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Author: jimit105

3630-total-characters-in-string-after-transformations-ii/README.md

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+<p>You are given a string <code>s</code> consisting of lowercase English letters, an integer <code>t</code> representing the number of <strong>transformations</strong> to perform, and an array <code>nums</code> of size 26. In one <strong>transformation</strong>, every character in <code>s</code> is replaced according to the following rules:</p>
+
+<ul>
+	<li>Replace <code>s[i]</code> with the <strong>next</strong> <code>nums[s[i] - &#39;a&#39;]</code> consecutive characters in the alphabet. For example, if <code>s[i] = &#39;a&#39;</code> and <code>nums[0] = 3</code>, the character <code>&#39;a&#39;</code> transforms into the next 3 consecutive characters ahead of it, which results in <code>&quot;bcd&quot;</code>.</li>
+	<li>The transformation <strong>wraps</strong> around the alphabet if it exceeds <code>&#39;z&#39;</code>. For example, if <code>s[i] = &#39;y&#39;</code> and <code>nums[24] = 3</code>, the character <code>&#39;y&#39;</code> transforms into the next 3 consecutive characters ahead of it, which results in <code>&quot;zab&quot;</code>.</li>
+</ul>
+
+<p>Return the length 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> <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, nums = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">7</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>
+	<p><strong>First Transformation (t = 1):</strong></p>
+
+	<ul>
+		<li><code>&#39;a&#39;</code> becomes <code>&#39;b&#39;</code> as <code>nums[0] == 1</code></li>
+		<li><code>&#39;b&#39;</code> becomes <code>&#39;c&#39;</code> as <code>nums[1] == 1</code></li>
+		<li><code>&#39;c&#39;</code> becomes <code>&#39;d&#39;</code> as <code>nums[2] == 1</code></li>
+		<li><code>&#39;y&#39;</code> becomes <code>&#39;z&#39;</code> as <code>nums[24] == 1</code></li>
+		<li><code>&#39;y&#39;</code> becomes <code>&#39;z&#39;</code> as <code>nums[24] == 1</code></li>
+		<li>String after the first transformation: <code>&quot;bcdzz&quot;</code></li>
+	</ul>
+	</li>
+	<li>
+	<p><strong>Second Transformation (t = 2):</strong></p>
+
+	<ul>
+		<li><code>&#39;b&#39;</code> becomes <code>&#39;c&#39;</code> as <code>nums[1] == 1</code></li>
+		<li><code>&#39;c&#39;</code> becomes <code>&#39;d&#39;</code> as <code>nums[2] == 1</code></li>
+		<li><code>&#39;d&#39;</code> becomes <code>&#39;e&#39;</code> as <code>nums[3] == 1</code></li>
+		<li><code>&#39;z&#39;</code> becomes <code>&#39;ab&#39;</code> as <code>nums[25] == 2</code></li>
+		<li><code>&#39;z&#39;</code> becomes <code>&#39;ab&#39;</code> as <code>nums[25] == 2</code></li>
+		<li>String after the second transformation: <code>&quot;cdeabab&quot;</code></li>
+	</ul>
+	</li>
+	<li>
+	<p><strong>Final Length of the string:</strong> The string is <code>&quot;cdeabab&quot;</code>, which has 7 characters.</p>
+	</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, nums = [2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">8</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>
+	<p><strong>First Transformation (t = 1):</strong></p>
+
+	<ul>
+		<li><code>&#39;a&#39;</code> becomes <code>&#39;bc&#39;</code> as <code>nums[0] == 2</code></li>
+		<li><code>&#39;z&#39;</code> becomes <code>&#39;ab&#39;</code> as <code>nums[25] == 2</code></li>
+		<li><code>&#39;b&#39;</code> becomes <code>&#39;cd&#39;</code> as <code>nums[1] == 2</code></li>
+		<li><code>&#39;k&#39;</code> becomes <code>&#39;lm&#39;</code> as <code>nums[10] == 2</code></li>
+		<li>String after the first transformation: <code>&quot;bcabcdlm&quot;</code></li>
+	</ul>
+	</li>
+	<li>
+	<p><strong>Final Length of the string:</strong> The string is <code>&quot;bcabcdlm&quot;</code>, which has 8 characters.</p>
+	</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>9</sup></code></li>
+	<li><code><font face="monospace">nums.length == 26</font></code></li>
+	<li><code><font face="monospace">1 &lt;= nums[i] &lt;= 25</font></code></li>
+</ul>

3630-total-characters-in-string-after-transformations-ii/solution.py

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+MOD = 10**9 + 7
+L = 26
+
+
+class Mat:
+    def __init__(self, copy_from: "Mat" = None) -> None:
+        self.a: List[List[int]] = [[0] * L for _ in range(L)]
+        if copy_from:
+            for i in range(L):
+                for j in range(L):
+                    self.a[i][j] = copy_from.a[i][j]
+
+    def __mul__(self, other: "Mat") -> "Mat":
+        result = Mat()
+        for i in range(L):
+            for j in range(L):
+                for k in range(L):
+                    result.a[i][j] = (
+                        result.a[i][j] + self.a[i][k] * other.a[k][j]
+                    ) % MOD
+        return result
+
+
+# identity matrix
+def I() -> Mat:
+    m = Mat()
+    for i in range(L):
+        m.a[i][i] = 1
+    return m
+
+
+# matrix exponentiation by squaring
+def quickmul(x: Mat, y: int) -> Mat:
+    ans = I()
+    cur = x
+    while y:
+        if y & 1:
+            ans = ans * cur
+        cur = cur * cur
+        y >>= 1
+    return ans
+
+
+class Solution:
+    def lengthAfterTransformations(
+        self, s: str, t: int, nums: List[int]
+    ) -> int:
+        T = Mat()
+        for i in range(26):
+            for j in range(1, nums[i] + 1):
+                T.a[(i + j) % 26][i] = 1
+
+        res = quickmul(T, t)
+
+        f = [0] * 26
+        for ch in s:
+            f[ord(ch) - ord("a")] += 1
+
+        ans = 0
+        for i in range(26):
+            for j in range(26):
+                ans = (ans + res.a[i][j] * f[j]) % MOD
+
+        return ans