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
Solution: HashTable
Compute the frequency of each characters.
Count the # of characters with odd frequency |odd|, each palindrome can consume at most one char with odd frequency. thus k must >= |odd|.
ans = k <= len(s) and k >= odd
Time complexity: O(n)
Space complexity: O(1)
C++
1 2 3 4 5 6 7 8 9 10 11 |
// Author: Huahua class Solution { public: bool canConstruct(string s, int k) { if (k > s.length()) return false; vector<int> f(26); for (char c : s) f[c - 'a'] ^= 1; int odd = accumulate(begin(f), end(f), 0); return k >= odd; } }; |
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