Add solution #466

Author: JoshCrozier

README.md

@@ -373,6 +373,7 @@
 462|[Minimum Moves to Equal Array Elements II](./0462-minimum-moves-to-equal-array-elements-ii.js)|Medium|
 463|[Island Perimeter](./0463-island-perimeter.js)|Medium|
 464|[Can I Win](./0464-can-i-win.js)|Medium|
+466|[Count The Repetitions](./0466-count-the-repetitions.js)|Hard|
 472|[Concatenated Words](./0472-concatenated-words.js)|Hard|
 476|[Number Complement](./0476-number-complement.js)|Easy|
 482|[License Key Formatting](./0482-license-key-formatting.js)|Easy|

solutions/0466-count-the-repetitions.js

@@ -0,0 +1,56 @@
+/**
+ * 466. Count The Repetitions
+ * https://leetcode.com/problems/count-the-repetitions/
+ * Difficulty: Hard
+ *
+ * We define str = [s, n] as the string str which consists of the string s concatenated n times.
+ *
+ * - For example, str == ["abc", 3] =="abcabcabc".
+ *
+ * We define that string s1 can be obtained from string s2 if we can remove some characters from
+ * s2 such that it becomes s1.
+ *
+ * - For example, s1 = "abc" can be obtained from s2 = "abdbec" based on our definition by
+ *   removing the bolded underlined characters.
+ *
+ * You are given two strings s1 and s2 and two integers n1 and n2. You have the two strings
+ * str1 = [s1, n1] and str2 = [s2, n2].
+ *
+ * Return the maximum integer m such that str = [str2, m] can be obtained from str1.
+ */
+
+/**
+ * @param {string} s1
+ * @param {number} n1
+ * @param {string} s2
+ * @param {number} n2
+ * @return {number}
+ */
+var getMaxRepetitions = function(s1, n1, s2, n2) {
+  const pattern = {};
+  let s2Count = 0;
+  let s2Index = 0;
+
+  for (let i = 0; i < n1; i++) {
+    for (let j = 0; j < s1.length; j++) {
+      if (s1[j] === s2[s2Index]) {
+        s2Index++;
+        if (s2Index === s2.length) {
+          s2Index = 0;
+          s2Count++;
+        }
+      }
+    }
+
+    if (pattern[s2Index]) {
+      const [prevIndex, prevCount] = pattern[s2Index];
+      const remainingCycles = Math.floor((n1 - i - 1) / (i - prevIndex));
+      i += remainingCycles * (i - prevIndex);
+      s2Count += remainingCycles * (s2Count - prevCount);
+    } else {
+      pattern[s2Index] = [i, s2Count];
+    }
+  }
+
+  return Math.floor(s2Count / n2);
+};