Add solution #2117

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,755 LeetCode solutions in JavaScript
+# 1,756 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1621,6 +1621,7 @@
 2114|[Maximum Number of Words Found in Sentences](./solutions/2114-maximum-number-of-words-found-in-sentences.js)|Easy|
 2115|[Find All Possible Recipes from Given Supplies](./solutions/2115-find-all-possible-recipes-from-given-supplies.js)|Medium|
 2116|[Check if a Parentheses String Can Be Valid](./solutions/2116-check-if-a-parentheses-string-can-be-valid.js)|Medium|
+2117|[Abbreviating the Product of a Range](./solutions/2117-abbreviating-the-product-of-a-range.js)|Hard|
 2127|[Maximum Employees to Be Invited to a Meeting](./solutions/2127-maximum-employees-to-be-invited-to-a-meeting.js)|Hard|
 2129|[Capitalize the Title](./solutions/2129-capitalize-the-title.js)|Easy|
 2130|[Maximum Twin Sum of a Linked List](./solutions/2130-maximum-twin-sum-of-a-linked-list.js)|Medium|

solutions/2117-abbreviating-the-product-of-a-range.js

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+/**
+ * 2117. Abbreviating the Product of a Range
+ * https://leetcode.com/problems/abbreviating-the-product-of-a-range/
+ * Difficulty: Hard
+ *
+ * You are given two positive integers left and right with left <= right. Calculate the product of
+ * all integers in the inclusive range [left, right].
+ *
+ * Since the product may be very large, you will abbreviate it following these steps:
+ * 1. Count all trailing zeros in the product and remove them. Let us denote this count as C.
+ *    - For example, there are 3 trailing zeros in 1000, and there are 0 trailing zeros in 546.
+ * 2. Denote the remaining number of digits in the product as d. If d > 10, then express the product
+ *    as <pre>...<suf> where <pre> denotes the first 5 digits of the product, and <suf> denotes the
+ *    last 5 digits of the product after removing all trailing zeros. If d <= 10, we keep it
+ *    unchanged.
+ *    - For example, we express 1234567654321 as 12345...54321, but 1234567 is represented as
+ *      1234567.
+ * 3. Finally, represent the product as a string "<pre>...<suf>eC".
+ *    - For example, 12345678987600000 will be represented as "12345...89876e5".
+ *
+ * Return a string denoting the abbreviated product of all integers in the inclusive range [left,
+ * right].
+ */
+
+/**
+ * @param {number} left
+ * @param {number} right
+ * @return {string}
+ */
+var abbreviateProduct = function(left, right) {
+  let zeros = 0;
+  let count2 = 0;
+  let count5 = 0;
+
+  for (let i = left; i <= right; i++) {
+    let n = i;
+    while (n % 2 === 0) {
+      count2++;
+      n = Math.floor(n / 2);
+    }
+    n = i;
+    while (n % 5 === 0) {
+      count5++;
+      n = Math.floor(n / 5);
+    }
+  }
+  zeros = Math.min(count2, count5);
+
+  let digits = 0;
+  for (let i = left; i <= right; i++) {
+    digits += Math.log10(i);
+  }
+  digits = Math.floor(digits) + 1;
+
+  if (digits - zeros <= 10) {
+    let product = 1n;
+    for (let i = left; i <= right; i++) {
+      product *= BigInt(i);
+    }
+    for (let i = 0; i < zeros; i++) {
+      product /= 10n;
+    }
+    return product.toString() + 'e' + zeros;
+  }
+
+  let prefix = 1;
+  for (let i = left; i <= right; i++) {
+    prefix *= i;
+    while (prefix >= 1e10) {
+      prefix /= 10;
+    }
+  }
+  prefix = prefix.toString().slice(0, 5);
+
+  let suffix = 1n;
+  for (let i = right; i >= left; i--) {
+    suffix = (suffix * BigInt(i));
+    while (suffix % 10n === 0n) {
+      suffix /= 10n;
+    }
+    suffix = suffix % (10n ** 15n);
+  }
+
+  suffix = suffix.toString();
+  while (suffix.length < 5) {
+    suffix = '0' + suffix;
+  }
+  suffix = suffix.slice(-5);
+
+  return prefix + '...' + suffix + 'e' + zeros;
+};