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+// 466. Count The Repetitions
+#include "leetcode.h"
+
+/*
+ * 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 strng 's1' can be obtained from string 's2' if we can remove some
+ * characters from 's2' such that it becomes 's1'. 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'
+ */
+
+int getMaxRepetitions(char *s1, int n1, char *s2, int n2) {
+ int a = 0, b = 0, c = 0, d = 0, e = -1, f = 0;
+ char s1r, s2r = s2[b];
+ while (c < n1) {
+ if (!(s1r = s1[a++])) {
+ c++;
+ a = 0;
+ continue;
+ }
+ if (s2r == s1r) {
+ if (e == -1)
+ e = a;
+ else if ((e == a) && (b == 0))
+ return ((((n1 * d) / c) - (f == c)) / n2);
+ s2r = s2[++b];
+ }
+ if (!s2r) {
+ s2r = s2[(b = 0)];
+ d++;
+ f = c;
+ }
+ }
+ return (d / n2);
+}
+
+int main() {
+ char *s11 = "acb", *s21 = "ab", *s12 = "acb", *s22 = "acb";
+ printf("%d\n", getMaxRepetitions(s11, 4, s21, 2)); // expect: 2
+ printf("%d\n", getMaxRepetitions(s12, 1, s22, 1)); // expect: 1
+}