1400. Construct K Palindrome Strings

1400. Construct K Palindrome Strings

Difficulty: Medium

Topics: Hash Table, String, Greedy, Counting

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

Hint:

1. If the s.length < k we cannot construct k strings from s and answer is false.
2. If the number of characters that have odd counts is > k then the minimum number of palindrome strings we can construct is > k and answer is false.
3. Otherwise you can construct exactly k palindrome strings and answer is true (why ?).

Solution:

We need to consider the following points:

Key Observations:

1. Palindrome Characteristics:

   • A palindrome is a string that reads the same forwards and backwards.
   • For an even-length palindrome, all characters must appear an even number of times.
   • For an odd-length palindrome, all characters except one must appear an even number of times (the character that appears an odd number of times will be in the center).

2. Necessary Conditions:

   • If the length of s is less than k, it's impossible to form k strings, so return false.
   • The total number of characters that appear an odd number of times must be at most k to form k palindromes. This is because each palindrome can have at most one character with an odd count (the center character for odd-length palindromes).

Approach:

1. Count the frequency of each character in the string.
2. Count how many characters have an odd frequency.
3. If the number of odd frequencies exceeds k, return false (because it's impossible to form k palindromes).

Let's implement this solution in PHP: 1400. Construct K Palindrome Strings

<?php
/**
 * @param String $s
 * @param Integer $k
 * @return Boolean
 */
function canConstruct($s, $k) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Test cases
var_dump(canConstruct("annabelle", 2)); // Output: true
var_dump(canConstruct("leetcode", 3)); // Output: false
var_dump(canConstruct("true", 4));      // Output: true
?>

Explanation:

1. Frequency Count: We use an associative array $freq to count the occurrences of each character in the string.
2. Odd Count: We check how many characters have odd occurrences. This will help us determine if we can form palindromes.
3. Condition Check: If the number of characters with odd frequencies is greater than k, it's impossible to form k palindromes, so we return false. Otherwise, we return true.

Time Complexity:

• Counting the frequencies takes O(n), where n is the length of the string.
• Checking the odd frequencies takes O(m), where m is the number of distinct characters (at most 26 for lowercase English letters).
• The overall time complexity is O(n + m), which simplifies to O(n) in this case.

Edge Cases:

1. If k is greater than the length of s, we return false.
2. If all characters have even frequencies, we can always form a palindrome, so the result depends on whether k is possible.

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