Longest Palindrome

To solve this problem, you can use a hashmap (in Python it’s a dictionary) to count the frequency of each character in the string. The longest palindrome that can be constructed from these characters is twice the sum of the even counts, plus one if there is any character with an odd count.

Here is the Python code:

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class Solution:
    def longestPalindrome(self, s: str) -> int:
        count = collections.Counter(s)
        length = 0
        has_odd = False
        for val in count.values():
            length += (val // 2) * 2
            if val % 2 == 1:
                has_odd = True
        return length + has_odd

In this solution:

  1. We use collections.Counter to count the occurrence of each character in s.
  2. We initialize the length of the longest palindrome that can be constructed to 0.
  3. We iterate through the values in the count dictionary (which represent the counts of each character). For each value, we add to length the largest even number less than or equal to the value (which is (val // 2) * 2).
  4. If the value is odd, we set has_odd to True.
  5. At the end, we return length + has_odd. This means if there’s any character with an odd count, we can put one of them in the middle of the palindrome. Otherwise, the length of the longest palindrome is just twice the sum of the even counts.

This is a frequency counting problem.

Yes, you’re correct. This problem revolves around counting the frequency of each character in the string. Since a palindrome is a string that reads the same backward as forward, it must be constructed from characters with even counts, plus possibly one character with an odd count in the middle.

By counting the frequency of each character, you can determine how many characters have even counts and whether any characters have odd counts. The longest palindrome that can be constructed is twice the sum of the even counts, plus one if there is any character with an odd count.

You don’t need to construct the palindrome, since the question only asks the lenght of the palindrome that can be constructed.

Absolutely right! We only need to calculate the length of the longest palindrome that can be constructed, not the actual palindrome itself. This significantly simplifies the problem. By calculating the frequency of each character, we can find the maximum length of a palindrome that can be built from these characters, without needing to actually construct the palindrome string.

Identifying Problem Isomorphism

“Longest Palindrome” can be approximately mapped to “Valid Palindrome”.

In “Longest Palindrome”, you are given a string and asked to determine the length of the longest palindrome that can be built with those letters. This means, you need to determine how the letters can be rearranged to form the longest possible palindrome.

In “Valid Palindrome”, you are given a string and you need to verify if it is a palindrome considering only alphanumeric characters and ignoring cases.

Both problems are related to the concept of palindromes, strings that read the same backwards and forwards. The complexity is in different aspects of the problem. For “Longest Palindrome”, you have to count occurrences and understand how to rearrange those occurrences to maximize the palindrome’s length. For “Valid Palindrome”, the problem is simpler as you simply need to check if the given string is a palindrome.

“Longest Palindrome” is more complex as it involves rearranging and maximizing the length of the palindrome, while “Valid Palindrome” is simpler and only involves verifying if a string is a palindrome or not.

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class Solution:
    def longestPalindrome(self, s: str) -> int:
        odd_count = 0
        d = {}
        for ch in s:
            if ch in d:
                d[ch] += 1
            else:
                d[ch] = 1
            if d[ch] % 2 == 1:
                odd_count += 1
            else:
                odd_count -= 1
        if odd_count > 1:
            return len(s) - odd_count + 1
        return len(s)

Problem Classification

This problem can be classified under:

  1. String Manipulation: The problem deals with a string input and requires operations to be performed on this string.

  2. Counting: The problem requires counting the frequency of each character in the string.

  3. Palindrome: A key part of the problem is understanding the properties of palindromes and how they can be constructed.

  4. Case Sensitivity: The problem specifically mentions that letters are case sensitive, meaning “A” and “a” are considered distinct.

  5. Mathematics: The problem involves finding the maximum length of a palindrome, which can be viewed as a kind of optimization problem.

  6. Data Structures: You might use data structures like Hash Table or Set to solve the problem.

Language Agnostic Coding Drills

Here’s a breakdown of the problem into smaller units of learning and their step-by-step approach:

  1. Understanding the problem statement: The problem requires us to construct a palindrome from the input string. A palindrome is a word or phrase that reads the same forwards and backwards. In this case, we need to return the length of the longest palindrome that can be created from the given characters. This will require an understanding of how palindromes work.

  2. Counting characters in a string: The solution to this problem involves counting the number of occurrences of each character in the string. We can maintain a dictionary (or hash table) where the keys are the characters in the string and the values are their counts.

  3. Dealing with odd and even counts: The properties of palindromes dictate that we can have at most one character that appears an odd number of times (in the center of the palindrome), while every other character must appear an even number of times. If a character has an even count, we can use all of them in the palindrome. If a character has an odd count, we can use one less than its count to maintain the palindrome property.

  4. Maintaining a count of characters with odd frequencies: As we iterate through the string, we can maintain a count of characters that have odd frequencies. If we encounter a character an odd number of times, we increment the odd_count. If we encounter a character an even number of times, we decrement the odd_count.

  5. Calculating the maximum length of the palindrome: Finally, if the odd_count is greater than 1, we subtract it from the total length of the string and add 1 (to account for the one character that can appear an odd number of times). If the odd_count is not greater than 1, the length of the string is the maximum length of the palindrome we can construct.

Combining all these concepts will provide the final solution to the problem.

Targeted Drills in Python

  1. Understanding the problem statement: This step does not involve any coding, it’s just about understanding what a palindrome is and what we are being asked to find.

  2. Counting characters in a string:

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def count_chars(s: str) -> dict:
    char_count = {}
    for ch in s:
        if ch in char_count:
            char_count[ch] += 1
        else:
            char_count[ch] = 1
    return char_count

print(count_chars('abba'))  # Output: {'a': 2, 'b': 2}
  1. Dealing with odd and even counts:
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def handle_counts(char_count: dict) -> dict:
    for char in char_count:
        if char_count[char] % 2 == 0:
            char_count[char] = char_count[char]
        else:
            char_count[char] = char_count[char] - 1
    return char_count

print(handle_counts({'a': 2, 'b': 3}))  # Output: {'a': 2, 'b': 2}
  1. Maintaining a count of characters with odd frequencies:
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def count_odd(char_count: dict) -> int:
    odd_count = 0
    for char in char_count:
        if char_count[char] % 2 != 0:
            odd_count += 1
    return odd_count

print(count_odd({'a': 2, 'b': 3}))  # Output: 1
  1. Calculating the maximum length of the palindrome:
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def max_palindrome_len(s: str, odd_count: int) -> int:
    if odd_count > 1:
        return len(s) - odd_count + 1
    return len(s)

print(max_palindrome_len('abba', 1))  # Output: 4

After understanding and coding these individual parts, the final problem is a combination of all these steps. You count the characters, handle odd and even counts, count the odd frequencies, and calculate the maximum length of the palindrome that can be built. This forms the final solution to the problem.