Sync LeetCode submission Runtime - 3 ms (20.66%), Memory - 17.7 MB (71.93%)

Author: jimit105

2050-count-good-numbers/README.md

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+<p>A digit string is <strong>good</strong> if the digits <strong>(0-indexed)</strong> at <strong>even</strong> indices are <strong>even</strong> and the digits at <strong>odd</strong> indices are <strong>prime</strong> (<code>2</code>, <code>3</code>, <code>5</code>, or <code>7</code>).</p>
+
+<ul>
+	<li>For example, <code>&quot;2582&quot;</code> is good because the digits (<code>2</code> and <code>8</code>) at even positions are even and the digits (<code>5</code> and <code>2</code>) at odd positions are prime. However, <code>&quot;3245&quot;</code> is <strong>not</strong> good because <code>3</code> is at an even index but is not even.</li>
+</ul>
+
+<p>Given an integer <code>n</code>, return <em>the <strong>total</strong> number of good digit strings of length </em><code>n</code>. Since the answer may be large, <strong>return it modulo </strong><code>10<sup>9</sup> + 7</code>.</p>
+
+<p>A <strong>digit string</strong> is a string consisting of digits <code>0</code> through <code>9</code> that may contain leading zeros.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 1
+<strong>Output:</strong> 5
+<strong>Explanation:</strong> The good numbers of length 1 are &quot;0&quot;, &quot;2&quot;, &quot;4&quot;, &quot;6&quot;, &quot;8&quot;.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 4
+<strong>Output:</strong> 400
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 50
+<strong>Output:</strong> 564908303
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 10<sup>15</sup></code></li>
+</ul>

2050-count-good-numbers/solution.py

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+# Approach 1: Fast Exponentiation
+
+# Time: O(log n)
+# Space: O(1)
+
+class Solution:
+    def countGoodNumbers(self, n: int) -> int:
+        mod = 10 ** 9 + 7
+
+        # Calculate x ^ y % mod
+        def quick_mul(x, y):
+            ret, mul = 1, x
+            while y > 0:
+                if y % 2 == 1:
+                    ret = ret * mul % mod
+                mul = mul * mul % mod
+                y //= 2
+            return ret
+
+        return quick_mul(5, (n + 1) // 2) * quick_mul(4, n // 2) % mod
+