edit 553

Author: fishercoder1534

README.md

@@ -87,7 +87,7 @@ Your ideas/fixes/algorithms are more than welcome!
 |556|[Next Greater Element III](https://leetcode.com/problems/next-greater-element-iii/)|[Solution](../master/src/main/java/com/fishercoder/solutions/NextGreaterElementIII.java) | O(n)|O(1)| Medium | String
 |555|[Split Concatenated Strings](https://leetcode.com/problems/split-concatenated-strings/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_555.java) | O(n^2) |O(n) | Medium | String
 |554|[Brick Wall](https://leetcode.com/problems/brick-wall/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_554.java) | O(n) (n is total number of bricks in the wall) |O(m) (m is width of the wall) | Medium | HashMap
-|553|[Optimal Division](https://leetcode.com/problems/optimal-division/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_553.java) | O(n) | O(n) | Medium | String
+|553|[Optimal Division](https://leetcode.com/problems/optimal-division/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_553.java) | O(n) | O(n) | Medium | String, Math
 |552|[Student Attendance Record II](https://leetcode.com/problems/student-attendance-record-ii/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_552.java) | O(n)| O(1) | Hard| DP
 |551|[Student Attendance Record I](https://leetcode.com/problems/student-attendance-record-i/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_551.java) | O(n)| O(1) | Easy| String 
 |549|[Binary Tree Longest Consecutive Sequence II](https://leetcode.com/problems/binary-tree-longest-consecutive-sequence-ii/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_549.java) | O(n) |O(n) | Medium | Tree

src/main/java/com/fishercoder/solutions/_553.java

@@ -44,7 +44,6 @@ public String optimalDivision(int[] nums) {
         if (nums.length == 1) return "" + nums[0];
         if (nums.length == 2) return nums[0] + "/" + nums[1];
 
-
         /**Tricky one: the solution is fixed: always wrap the one from the second until the last.
          * Another important thing to note that such way could work is that:
          * the prerequisite is: Elements will be in range [2,1000], so no elements are smaller than 1.*/