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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;
    }
};

  • 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;
    }
}

############

class Solution {
    public boolean canConstruct(String s, int k) {
        if (s.length() < k) {
            return false;
        }
        int[] counter = new int[26];
        for (char c : s.toCharArray()) {
            ++counter[c - 'a'];
        }
        int cnt = 0;
        for (int v : counter) {
            if (v % 2 == 1) {
                ++cnt;
            }
        }
        return cnt <= 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;
    }
};

  • class Solution:
    def canConstruct(self, s: str, k: int) -> bool:
        if len(s) < k:
            return False
        counter = Counter(s)
        cnt = sum(1 for n in counter.values() if n % 2 == 1)
        return cnt <= k



  • func canConstruct(s string, k int) bool {
	if len(s) < k {
		return false
	}
	counter := make([]int, 26)
	for _, c := range s {
		counter[c-'a']++
	}
	cnt := 0
	for _, v := range counter {
		if v%2 == 1 {
			cnt++
		}
	}
	return cnt <= 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;
}

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