<p>Given two strings <code>ransomNote</code> and <code>magazine</code>, return <code>true</code><em> if </em><code>ransomNote</code><em> can be constructed by using the letters from </em><code>magazine</code><em> and </em><code>false</code><em> otherwise</em>.</p>

<p>Each letter in <code>magazine</code> can only be used once in <code>ransomNote</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> ransomNote = "a", magazine = "b"
<strong>Output:</strong> false
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> ransomNote = "aa", magazine = "ab"
<strong>Output:</strong> false
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> ransomNote = "aa", magazine = "aab"
<strong>Output:</strong> true
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= ransomNote.length, magazine.length &lt;= 10<sup>5</sup></code></li>
	<li><code>ransomNote</code> and <code>magazine</code> consist of lowercase English letters.</li>
</ul>

class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        ransom_note_counter = Counter(ransomNote)
        magazine_counter = Counter(magazine)

        for i in range(len(ransomNote)):
            if ransomNote[i] in magazine_counter:
                ransom_note_counter[ransomNote[i]] -= 1
                magazine_counter[ransomNote[i]] -= 1
                if magazine_counter[ransomNote[i]] == 0:
                    del magazine_counter[ransomNote[i]]
                if ransom_note_counter[ransomNote[i]] == 0:
                    del ransom_note_counter[ransomNote[i]]
            else:
                return False

        return len(ransom_note_counter) == 0
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