Add solution #902

Author: JoshCrozier

README.md

@@ -712,6 +712,7 @@
 899|[Orderly Queue](./0899-orderly-queue.js)|Hard|
 900|[RLE Iterator](./0900-rle-iterator.js)|Medium|
 901|[Online Stock Span](./0901-online-stock-span.js)|Medium|
+902|[Numbers At Most N Given Digit Set](./0902-numbers-at-most-n-given-digit-set.js)|Hard|
 905|[Sort Array By Parity](./0905-sort-array-by-parity.js)|Easy|
 909|[Snakes and Ladders](./0909-snakes-and-ladders.js)|Medium|
 912|[Sort an Array](./0912-sort-an-array.js)|Medium|

solutions/0902-numbers-at-most-n-given-digit-set.js

@@ -0,0 +1,48 @@
+/**
+ * 902. Numbers At Most N Given Digit Set
+ * https://leetcode.com/problems/numbers-at-most-n-given-digit-set/
+ * Difficulty: Hard
+ *
+ * Given an array of digits which is sorted in non-decreasing order. You can write numbers using
+ * each digits[i] as many times as we want. For example, if digits = ['1','3','5'], we may write
+ * numbers such as '13', '551', and '1351315'.
+ *
+ * Return the number of positive integers that can be generated that are less than or equal to
+ * a given integer n.
+ */
+
+/**
+ * @param {string[]} digits
+ * @param {number} n
+ * @return {number}
+ */
+var atMostNGivenDigitSet = function(digits, n) {
+  const str = n.toString();
+  const digitCount = digits.length;
+  let result = 0;
+
+  for (let i = 1; i < str.length; i++) {
+    result += Math.pow(digitCount, i);
+  }
+
+  function countValid(prefixLen) {
+    if (prefixLen === str.length) return 1;
+
+    const currentDigit = str[prefixLen];
+    let valid = 0;
+
+    for (const digit of digits) {
+      if (digit < currentDigit) {
+        valid += Math.pow(digitCount, str.length - prefixLen - 1);
+      } else if (digit === currentDigit) {
+        valid += countValid(prefixLen + 1);
+      } else {
+        break;
+      }
+    }
+
+    return valid;
+  }
+
+  return result + countValid(0);
+};