diff --git a/InfinitString/InfinityString.class b/InfinitString/InfinityString.class
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diff --git a/InfinitString/InfinityString.java b/InfinitString/InfinityString.java
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+import java.util.Scanner;
+
+/**
+ * InfinityString
+ * There is a string, , of lowercase English letters that is repeated infinitely
+ * many times. Given an integer, , find and print the number of letter a's in
+ * the first
+ * 
+ * letters of the infinite string.
+ * 
+ * Example
+ * 
+ * The substring we consider is , the first characters of the infinite string.
+ * There are
+ * 
+ * occurrences of a in the substring.
+ * 
+ * Function Description
+ * 
+ * Complete the repeatedString function in the editor below.
+ * 
+ * repeatedString has the following parameter(s):
+ * 
+ * s: a string to repeat
+ * n: the number of characters to consider
+ * 
+ * Returns
+ * 
+ * int: the frequency of a in the substring
+ * 
+ * Input Format
+ * 
+ * The first line contains a single string,
+ * .
+ * The second line contains an integer,
+ * 
+ * .
+ * 
+ * Constraints
+ * 
+ * For of the test cases,
+ * 
+ * .
+ * 
+ * Sample Input
+ * 
+ * Sample Input 0
+ * 
+ * aba
+ * 10
+ * 
+ * Sample Output 0
+ * 
+ * 7
+ * 
+ */
+public class InfinityString {
+
+  public static void main(String[] args) {
+
+    // MY SOLUTION IS SLOW BUT IT WORKS
+    Scanner sc = new Scanner(System.in);
+    System.out.println("Please enter your 'String' here and it must contains the letter A: ");
+    String str = sc.nextLine();
+    System.out.println("Please enter how many times you want to repeat that: ");
+    int x = sc.nextInt();
+    System.out.println(numsOfAs(str, x));
+    sc.close();
+  }
+
+  //
+  // private static int numsOfAs(String str, int n) {
+  // int numberOfAs = 0;
+  // int x = n;
+  // String newSTR = "";
+  // int remeinder = n / str.length();
+  // while (n > 0 && newSTR.toCharArray().length < n + remeinder) {
+  // newSTR += str;
+  // n--;
+  // }
+  //
+  // char[] charNewstr = newSTR.toCharArray();
+  // for (int i = 0; i < x; i++) {
+  // if (charNewstr[i] == 'a') {
+  // numberOfAs++;
+  // }
+  // }
+  // return numberOfAs;
+  //
+  //
+  private static int numsOfAs(String str, int n) {
+    int numberOfAs = 0;
+    int numRepeats = n / str.length();
+    int remainder = n % str.length();
+
+    // Count the number of 'a' characters in the repeated string
+    numberOfAs = countAs(str) * numRepeats;
+
+    // Count the number of 'a' characters in the remaining portion
+    if (remainder > 0) {
+      numberOfAs += countAs(str.substring(0, remainder));
+    }
+
+    return numberOfAs;
+  }
+
+  private static int countAs(String str) {
+    int count = 0;
+    for (char c : str.toCharArray()) {
+      if (c == 'a') {
+        count++;
+      }
+    }
+    return count;
+  }
+
+}