Add solution #793

Author: JoshCrozier

README.md

@@ -603,6 +603,7 @@
 790|[Domino and Tromino Tiling](./0790-domino-and-tromino-tiling.js)|Medium|
 791|[Custom Sort String](./0791-custom-sort-string.js)|Medium|
 792|[Number of Matching Subsequences](./0792-number-of-matching-subsequences.js)|Medium|
+793|[Preimage Size of Factorial Zeroes Function](./0793-preimage-size-of-factorial-zeroes-function.js)|Hard|
 796|[Rotate String](./0796-rotate-string.js)|Easy|
 802|[Find Eventual Safe States](./0802-find-eventual-safe-states.js)|Medium|
 804|[Unique Morse Code Words](./0804-unique-morse-code-words.js)|Easy|

solutions/0793-preimage-size-of-factorial-zeroes-function.js

@@ -0,0 +1,69 @@
+/**
+ * 793. Preimage Size of Factorial Zeroes Function
+ * https://leetcode.com/problems/preimage-size-of-factorial-zeroes-function/
+ * Difficulty: Hard
+ *
+ * Let f(x) be the number of zeroes at the end of x!. Recall that x! = 1 * 2 * 3 * ... * x
+ * and by convention, 0! = 1.
+ *
+ * For example, f(3) = 0 because 3! = 6 has no zeroes at the end, while f(11) = 2 because
+ * 11! = 39916800 has two zeroes at the end.
+ *
+ * Given an integer k, return the number of non-negative integers x have the property that f(x) = k.
+ */
+
+/**
+ * @param {number} k
+ * @return {number}
+ */
+var preimageSizeFZF = function(k) {
+  return findUpperBound(k) - findLowerBound(k);
+};
+
+function findLowerBound(k) {
+  let left = 0;
+  let right = 5 * (10 ** 10);
+
+  while (left < right) {
+    const mid = Math.floor((left + right) / 2);
+    const zeroes = countTrailingZeroes(mid);
+
+    if (zeroes < k) {
+      left = mid + 1;
+    } else {
+      right = mid;
+    }
+  }
+
+  return left;
+}
+
+function findUpperBound(k) {
+  let left = 0;
+  let right = 5 * (10 ** 10);
+
+  while (left < right) {
+    const mid = Math.floor((left + right) / 2);
+    const zeroes = countTrailingZeroes(mid);
+
+    if (zeroes <= k) {
+      left = mid + 1;
+    } else {
+      right = mid;
+    }
+  }
+
+  return left;
+}
+
+function countTrailingZeroes(n) {
+  let count = 0;
+  let powerOfFive = 5;
+
+  while (n >= powerOfFive) {
+    count += Math.floor(n / powerOfFive);
+    powerOfFive *= 5;
+  }
+
+  return count;
+}