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I did not find a solution to problem number 1400 in the Python language - that's why I added a solution to this problem, in addition I updated the readme file.
from collections import Counter
def can_construct_palindrome(s, k):
# Count the frequency of each character in the string
char_count = Counter(s)
# Count the number of characters with odd frequency
odd_count = sum(count % 2 for count in char_count.values())
# If k is greater than or equal to the number of odd-count characters,
# it's possible to construct k palindromic strings
return k >= odd_count and k <= len(s)
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welcome commentedon May 3, 2024
Thanks for helping us improve and opening your first issue here! Don't forget to give us a 🌟 to support us.
While you're waiting, I just wanted to make sure you've had a chance to look at our Readme and Pull Request Guidelines.
crazyGru commentedon Jun 12, 2024