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;
    }
}

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