[leetcode] 1400. Construct K Palindrome Strings

Description

Given a string s and an integer k. You should construct k non-empty palindrome strings using all the characters in s.

Return True if you can use all the characters in s to construct k palindrome strings or False otherwise.

Example 1:

Input: s = "annabelle", k = 2
Output: true
Explanation: You can construct two palindromes using all characters in s.
Some possible constructions "anna" + "elble", "anbna" + "elle", "anellena" + "b"

Example 2:

Input: s = "leetcode", k = 3
Output: false
Explanation: It is impossible to construct 3 palindromes using all the characters of s.

Example 3:

Input: s = "true", k = 4
Output: true
Explanation: The only possible solution is to put each character in a separate string.

Example 4:

Input: s = "yzyzyzyzyzyzyzy", k = 2
Output: true
Explanation: Simply you can put all z's in one string and all y's in the other string. Both strings will be palindrome.

Example 5:

Input: s = "cr", k = 7
Output: false
Explanation: We don't have enough characters in s to construct 7 palindromes.

Constraints:

• 1 <= s.length <= 10^5
• All characters in s are lower-case English letters.
• 1 <= k <= 10^5

分析

题目的意思是：给定一个字符串，问能否构成K个回文子串。
这道题我的思路很直接：

• 首先字符串长度小于K的话，肯定不能构成K个回文字串。
• 其次，统计字符的频率，如果字符的奇数项的个数大于K的话，则不能构成，因为回文字串最多智能有一个奇数的字母。小于等于K则能构成了哈。

代码

class Solution:
    def canConstruct(self, s: str, k: int) -> bool:
        if(len(s)<k):
            return False
        d=defaultdict(int)
        for ch in s:
            d[ch]+=1
        res=0
        for key,v in d.items():
            if(v%2==1):
                res+=1
        if(res<=k):
            return True
        else:
            return False