11/30/2023 0 Comments Palindrome permutation![]() ![]() Return sum(freq%2 for freq in Counter(data).values()) bool:ĭata = Counter(data.replace(' ', ''). You can return the results of expressions directly.So really all you care about is if there is more than 1 odd count. A simple solution is to run two loops, the outer loop picks all characters one by one, and the inner loop counts the number of occurrences of the picked character. If num_odd is 1, then there must be an odd number of elements. A set of characters can form a palindrome if at most one character occurs an odd number of times and all characters occur an even number of times. If num_odd is 0, then there must be an even number of elements.The only possible values for freq % 2 are 0 and 1, so num_odd = sum(1 for char, freq in Counter(data).items() if char != ' ' and freq % 2 = 1) can be simplified as num_odd = sum(freq % 2 for Counter(data).values()) 266 - Palindrome Permutation Leetcode 266 - Palindrome Permutation Posted on Aug2 minute read Welcome to Subscribe On Youtube Question Formatted question description: Given a string s, return true if a permutation of the string could form a palindrome and false otherwise.You strip out spaces, then on the next line check if each character is a space.If num_odd = 1 and len(data) % 2 = 1 or num_odd = 0 and len(data) % 2 = 0: # Check for two types of palindromes, 1) Odd Length (e.g. Num_odd = sum(1 for char, freq in Counter(data).items() if char != ' ' and freq % 2 = 1) """Given a string, check if it is a permutation of a palindrome.""" ![]() If length is even no letter should have an odd frequency. We will generate the permutations for half of the frequency of each character and then append the reverse of the permutation generated. If length is odd then only one character must contain an odd count. Fill with the frequency with the count of characters. I believe my code has time complexity \$O(n)\$ and space complexity \$ O(n)\$ is that correct? If so, is there any way to improve on either the space or time complexity? from collections import Counterĭef is_palindrome_permutation(data: str) -> bool: Check whether letters of string can make a palindrome or not if it cant form a palindrome return. I wanted to get feedback on making my code clearer and more pythonic. Output: True (permutations: "taco cat", "atco cta", etc.) Given a string, write a function to check if it is a permutation of a palindrome. ![]() In simple terms, the problem states that: Make the string a palindrome by rearranging it in any way by removing the minimum number of characters including removing 0 number of character if possible. This is a solution to exercise 1.4 from Cracking the Coding Interview written using Python 3. Practice Given a string S, we have to find minimum characters that we can remove to make any permutation of the string S a palindrome. ![]()
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