TopCoder Problem Statement - InfiniteString |

Single Round Match 658 Round 1 - Division II, Level One |

Determine if two strings, when repeated infinitely, are equal.

1 public class InfiniteString { 2 3 public String equal(String s, String t) { 4 5 int ptrS = 0; 6 int ptrT = 0; 7 8 /* 9 * Continue looping one character at a time until 10 * we reach both the end of String s and the end 11 * of String t at the same time. 12 */ 13 while ((ptrS < s.length()) || (ptrT < t.length())) { 14 15 /* 16 * If at the end of a String, start over at 17 * the beginning. 18 */ 19 ptrS %= s.length(); 20 ptrT %= t.length(); 21 22 // Return "Not Equal" if characters do not match 23 if (s.charAt(ptrS) != t.charAt(ptrT)) return "Not equal"; 24 25 ptrS++; 26 ptrT++; 27 } 28 29 return "Equal"; 30 } 31 }

Imagine creating two Strings s1 and t1 such that s1 = s + s + s ... and t1 = t + t + t + t + ... and the length of both s1 and t1 is equal to the least common multiple of the lengths of s and t.

For example if s = "ababab" and t = "abab", then the length of s is 6, the length of t is 4, and their least common multiple is 12. Therefore, s1 = s + s = "abababababab" and t1 = t + t + t = "abababababab". Since s1 == t1 we should return "Equal"

The important point is that if we step through the Strings s and t one character at a time, and loop back to the beginning of the String when we reach the end; then we can stop when we reach the end of both s and t at the same time. That's the condition that is checked for in the while loop.

The rest is just a matter of checking that the characters match at each position, and incrementing and resetting the pointer to the current character.

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