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 $C=26$.

Python3JavaC++GoTypeScript

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

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;
    }
};

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;
}

1400. Construct K Palindrome Strings

Description

Solutions

Solution 1: Counting

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