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Problem: Can Construct K Palindromes

This document explains step-by-step how the solution is implemented in various programming languages (C++, Java, JavaScript, Python, Go) to determine whether it is possible to construct k palindrome strings using all the characters of a given string s.

General Steps

Each implementation follows these steps:

  1. Initial Check:

    • If k (the number of palindromes to construct) is greater than the length of the string s, it is impossible to construct the palindromes. Return false.

  2. Count Character Frequencies:

    • Use a frequency array (or hashmap) to count how many times each character appears in the string.

  3. Count Odd Frequencies:

    • Palindromes allow at most one character to have an odd frequency. Traverse the frequency array and count the characters that appear an odd number of times.

  4. Decision:

    • If the number of odd frequencies is less than or equal to k, return true. Otherwise, return false.

C++ Implementation Explanation

  1. Initial Check:
    • Verify if k > s.length(). If true, return false.
  2. Create Frequency Array:
    • Use a vector<int> of size 26 to count the occurrences of each character in the string.
  3. Iterate Through the String:
    • For every character in s, increment its corresponding frequency in the array.
  4. Count Odd Frequencies:
    • Traverse the frequency array and count how many values are odd.
  5. Return Result:
    • Check if the number of odd frequencies is less than or equal to k. Return true or false accordingly.

Java Implementation Explanation

  1. Initial Check:
    • If k > s.length(), return false because more palindromes than characters cannot be created.
  2. Create Frequency Array:
    • Use an int[] array of size 26 to store character frequencies.
  3. Iterate Through the String:
    • Convert the string into a character array and update the frequency for each character in the array.
  4. Count Odd Frequencies:
    • Traverse the frequency array and count characters that have odd occurrences.
  5. Return Result:
    • If the number of odd frequencies is less than or equal to k, return true. Otherwise, return false.

JavaScript Implementation Explanation

  1. Initial Check:
    • If k > s.length, return false because we cannot have more palindromes than the total number of characters in the string.
  2. Create Frequency Array:
    • Use an array of size 26 initialized with zeros to count the occurrences of each character in the string.
  3. Iterate Through the String:
    • For each character in the string, calculate its ASCII code, map it to the array index, and increment the corresponding frequency.
  4. Count Odd Frequencies:
    • Traverse the frequency array and count the number of characters with odd occurrences.
  5. Return Result:
    • Compare the odd frequency count with k. Return true if the count is less than or equal to k, otherwise return false.

Python Implementation Explanation

  1. Initial Check:
    • If k > len(s), return False. This ensures the number of palindromes does not exceed the total number of characters.
  2. Create Frequency Array:
    • Use a list of size 26 initialized with zeros to count the occurrences of each character.
  3. Iterate Through the String:
    • Convert each character to its corresponding index (using ord(char) - ord('a')) and increment its frequency in the list.
  4. Count Odd Frequencies:
    • Use a generator expression to count the number of characters with odd frequencies in the list.
  5. Return Result:
    • If the odd frequency count is less than or equal to k, return True. Otherwise, return False.

Go Implementation Explanation

  1. Initial Check:
    • If k > len(s), return false because constructing more palindromes than available characters is impossible.
  2. Create Frequency Array:
    • Use a slice of size 26 initialized with zeros to store the frequencies of each character in the string.
  3. Iterate Through the String:
    • Convert each character to its corresponding index (char - 'a') and increment the frequency at that index.
  4. Count Odd Frequencies:
    • Iterate through the frequency slice and count how many characters have odd frequencies.
  5. Return Result:
    • Return true if the odd frequency count is less than or equal to k, otherwise return false.

Key Notes

  • All implementations share the same logic but are adapted to each language’s syntax and features.
  • The key is understanding the connection between character frequencies and palindrome formation.