Welcome to Subscribe On Youtube

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
• C++
• Python
• Go

• 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
  }
  

All Problems

All Solutions