Formatted question description: https://leetcode.ca/all/1400.html

# 1400. Construct K Palindrome Strings (Medium)

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

Related Topics:
Greedy

## Solution 1.

Trivial cases:

• k > N, false
• k == N, true

Then just count the characters with odd frequency. Return true is the frequency <= k.

// OJ: https://leetcode.com/problems/construct-k-palindrome-strings/

// Time: O(N)
// Space: O(1)
class Solution {
public:
bool canConstruct(string s, int k) {
int N = s.size();
if (k > N) return false;
if (k == N) return true;
int cnt[26] = {0};
for (char c : s) cnt[c - 'a']++;
int sum = 0;
for (int i = 0; i < 26; ++i) if (cnt[i] % 2) ++sum;
return sum <= k;
}
};


Java

class Solution {
public boolean canConstruct(String s, int k) {
int length = s.length();
if (length < k)
return false;
if (length == k)
return true;
int[] counts = new int[26];
for (int i = 0; i < length; i++) {
char c = s.charAt(i);
counts[c - 'a']++;
}
int oddCount = 0;
for (int i = 0; i < 26; i++) {
if (counts[i] % 2 == 1)
oddCount++;
}
return oddCount <= k;
}
}