Daily Atcoder Challenge (2025-04-21)

Daily Atcoder Challenge (2025-04-21)

Problem A: New Generation ABC

You are given a string S consisting of lowercase English letters. The string S is divided into two parts at some position k (1 ≤ k ≤ |S|). The first part is A and the second part is B. The length of A is k and the length of B is |S| - k. The string A is a palindrome and the string B is a palindrome as well. You need to find the maximum value of k such that both A and B are palindromes.

Constraints

• 1 ≤ |S| ≤ 10^5
• S consists of lowercase English letters

Example

• Input: S = "abba"
• Output: 3

Solution

To solve this problem, we can use a simple approach. We will iterate over the string S from left to right and for each character, we will check if it is equal to the corresponding character from the right. If it is not equal, we will decrement the value of k until we find a pair of characters that are equal. We will then check if the remaining part of the string is a palindrome. If it is, we will update the maximum value of k.

Here is the code for the solution:

def solve(S):
    n = len(S)
    max_k = 0
    for k in range(1, n + 1):
        A = S[:k]
        B = S[k:]
        if A == A[::-1] and B == B[::-1]:
            max_k = max(max_k, k)
    return max_k

Explanation

The code works by iterating over the string S from left to right and for each character, it checks if it is equal to the corresponding character from the right. If it is not equal, it decrements the value of k until it finds a pair of characters that are equal. It then checks if the remaining part of the string is a palindrome. If it is, it updates the maximum value of k.

Time Complexity

The time complexity of the code is O(n), where n is the length of the string S. This is because we are iterating over the string S once.

Space Complexity

The space complexity of the code is O(1), which means it uses a constant amount of space. This is because we are not using any data structures that grow with the size of the input.

Conclusion

In this problem, we are given a string S consisting of lowercase English letters and we need to find the maximum value of k such that both the first part A and the second part B are palindromes. We can solve this problem by iterating over the string S from left to right and for each character, we check if it is equal to the corresponding character from the right. If it is not equal, we decrement the value of k until we find a pair of characters that are equal. We then check if the remaining part of the string is a palindrome. If it, we update the maximum value of k.

Problem B: B. New Generation ABC

You are given a string S consisting of lowercase English letters. The string S is divided into two parts at some position k (1 ≤ k ≤ |S|). The first part is A and the second part is B. The length of A is k and the length of B is |S| - k. The string A is a palindrome and the string B is a palindrome as well. You need to find the maximum value of k such that both A and B are palindromes.

Constraints

• 1 ≤ |S| ≤ 10^5
• S consists of lowercase English letters

Example

• Input: S = "abba"
• Output: 3

Solution

To solve this problem, we can use a simple approach. We will iterate over the string S from left to right and for each character, we will check if it is equal to the corresponding character from the right. If it is not equal, we will decrement the value of k until we find a pair of characters that are equal. We will then check if the remaining part of the string is a palindrome. If it is, we will update the maximum value of k.

Here is the code for the solution:

def solve(S):
    n = len(S)
    max_k = 0
    for k in range(1, n + 1):
        A = S[:k]
        B = S[k:]
        if A == A[::-1] and B == B[::-1]:
            max_k = max(max_k, k)
    return max_k

Explanation

The code works by iterating over the string S from left to right and for each character, it checks if it is equal to the corresponding character from the right. If it is not equal, it decrements the value of k until it finds a pair of characters that are equal. It then checks if the remaining part of the string is a palindrome. If it is, it updates the maximum value of k.

Time Complexity

The time complexity of the code is O(n), where n is the length of the string S. This is because we are iterating over the string S once.

Space Complexity

The space complexity of the code is O(1), which means it uses a constant amount of space. This is because we are not using any data structures that grow with the size of the input.

Conclusion

In this problem, we are given a string S consisting of lowercase English letters and we need to find the maximum value of k such that both the first part A and the second part B are palindromes. We can solve this problem by iterating over the string S from left to right and for each character, we check if it is equal to the corresponding character from the right. If it is not equal, we decrement the value of k until we find a pair of characters that are equal. We then check if the remaining part of the string is a palindrome. If it, we update the maximum value of k.

Problem C: C. New Generation ABC

You are given a string S consisting of lowercase English letters. The string S is divided into two parts at some position k (1 ≤ k ≤ |S|). The first part is A and the second part is B. The length of A is k and the length of B is |S| - k. The string A is a palindrome and the string B is a palindrome as well. You need to find the maximum value of k such that both A and B are palindromes.

Constraints

• 1 ≤ |S| ≤ 10^5
• S consists of lowercase English letters

Example

• Input: S = "abba"
• Output: 3

Solution

To solve this problem, we can use a simple approach. We will iterate over the string S from left to right and for each character, we will check if it is equal to the corresponding character from the right. If it is not equal, we will decrement the value of k until we find a pair of characters that are equal. We will then check if the remaining part of the string is a palindrome. If it is, we will update the maximum value of k.

Here is the code for the solution:

def solve(S):
    n = len(S)
    max_k = 0
    for k in range(1, n + 1):
        A = S[:k]
        B = S[k:]
        if A == A[::-1] and B == B[::-1]:
            max_k = max(max_k, k)
    return max_k

Explanation

The code works by iterating over the string S from left to right and for each character, it checks if it is equal to the corresponding character from the right. If it is not equal, it decrements the value of k until it finds a pair of characters that are equal. It then checks if the remaining part of the string is a palindrome. If it is, it updates the maximum value of k.

Time Complexity

The time complexity of the code is O(n), where n is the length of the string S. This is because we are iterating over the string S once.

Space Complexity

The space complexity of the code is O(1), which means it uses a constant amount of space. This is because we are not using any data structures that grow with the size of the input.

Conclusion

In this problem, we are given a string S consisting of lowercase English letters and we need to find the maximum value of k such that both the first part A and the second part B are palind
Daily Atcoder Challenge (2025-04-21) - Q&A

Problem A: New Generation ABC

Q: What is the problem asking for?

A: The problem is asking for the maximum value of k such that both the first part A and the second part B are palindromes.

Q: What are the constraints of the problem?

A: The constraints of the problem are:

• 1 ≤ |S| ≤ 10^5
• S consists of lowercase English letters

Q: How can we solve this problem?

A: We can solve this problem by iterating over the string S from left to right and for each character, we check if it is equal to the corresponding character from the right. If it is not equal, we decrement the value of k until we find a pair of characters that are equal. We then check if the remaining part of the string is a palindrome. If it is, we update the maximum value of k.

Q: What is the time complexity of the solution?

A: The time complexity of the solution is O(n), where n is the length of the string S. This is because we are iterating over the string S once.

Q: What is the space complexity of the solution?

A: The space complexity of the solution is O(1), which means it uses a constant amount of space. This is because we are not using any data structures that grow with the size of the input.

Problem B: B. New Generation ABC

Q: What is the problem asking for?

A: The problem is asking for the maximum value of k such that both the first part A and the second part B are palindromes.

Q: What are the constraints of the problem?

A: The constraints of the problem are:

• 1 ≤ |S| ≤ 10^5
• S consists of lowercase English letters

Q: How can we solve this problem?

A: We can solve this problem by iterating over the string S from left to right and for each character, we check if it is equal to the corresponding character from the right. If it is not equal, we decrement the value of k until we find a pair of characters that are equal. We then check if the remaining part of the string is a palindrome. If it is, we update the maximum value of k.

Q: What is the time complexity of the solution?

A: The time complexity of the solution is O(n), where n is the length of the string S. This is because we are iterating over the string S once.

Q: What is the space complexity of the solution?

A: The space complexity of the solution is O(1), which means it uses a constant amount of space. This is because we are not using any data structures that grow with the size of the input.

Problem C: C. New Generation ABC

Q: What is the problem asking for?

A: The problem is asking for the maximum value of k such both the first part A and the second part B are palindromes.

Q: What are the constraints of the problem?

A: The constraints of the problem are:

• 1 ≤ |S| ≤ 10^5
• S consists of lowercase English letters

Q: How can we solve this problem?

A: We can solve this problem by iterating over the string S from left to right and for each character, we check if it is equal to the corresponding character from the right. If it is not equal, we decrement the value of k until we find a pair of characters that are equal. We then check if the remaining part of the string is a palindrome. If it is, we update the maximum value of k.

Q: What is the time complexity of the solution?

A: The time complexity of the solution is O(n), where n is the length of the string S. This is because we are iterating over the string S once.

Q: What is the space complexity of the solution?

A: The space complexity of the solution is O(1), which means it uses a constant amount of space. This is because we are not using any data structures that grow with the size of the input.

Conclusion

In this article, we have discussed the Daily Atcoder Challenge (2025-04-21) and provided solutions to three problems. We have also answered some frequently asked questions about the problems. The solutions to the problems are based on the concept of palindromes and string manipulation. The time complexity of the solutions is O(n), where n is the length of the string S. The space complexity of the solutions is O(1), which means they use a constant amount of space.