<p>Given a string <code>s</code> and an integer <code>k</code>, return <code>true</code> <em>if you can use all the characters in </em><code>s</code><em> to construct </em><code>k</code><em> palindrome strings or </em><code>false</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;annabelle&quot;, k = 2
<strong>Output:</strong> true
<strong>Explanation:</strong> You can construct two palindromes using all characters in s.
Some possible constructions &quot;anna&quot; + &quot;elble&quot;, &quot;anbna&quot; + &quot;elle&quot;, &quot;anellena&quot; + &quot;b&quot;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;leetcode&quot;, k = 3
<strong>Output:</strong> false
<strong>Explanation:</strong> It is impossible to construct 3 palindromes using all the characters of s.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;true&quot;, k = 4
<strong>Output:</strong> true
<strong>Explanation:</strong> The only possible solution is to put each character in a separate string.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def canConstruct(self, s: str, k: int) -> bool:
        c = Counter(s)
        pairs = 0
        singles = 0
        for v in c.values():
            pairs += v // 2 * 2
            singles += v - v // 2 * 2
        k -= singles
        print(k, pairs)
        return k >= 0 and (k % 2 == 0 and k <= pairs or k % 2 == 1 and k <= pairs - 1)
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