# 1400. Construct K Palindrome Strings

## Description

Given a string s and an integer k, 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.


Constraints:

• 1 <= s.length <= 105
• s consists of lowercase English letters.
• 1 <= k <= 105

## Solutions

Solution 1: Counting

First, we check if the length of string $s$ is less than $k$. If it is, we cannot construct $k$ palindrome strings, so we can directly return false.

Otherwise, we use a hash table or an array $cnt$ to count the occurrences of each character in string $s$. Finally, we only need to count the number of characters $x$ that appear an odd number of times in $cnt$. If $x$ is greater than $k$, we cannot construct $k$ palindrome strings, so we return false; otherwise, we return true.

The time complexity is $O(n)$, and the space complexity is $O(C)$. Where $n$ is the length of string $s$, and $C$ is the size of the character set, here $C=26$.

• class Solution {
public boolean canConstruct(String s, int k) {
int n = s.length();
if (n < k) {
return false;
}
int[] cnt = new int[26];
for (int i = 0; i < n; ++i) {
++cnt[s.charAt(i) - 'a'];
}
int x = 0;
for (int v : cnt) {
x += v & 1;
}
return x <= k;
}
}

• class Solution {
public:
bool canConstruct(string s, int k) {
if (s.size() < k) {
return false;
}
int cnt[26]{};
for (char& c : s) {
++cnt[c - 'a'];
}
int x = 0;
for (int v : cnt) {
x += v & 1;
}
return x <= k;
}
};

• class Solution:
def canConstruct(self, s: str, k: int) -> bool:
if len(s) < k:
return False
cnt = Counter(s)
return sum(v & 1 for v in cnt.values()) <= k


• func canConstruct(s string, k int) bool {
if len(s) < k {
return false
}
cnt := [26]int{}
for _, c := range s {
cnt[c-'a']++
}
x := 0
for _, v := range cnt {
x += v & 1
}
return x <= k
}

• function canConstruct(s: string, k: number): boolean {
if (s.length < k) {
return false;
}
const cnt: number[] = new Array(26).fill(0);
for (const c of s) {
++cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)];
}
let x = 0;
for (const v of cnt) {
x += v & 1;
}
return x <= k;
}