<p>You are given two <strong>0-indexed</strong> strings <code>str1</code> and <code>str2</code>.</p>

<p>In an operation, you select a <strong>set</strong> of indices in <code>str1</code>, and for each index <code>i</code> in the set, increment <code>str1[i]</code> to the next character <strong>cyclically</strong>. That is <code>&#39;a&#39;</code> becomes <code>&#39;b&#39;</code>, <code>&#39;b&#39;</code> becomes <code>&#39;c&#39;</code>, and so on, and <code>&#39;z&#39;</code> becomes <code>&#39;a&#39;</code>.</p>

<p>Return <code>true</code> <em>if it is possible to make </em><code>str2</code> <em>a subsequence of </em><code>str1</code> <em>by performing the operation <strong>at most once</strong></em>, <em>and</em> <code>false</code> <em>otherwise</em>.</p>

<p><strong>Note:</strong> A subsequence of a string is a new string that is formed from the original string by deleting some (possibly none) of the characters without disturbing the relative positions of the remaining characters.</p>

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

<pre>
<strong>Input:</strong> str1 = &quot;abc&quot;, str2 = &quot;ad&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> Select index 2 in str1.
Increment str1[2] to become &#39;d&#39;. 
Hence, str1 becomes &quot;abd&quot; and str2 is now a subsequence. Therefore, true is returned.</pre>

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

<pre>
<strong>Input:</strong> str1 = &quot;zc&quot;, str2 = &quot;ad&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> Select indices 0 and 1 in str1. 
Increment str1[0] to become &#39;a&#39;. 
Increment str1[1] to become &#39;d&#39;. 
Hence, str1 becomes &quot;ad&quot; and str2 is now a subsequence. Therefore, true is returned.</pre>

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

<pre>
<strong>Input:</strong> str1 = &quot;ab&quot;, str2 = &quot;d&quot;
<strong>Output:</strong> false
<strong>Explanation:</strong> In this example, it can be shown that it is impossible to make str2 a subsequence of str1 using the operation at most once. 
Therefore, false is returned.</pre>

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

<ul>
	<li><code>1 &lt;= str1.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= str2.length &lt;= 10<sup>5</sup></code></li>
	<li><code>str1</code> and <code>str2</code> consist of only lowercase English letters.</li>
</ul>

class Solution:
    def canMakeSubsequence(self, str1: str, str2: str) -> bool:
        i, j = 0, 0  # Pointers for str1 and str2
        
        while i < len(str1) and j < len(str2):
            # Check if current character in str1 or its next cyclic character matches current character in str2
            if str1[i] == str2[j] or chr((ord(str1[i]) - ord('a') + 1) % 26 + ord('a')) == str2[j]:
                j += 1  # Move pointer for str2
            i += 1  # Always move pointer for str1
        
        return j == len(str2)  # Return True if all characters of str2 have been matched

# Example usage:
solution = Solution()
print(solution.canMakeSubsequence("abc", "ad"))  # Output: True
print(solution.canMakeSubsequence("zc", "ad"))   # Output: True
print(solution.canMakeSubsequence("ab", "d"))    # Output: False
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