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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;
    }
    
    

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