Sync LeetCode submission Runtime - 2187 ms (64.05%), Memory - 18 MB (91.50%)

Author: jimit105

3629-total-characters-in-string-after-transformations-i/README.md

@@ -0,0 +1,79 @@
+<p>You are given a string <code>s</code> and an integer <code>t</code>, representing the number of <strong>transformations</strong> to perform. In one <strong>transformation</strong>, every character in <code>s</code> is replaced according to the following rules:</p>
+
+<ul>
+	<li>If the character is <code>&#39;z&#39;</code>, replace it with the string <code>&quot;ab&quot;</code>.</li>
+	<li>Otherwise, replace it with the <strong>next</strong> character in the alphabet. For example, <code>&#39;a&#39;</code> is replaced with <code>&#39;b&#39;</code>, <code>&#39;b&#39;</code> is replaced with <code>&#39;c&#39;</code>, and so on.</li>
+</ul>
+
+<p>Return the <strong>length</strong> of the resulting string after <strong>exactly</strong> <code>t</code> transformations.</p>
+
+<p>Since the answer may be very large, return it <strong>modulo</strong><!-- notionvc: eb142f2b-b818-4064-8be5-e5a36b07557a --> <code>10<sup>9</sup> + 7</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot;abcyy&quot;, t = 2</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">7</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li><strong>First Transformation (t = 1)</strong>:
+
+	<ul>
+		<li><code>&#39;a&#39;</code> becomes <code>&#39;b&#39;</code></li>
+		<li><code>&#39;b&#39;</code> becomes <code>&#39;c&#39;</code></li>
+		<li><code>&#39;c&#39;</code> becomes <code>&#39;d&#39;</code></li>
+		<li><code>&#39;y&#39;</code> becomes <code>&#39;z&#39;</code></li>
+		<li><code>&#39;y&#39;</code> becomes <code>&#39;z&#39;</code></li>
+		<li>String after the first transformation: <code>&quot;bcdzz&quot;</code></li>
+	</ul>
+	</li>
+	<li><strong>Second Transformation (t = 2)</strong>:
+	<ul>
+		<li><code>&#39;b&#39;</code> becomes <code>&#39;c&#39;</code></li>
+		<li><code>&#39;c&#39;</code> becomes <code>&#39;d&#39;</code></li>
+		<li><code>&#39;d&#39;</code> becomes <code>&#39;e&#39;</code></li>
+		<li><code>&#39;z&#39;</code> becomes <code>&quot;ab&quot;</code></li>
+		<li><code>&#39;z&#39;</code> becomes <code>&quot;ab&quot;</code></li>
+		<li>String after the second transformation: <code>&quot;cdeabab&quot;</code></li>
+	</ul>
+	</li>
+	<li><strong>Final Length of the string</strong>: The string is <code>&quot;cdeabab&quot;</code>, which has 7 characters.</li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot;azbk&quot;, t = 1</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">5</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li><strong>First Transformation (t = 1)</strong>:
+
+	<ul>
+		<li><code>&#39;a&#39;</code> becomes <code>&#39;b&#39;</code></li>
+		<li><code>&#39;z&#39;</code> becomes <code>&quot;ab&quot;</code></li>
+		<li><code>&#39;b&#39;</code> becomes <code>&#39;c&#39;</code></li>
+		<li><code>&#39;k&#39;</code> becomes <code>&#39;l&#39;</code></li>
+		<li>String after the first transformation: <code>&quot;babcl&quot;</code></li>
+	</ul>
+	</li>
+	<li><strong>Final Length of the string</strong>: The string is <code>&quot;babcl&quot;</code>, which has 5 characters.</li>
+</ul>
+</div>
+
+<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 only of lowercase English letters.</li>
+	<li><code>1 &lt;= t &lt;= 10<sup>5</sup></code></li>
+</ul>

3629-total-characters-in-string-after-transformations-i/solution.py

@@ -0,0 +1,23 @@
+# Approach: Recurrence
+
+class Solution:
+    def lengthAfterTransformations(self, s: str, t: int) -> int:
+        MOD = 10 ** 9 + 7
+        count = [0] * 26
+
+        for ch in s:
+            count[ord(ch) - ord('a')] += 1
+
+        for round in range(t):
+            nxt = [0] * 26
+            nxt[0] = count[25]
+            nxt[1] = (count[25] + count[0]) % MOD
+
+            for i in range(2, 26):
+                nxt[i] = count[i - 1]
+
+            count = nxt
+
+        ans = sum(count) % MOD
+        return ans
+