Add solution #1922

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,270 LeetCode solutions in JavaScript
+# 1,271 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1130,6 +1130,7 @@
 1886|[Determine Whether Matrix Can Be Obtained By Rotation](./solutions/1886-determine-whether-matrix-can-be-obtained-by-rotation.js)|Easy|
 1910|[Remove All Occurrences of a Substring](./solutions/1910-remove-all-occurrences-of-a-substring.js)|Medium|
 1920|[Build Array from Permutation](./solutions/1920-build-array-from-permutation.js)|Easy|
+1922|[Count Good Numbers](./solutions/1922-count-good-numbers.js)|Medium|
 1926|[Nearest Exit from Entrance in Maze](./solutions/1926-nearest-exit-from-entrance-in-maze.js)|Medium|
 1929|[Concatenation of Array](./solutions/1929-concatenation-of-array.js)|Easy|
 1930|[Unique Length-3 Palindromic Subsequences](./solutions/1930-unique-length-3-palindromic-subsequences.js)|Medium|

solutions/1922-count-good-numbers.js

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+/**
+ * 1922. Count Good Numbers
+ * https://leetcode.com/problems/count-good-numbers/
+ * Difficulty: Medium
+ *
+ * A digit string is good if the digits (0-indexed) at even indices are even and the digits
+ * at odd indices are prime (2, 3, 5, or 7).
+ *
+ * For example, "2582" is good because the digits (2 and 8) at even positions are even and
+ * the digits (5 and 2) at odd positions are prime. However, "3245" is not good because 3
+ * is at an even index but is not even.
+ *
+ * Given an integer n, return the total number of good digit strings of length n. Since the
+ * answer may be large, return it modulo 109 + 7.
+ *
+ * A digit string is a string consisting of digits 0 through 9 that may contain leading zeros.
+ */
+
+/**
+ * @param {number} n
+ * @return {number}
+ */
+var countGoodNumbers = function(n) {
+  const MOD = 1e9 + 7;
+  const evenCount = 5;
+  const primeCount = 4;
+  const evenPositions = Math.ceil(n / 2);
+  const oddPositions = Math.floor(n / 2);
+  const evenResult = power(evenCount, evenPositions);
+  const oddResult = power(primeCount, oddPositions);
+
+  return Number(BigInt(evenResult) * BigInt(oddResult) % BigInt(MOD));
+
+  function power(base, exponent) {
+    if (exponent === 0) return 1;
+    let half = power(base, Math.floor(exponent / 2));
+    half = BigInt(half) * BigInt(half) % BigInt(MOD);
+    if (exponent % 2) half = half * BigInt(base) % BigInt(MOD);
+    return Number(half);
+  }
+};