edit 357

Author: fishercoder1534

README.md

@@ -252,7 +252,7 @@ Your ideas/fixes/algorithms are more than welcome!
 |360|[Sort Transformed Array](https://leetcode.com/problems/sort-transformed-array/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_360.java)| O(n)|O(1) | Medium| Two Pointers, Math
 |359|[Logger Rate Limiter](https://leetcode.com/problems/logger-rate-limiter/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_359.java)| amortized O(1)|O(k) | Easy| HashMap
 |358|[Rearrange String k Distance Apart](https://leetcode.com/problems/rearrange-string-k-distance-apart/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_358.java)| O(n)|O(n) | Hard| HashMap, Heap, Greedy
-|357|[Count Numbers with Unique Digits](https://leetcode.com/problems/count-numbers-with-unique-digits/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_357.java)| O(?)|O(?) | Medium|
+|357|[Count Numbers with Unique Digits](https://leetcode.com/problems/count-numbers-with-unique-digits/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_357.java)| O(n)|O(1) | Medium| DP, Math
 |356|[Line Reflection](https://leetcode.com/problems/line-reflection/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_356.java)| O(n)|O(n) | Medium| HashSet
 |355|[Design Twitter](https://leetcode.com/problems/design-twitter/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_355.java)| O(n)|O(n) | Medium| Design, HashMap, Heap
 |354|[Russian Doll Envelopes](https://leetcode.com/problems/russian-doll-envelopes/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_354.java)| O(nlogn)|O(1) | Hard| DP, Binary Search

src/main/java/com/fishercoder/solutions/_357.java

@@ -1,23 +1,28 @@
 package com.fishercoder.solutions;
 
 /**
+ * 357. Count Numbers with Unique Digits
  * Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10n.
 
  Example:
  Given n = 2, return 91. (The answer should be the total numbers in the range of 0 ≤ x < 100, excluding [11,22,33,44,55,66,77,88,99])
 
  Hint:
-
  A direct way is to use the backtracking approach.
- Backtracking should contains three states which are (the current number, number of steps to get that number and a bitmask which represent which number is marked as visited so far in the current number). Start with state (0,0,0) and count all valid number till we reach number of steps equals to 10n.
+ Backtracking should contains three states which are
+ (the current number, number of steps to get that number and a bitmask which
+ represent which number is marked as visited so far in the current number).
+ Start with state (0,0,0) and count all valid number till we reach number of steps equals to 10n.
  This problem can also be solved using a dynamic programming approach and some knowledge of combinatorics.
  Let f(k) = count of numbers with unique digits with length equals k.
  f(1) = 10, ..., f(k) = 9 * 9 * 8 * ... (9 - k + 2) [The first factor is 9 because a number cannot start with 0].
  */
 public class _357 {
 
     public int countNumbersWithUniqueDigits(int n) {
-        if (n == 0)     return 1;
+        if (n == 0) {
+            return 1;
+        }
         int res = 10;
         int uniqueDigits = 9;
         int availableNumber = 9;