feat: add solution 1808. Maximize Number of Nice Divisors

Author: tzuyi0817

solutions/1808. Maximize Number of Nice Divisors/README.md

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+# [1808. Maximize Number of Nice Divisors](https://leetcode.com/problems/maximize-number-of-nice-divisors)
+
+## Description
+
+<div class="elfjS" data-track-load="description_content"><p>You are given a positive integer <code>primeFactors</code>. You are asked to construct a positive integer <code>n</code> that satisfies the following conditions:</p>
+
+<ul>
+  <li>The number of prime factors of <code>n</code> (not necessarily distinct) is <strong>at most</strong> <code>primeFactors</code>.</li>
+  <li>The number of nice divisors of <code>n</code> is maximized. Note that a divisor of <code>n</code> is <strong>nice</strong> if it is divisible by every prime factor of <code>n</code>. For example, if <code>n = 12</code>, then its prime factors are <code>[2,2,3]</code>, then <code>6</code> and <code>12</code> are nice divisors, while <code>3</code> and <code>4</code> are not.</li>
+</ul>
+
+<p>Return <em>the number of nice divisors of</em> <code>n</code>. Since that number can be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>
+
+<p>Note that a prime number is a natural number greater than <code>1</code> that is not a product of two smaller natural numbers. The prime factors of a number <code>n</code> is a list of prime numbers such that their product equals <code>n</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre><strong>Input:</strong> primeFactors = 5
+<strong>Output:</strong> 6
+<strong>Explanation:</strong> 200 is a valid value of n.
+It has 5 prime factors: [2,2,2,5,5], and it has 6 nice divisors: [10,20,40,50,100,200].
+There is not other value of n that has at most 5 prime factors and more nice divisors.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre><strong>Input:</strong> primeFactors = 8
+<strong>Output:</strong> 18
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= primeFactors &lt;= 10<sup>9</sup></code></li>
+</ul></div>
+
+<p>&nbsp;</p>
+
+## Solutions
+
+**Solution: `Math`**
+
+- Time complexity: <em>O(logn)</em>
+- Space complexity: <em>O(1)</em>
+
+<p>&nbsp;</p>
+
+### **JavaScript**
+
+```js
+/**
+ * @param {number} primeFactors
+ * @return {number}
+ */
+const maxNiceDivisors = function (primeFactors) {
+  if (primeFactors <= 3) return primeFactors;
+  const MODULO = BigInt(10 ** 9 + 7);
+
+  const powMod = (base, exponent) => {
+    let result = 1n;
+
+    while (exponent) {
+      if (exponent % 2n) {
+        result = (result * base) % MODULO;
+      }
+
+      base = (base * base) % MODULO;
+      exponent /= 2n;
+    }
+
+    return result;
+  };
+
+  const count = BigInt(Math.floor(primeFactors / 3));
+  const remainder = primeFactors % 3;
+
+  if (remainder === 0) return Number(powMod(3n, count));
+  if (remainder === 1) {
+    const result = powMod(3n, count - 1n);
+
+    return Number((result * 4n) % MODULO);
+  }
+
+  return Number((powMod(3n, count) * 2n) % MODULO);
+};
+```

solutions/1808. Maximize Number of Nice Divisors/solution.js

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+/**
+ * @param {number} primeFactors
+ * @return {number}
+ */
+const maxNiceDivisors = function (primeFactors) {
+  if (primeFactors <= 3) return primeFactors;
+  const MODULO = BigInt(10 ** 9 + 7);
+
+  const powMod = (base, exponent) => {
+    let result = 1n;
+
+    while (exponent) {
+      if (exponent % 2n) {
+        result = (result * base) % MODULO;
+      }
+
+      base = (base * base) % MODULO;
+      exponent /= 2n;
+    }
+
+    return result;
+  };
+
+  const count = BigInt(Math.floor(primeFactors / 3));
+  const remainder = primeFactors % 3;
+
+  if (remainder === 0) return Number(powMod(3n, count));
+  if (remainder === 1) {
+    const result = powMod(3n, count - 1n);
+
+    return Number((result * 4n) % MODULO);
+  }
+
+  return Number((powMod(3n, count) * 2n) % MODULO);
+};