brute force.

Recurrence:dp(i,j) = (s[i] == s[j]) && (j - i < 3 || dp(i+1,j-1))

</>

Code

def countSubstrings(s: str) -> int:
    n = len(s)
    memo = [[None]*n for _ in range(n)]

    def is_palindrome(i, j):
        if i >= j:
            return True
        if memo[i][j] is not None:
            return memo[i][j]
        memo[i][j] = (s[i] == s[j]) and is_palindrome(i+1, j-1)
        return memo[i][j]

    count = 0
    for i in range(n):
        for j in range(i, n):
            if is_palindrome(i, j):
                count += 1
    return count

# Driver code
if __name__ == '__main__':
    print(countSubstrings('aaa'))  # Expected output: 6

Line Notes

memo = [[None]*n for _ in range(n)]Initialize memo table to store palindrome results

def is_palindrome(i, j):Recursive function to check palindrome

if memo[i][j] is not None:Return cached result to avoid recomputation

memo[i][j] = (s[i] == s[j]) and is_palindrome(i+1, j-1)Store result based on ends and inner substring

public class Solution {
    private Boolean[][] memo;
    private String s;

    public boolean isPalindrome(int i, int j) {
        if (i >= j) return true;
        if (memo[i][j] != null) return memo[i][j];
        memo[i][j] = (s.charAt(i) == s.charAt(j)) && isPalindrome(i + 1, j - 1);
        return memo[i][j];
    }

    public int countSubstrings(String s) {
        this.s = s;
        int n = s.length();
        memo = new Boolean[n][n];
        int count = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                if (isPalindrome(i, j)) count++;
            }
        }
        return count;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.countSubstrings("aaa")); // Expected output: 6
    }
}

Line Notes

private Boolean[][] memo;Memo table to cache palindrome checks

if (memo[i][j] != null) return memo[i][j];Return cached result if available

memo[i][j] = (s.charAt(i) == s.charAt(j)) && isPalindrome(i + 1, j - 1);Store result based on ends and inner substring

for (int i = 0; i < n; i++) {Iterate over all substrings

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

class Solution {
    vector<vector<int>> memo;
    string s;

    bool isPalindrome(int i, int j) {
        if (i >= j) return true;
        if (memo[i][j] != -1) return memo[i][j];
        memo[i][j] = (s[i] == s[j]) && isPalindrome(i + 1, j - 1);
        return memo[i][j];
    }

public:
    int countSubstrings(string str) {
        s = str;
        int n = s.size();
        memo.assign(n, vector<int>(n, -1));
        int count = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                if (isPalindrome(i, j)) count++;
            }
        }
        return count;
    }
};

int main() {
    Solution sol;
    cout << sol.countSubstrings("aaa") << endl; // Expected output: 6
    return 0;
}

Line Notes

vector<vector<int>> memo;Memo table to store palindrome results

if (memo[i][j] != -1) return memo[i][j];Return cached result if computed

memo[i][j] = (s[i] == s[j]) && isPalindrome(i + 1, j - 1);Store result based on ends and inner substring

for (int i = 0; i < n; i++) {Loop over all substrings

function countSubstrings(s) {
    const n = s.length;
    const memo = Array.from({ length: n }, () => Array(n).fill(undefined));

    function isPalindrome(i, j) {
        if (i >= j) return true;
        if (memo[i][j] !== undefined) return memo[i][j];
        memo[i][j] = (s[i] === s[j]) && isPalindrome(i + 1, j - 1);
        return memo[i][j];
    }

    let count = 0;
    for (let i = 0; i < n; i++) {
        for (let j = i; j < n; j++) {
            if (isPalindrome(i, j)) count++;
        }
    }
    return count;
}

// Test
console.log(countSubstrings('aaa')); // Expected output: 6

Line Notes

const memo = Array.from({ length: n }, () => Array(n).fill(undefined));Initialize memo table

if (memo[i][j] !== undefined) return memo[i][j];Return cached palindrome result

memo[i][j] = (s[i] === s[j]) && isPalindrome(i + 1, j - 1);Store result based on ends and inner substring

for (let i = 0; i < n; i++) {Iterate over all substrings

⏱

Complexity

TimeO(n^2)

SpaceO(n^2)

Memoization avoids repeated palindrome checks, reducing time to O(n^2). Space is used for memo table.

💡 For n=20, about 400 palindrome checks are done, much faster than brute force.

Interview Verdict: Accepted

This approach is efficient enough for moderate input sizes and introduces DP concepts well.

🧠

Dynamic Programming (Bottom-Up Tabulation)

💡 This approach uses a DP table filled iteratively, which is often easier to understand and implement than recursion with memoization.

Intuition

Build a 2D table where dp[i][j] indicates if s[i..j] is palindrome, filling from smaller substrings to larger.

Algorithm

1. Initialize dp table with False values.
2. Mark all single characters as palindrome (dp[i][i] = True).
3. Iterate over substring lengths from 2 to n.
4. For each substring, check ends and inner substring dp to set dp[i][j].
5. Count all dp[i][j] that are True.

💡 Filling the table by substring length ensures dependencies are resolved before use.

Recurrence:dp[i][j] = (s[i] == s[j]) && (j - i < 3 || dp[i+1][j-1])

</>

Code

def countSubstrings(s: str) -> int:
    n = len(s)
    dp = [[False]*n for _ in range(n)]
    count = 0

    for length in range(1, n+1):
        for i in range(n - length + 1):
            j = i + length - 1
            if s[i] == s[j]:
                if length <= 2:
                    dp[i][j] = True
                else:
                    dp[i][j] = dp[i+1][j-1]
            if dp[i][j]:
                count += 1
    return count

# Driver code
if __name__ == '__main__':
    print(countSubstrings('aaa'))  # Expected output: 6

Line Notes

dp = [[False]*n for _ in range(n)]Initialize DP table with False

for length in range(1, n+1):Iterate over substring lengths

if s[i] == s[j]:Check if substring ends match

dp[i][j] = dp[i+1][j-1]Use inner substring palindrome result

public class Solution {
    public int countSubstrings(String s) {
        int n = s.length();
        boolean[][] dp = new boolean[n][n];
        int count = 0;
        for (int length = 1; length <= n; length++) {
            for (int i = 0; i <= n - length; i++) {
                int j = i + length - 1;
                if (s.charAt(i) == s.charAt(j)) {
                    if (length <= 2) {
                        dp[i][j] = true;
                    } else {
                        dp[i][j] = dp[i + 1][j - 1];
                    }
                }
                if (dp[i][j]) count++;
            }
        }
        return count;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.countSubstrings("aaa")); // Expected output: 6
    }
}

Line Notes

boolean[][] dp = new boolean[n][n];DP table initialization

for (int length = 1; length <= n; length++) {Loop over substring lengths

if (s.charAt(i) == s.charAt(j)) {Check substring ends

dp[i][j] = dp[i + 1][j - 1];Refer to inner substring palindrome

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

int countSubstrings(string s) {
    int n = s.size();
    vector<vector<bool>> dp(n, vector<bool>(n, false));
    int count = 0;
    for (int length = 1; length <= n; length++) {
        for (int i = 0; i <= n - length; i++) {
            int j = i + length - 1;
            if (s[i] == s[j]) {
                if (length <= 2) {
                    dp[i][j] = true;
                } else {
                    dp[i][j] = dp[i + 1][j - 1];
                }
            }
            if (dp[i][j]) count++;
        }
    }
    return count;
}

int main() {
    cout << countSubstrings("aaa") << endl; // Expected output: 6
    return 0;
}

Line Notes

vector<vector<bool>> dp(n, vector<bool>(n, false));Initialize DP table

for (int length = 1; length <= n; length++) {Iterate over substring lengths

if (s[i] == s[j]) {Check if substring ends match

dp[i][j] = dp[i + 1][j - 1];Use inner substring palindrome info

function countSubstrings(s) {
    const n = s.length;
    const dp = Array.from({ length: n }, () => Array(n).fill(false));
    let count = 0;

    for (let length = 1; length <= n; length++) {
        for (let i = 0; i <= n - length; i++) {
            let j = i + length - 1;
            if (s[i] === s[j]) {
                if (length <= 2) {
                    dp[i][j] = true;
                } else {
                    dp[i][j] = dp[i + 1][j - 1];
                }
            }
            if (dp[i][j]) count++;
        }
    }
    return count;
}

// Test
console.log(countSubstrings('aaa')); // Expected output: 6

Line Notes

const dp = Array.from({ length: n }, () => Array(n).fill(false));Initialize DP table

for (let length = 1; length <= n; length++) {Loop over substring lengths

if (s[i] === s[j]) {Check substring ends

dp[i][j] = dp[i + 1][j - 1];Refer to inner substring palindrome

⏱

Complexity

TimeO(n^2)

SpaceO(n^2)

We fill an n by n table once, each cell computed in O(1).

💡 For n=20, 400 table entries are computed, efficient for typical interview constraints.

Interview Verdict: Accepted

This is the standard DP solution for this problem, balancing clarity and efficiency.

🧠

Expand Around Center (Optimal, Space Efficient)

💡 This approach uses the insight that palindromes expand from centers, avoiding DP tables and reducing space to O(1). It is optimal and often preferred in interviews.

Intuition

Every palindrome can be expanded from its center (single character or between two characters). Count palindromes by expanding around each center.

Algorithm

1. Iterate over each index as a center for odd-length palindromes.
2. Expand around center while characters match, count palindromes.
3. Repeat for even-length palindromes (center between two characters).
4. Sum all counts and return.

💡 This method avoids extra space and uses two-pointer expansion to count palindromes efficiently.

</>

Code

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

    def expand(left, right):
        nonlocal count
        while left >= 0 and right < n and s[left] == s[right]:
            count += 1
            left -= 1
            right += 1

    for i in range(n):
        expand(i, i)      # Odd length palindromes
        expand(i, i + 1)  # Even length palindromes

    return count

# Driver code
if __name__ == '__main__':
    print(countSubstrings('aaa'))  # Expected output: 6

Line Notes

def expand(left, right):Helper to expand around center and count palindromes

while left >= 0 and right < n and s[left] == s[right]:Expand while substring is palindrome

count += 1Increment count for each palindrome found

for i in range(n):Check all possible centers

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

        for (int i = 0; i < n; i++) {
            count += expandAroundCenter(s, i, i);     // Odd length
            count += expandAroundCenter(s, i, i + 1); // Even length
        }
        return count;
    }

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

    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.countSubstrings("aaa")); // Expected output: 6
    }
}

Line Notes

for (int i = 0; i < n; i++) {Iterate over all centers

count += expandAroundCenter(s, i, i);Count odd length palindromes

count += expandAroundCenter(s, i, i + 1);Count even length palindromes

while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) {Expand while palindrome condition holds

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

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

int countSubstrings(string s) {
    int n = s.size();
    int count = 0;
    for (int i = 0; i < n; i++) {
        count += expandAroundCenter(s, i, i);     // Odd length
        count += expandAroundCenter(s, i, i + 1); // Even length
    }
    return count;
}

int main() {
    cout << countSubstrings("aaa") << endl; // Expected output: 6
    return 0;
}

Line Notes

for (int i = 0; i < n; i++) {Iterate over all centers

count += expandAroundCenter(s, i, i);Count odd length palindromes

count += expandAroundCenter(s, i, i + 1);Count even length palindromes

while (left >= 0 && right < (int)s.size() && s[left] == s[right]) {Expand while palindrome condition holds

function countSubstrings(s) {
    const n = s.length;
    let count = 0;

    function expand(left, right) {
        while (left >= 0 && right < n && s[left] === s[right]) {
            count++;
            left--;
            right++;
        }
    }

    for (let i = 0; i < n; i++) {
        expand(i, i);      // Odd length
        expand(i, i + 1);  // Even length
    }

    return count;
}

// Test
console.log(countSubstrings('aaa')); // Expected output: 6

Line Notes

function expand(left, right) {Expand around center to count palindromes

while (left >= 0 && right < n && s[left] === s[right]) {Expand while substring is palindrome

count++;Increment count for each palindrome found

for (let i = 0; i < n; i++) {Check all possible centers

⏱

Complexity

TimeO(n^2)

SpaceO(1)

Each center expansion takes O(n) in worst case, and there are O(n) centers.

💡 For n=20, about 400 expansions are done, very efficient and uses minimal memory.

Interview Verdict: Accepted

This is the most optimal and elegant solution, preferred in interviews for its simplicity and efficiency.

📊

All Approaches - One-Glance Tradeoffs

💡 Expand Around Center is best for interviews due to simplicity and efficiency; DP tabulation is good for clarity; memoization introduces DP concepts; brute force is only for understanding.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(n^3)	O(1)	No	N/A	Mention only - never code
2. DP Memoization	O(n^2)	O(n^2)	Possible (deep recursion)	Yes	Good to explain DP concepts
3. DP Tabulation	O(n^2)	O(n^2)	No	Yes	Standard DP solution
4. Expand Around Center	O(n^2)	O(1)	No	Yes	Preferred solution to code

💼

Interview Strategy

💡 Use this guide to understand the problem deeply, starting from brute force to optimal solutions. Practice explaining each approach clearly and know when to use them.

How to Present

Step 1: Clarify problem and constraints.Step 2: Describe brute force approach and its inefficiency.Step 3: Introduce DP memoization to optimize repeated checks.Step 4: Present bottom-up DP for clarity and efficiency.Step 5: Show expand around center as optimal space-efficient solution.Step 6: Code the chosen approach and test with examples.

Time Allocation

Clarify: 2min → Approach: 5min → Code: 10min → Test: 3min. Total ~20min

What the Interviewer Tests

Understanding of palindrome properties, ability to optimize naive solutions, knowledge of DP and space optimization, and clean coding.

Common Follow-ups

• How to print all palindromic substrings → Use DP table or expand centers with storage.
• How to find longest palindromic substring → Modify DP or expand center to track max length.

💡 These follow-ups test your ability to extend the solution and handle related palindrome problems.

🔍

Pattern Recognition

When to Use

1) Problem asks for counting palindromic substrings or partitions, 2) Input is a string, 3) Overlapping subproblems exist in substring checks, 4) Optimal substructure based on substring properties.

Signature Phrases

count palindromic substringssubstring is palindromeexpand around center

NOT This Pattern When

Problems about subsequences or non-contiguous palindromes use different DP patterns.

Similar Problems

Longest Palindromic Substring - find max length palindromePalindrome Partitioning - partition string into palindromes

Practice

1. You are given two strings and want to find the minimum number of delete operations needed to make them equal. Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. Dynamic Programming based on Longest Common Subsequence (LCS)

B. Greedy approach deleting mismatched characters as soon as found

C. Pure brute force recursion without memoization

D. Sorting both strings and comparing characters sequentially

Solution

1. Step 1: Understand problem requires minimal deletions to equalize strings

   This is equivalent to finding the longest common subsequence (LCS) and deleting characters not in it.

2. Step 2: Recognize DP approach based on LCS guarantees optimal solution

   Greedy or sorting approaches fail because they do not consider subsequence order; brute force is correct but inefficient.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   LCS-based DP is classic for minimal delete operations [OK]

Hint: Minimal deletes -> LCS-based DP [OK]

Common Mistakes:

• Thinking greedy deletion works optimally
• Ignoring subsequence order
• Confusing sorting with subsequence

2. You are given a string s and a dictionary of words. You need to find all possible sentences where s can be segmented into a sequence of dictionary words. Which approach guarantees finding all valid sentences efficiently without redundant recomputation?

easy

A. Dynamic programming with bottom-up tabulation and backtracking to build all sentences

B. Greedy algorithm that picks the longest matching word at each step

C. Pure brute force recursion without memoization

D. Top-down recursion without caching intermediate results

Solution

1. Step 1: Understand problem requirements

   The problem requires enumerating all possible valid sentences, not just checking existence or counting.

2. Step 2: Evaluate approaches

   Greedy fails because it may miss valid segmentations. Pure recursion without memoization leads to exponential recomputation. Top-down recursion without caching also recomputes subproblems. Bottom-up DP with backtracking efficiently builds all solutions by storing intermediate results.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Bottom-up DP with backtracking enumerates all sentences efficiently [OK]

Hint: Bottom-up DP with backtracking enumerates all solutions [OK]

Common Mistakes:

• Thinking greedy or pure recursion suffices for all sentences

3. Suppose you want to find the length of the longest common subsequence between two strings, but now characters from the second string can be reused multiple times. Which modification to the standard LCS algorithm correctly handles this scenario?

hard

A. Use the same DP but allow dp[j] to depend on dp[j] itself without decrementing j in inner loop

B. Change the inner loop to iterate forwards and update dp[j] using dp[j] and dp[j-1] to allow reuse

C. Use a greedy approach matching characters from left to right allowing reuse

D. Use a bottom-up DP with a 2D table but ignore previous matches to allow reuse

Solution

1. Step 1: Understand reuse impact

   Allowing reuse of characters in the second string means dp[j] can be updated based on dp[j] itself (not only dp[j-1]) and the inner loop must run forwards to propagate reuse correctly.

2. Step 2: Identify correct DP iteration

   Iterating forwards in the inner loop allows dp[j] to accumulate matches with reuse, unlike standard LCS which iterates backwards to avoid reuse.

3. Step 3: Evaluate other options

   Greedy fails to find optimal subsequence. Ignoring previous matches breaks DP correctness. Using 2D DP without reuse logic is insufficient.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Forward iteration in inner loop enables reuse of characters [OK]

Hint: Forward inner loop iteration enables reuse in DP [OK]

Common Mistakes:

• Using backward iteration preventing reuse
• Trying greedy approach
• Ignoring DP state dependencies

4. Suppose the problem is modified so that you want to find the minimum cuts for palindrome partitioning but now the input string can contain wildcard characters '?' that can match any single character. Which of the following approaches correctly adapts the optimal DP solution to handle wildcards?

hard

A. Replace all '?' with all possible characters and run DP multiple times, taking minimum cuts over all results

B. Precompute all palindrome substrings ignoring '?', then run DP as usual

C. Use pure recursion with memoization since DP cannot handle wildcards efficiently

D. Modify palindrome checks to treat '?' as matching any character during center expansions, keeping the same DP structure

Solution

1. Step 1: Understand wildcard impact on palindrome checks

   Wildcards '?' can match any character, so palindrome checks must be adapted to treat '?' as a match with any character.

2. Step 2: Adapt DP palindrome center expansion

   Modify the center expansion condition to allow s[left] == s[right] or either is '?', preserving O(n^2) time and DP structure.

3. Step 3: Evaluate other options

   Precomputing ignoring '?' is incorrect as it misses wildcard matches. Pure recursion is inefficient. Replacing '?' with all characters is exponential and impractical.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   Adapting palindrome checks for '?' preserves DP correctness and efficiency [OK]

Hint: Adapt palindrome checks, keep DP structure for wildcards [OK]

Common Mistakes:

• Ignoring wildcard impact on palindrome checks
• Trying exponential brute force over all replacements
• Assuming DP cannot handle wildcards

5. Suppose you want to find the shortest common supersequence when characters from the input strings can be reused unlimited times. Which modification to the DP approach correctly handles this variant?

hard

A. Change the recurrence to allow staying at the same index in one string while advancing in the other

B. Modify the DP to consider infinite loops by adding memoization to avoid repeated states

C. Use the same DP but allow indices i and j to reset to zero after reaching the end, enabling reuse

D. No change needed; the original DP already accounts for unlimited reuse of characters

Solution

1. Step 1: Understand that unlimited reuse means characters can be used multiple times, so indices may not always advance

   Original DP always advances both indices or one index forward, disallowing reuse.

2. Step 2: Modify recurrence to allow staying at the same index in one string while advancing in the other

   This models reuse by allowing repeated characters from one string to be matched multiple times.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Allowing index to stay while advancing the other correctly models reuse [OK]

Hint: Reuse means indices may not always advance simultaneously [OK]

Common Mistakes:

• Assuming original DP handles reuse
• Trying to reset indices naively
• Adding memoization without changing recurrence