add 625

fishercoder1534 · fishercoder1534 · commit 6ae20594645a · 2017-06-20T04:12:39.000+09:00
diff --git a/README.md b/README.md
@@ -20,6 +20,7 @@ Your ideas/fixes/algorithms are more than welcome!
 
 |  #  |      Title     |   Solutions   | Time          | Space         | Difficulty  | Tag          | Notes
 |-----|----------------|---------------|---------------|---------------|-------------|--------------|-----
+|625|[Minimum Factorization](https://leetcode.com/problems/minimum-factorization/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_625.java) | O(?) |O(?) | Medium | 
 |624|[Maximum Distance in Arrays](https://leetcode.com/problems/maximum-distance-in-arrays/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_624.java) | O(nlogn) |O(1) | Easy | Sort, Array
 |623|[Add One Row to Tree](https://leetcode.com/problems/add-one-row-to-tree/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_623.java) | O(n) |O(h) | Medium | Tree
 |617|[Merge Two Binary Trees](https://leetcode.com/problems/merge-two-binary-trees/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_617.java) | O(n) |O(h) | Easy | Tree, Recursion
diff --git a/src/main/java/com/fishercoder/solutions/_625.java b/src/main/java/com/fishercoder/solutions/_625.java
@@ -0,0 +1,55 @@
+package com.fishercoder.solutions;
+
+import java.util.ArrayList;
+import java.util.List;
+
+/**
+ * 625. Minimum Factorization
+ *
+ * Given a positive integer a, find the smallest positive integer b whose multiplication of each digit equals to a.
+
+ If there is no answer or the answer is not fit in 32-bit signed integer, then return 0.
+
+ Example 1
+ Input:
+ 48
+ Output:
+ 68
+
+ Example 2
+ Input:
+ 15
+ Output:
+ 35
+ */
+public class _625 {
+
+    /**reference: https://discuss.leetcode.com/topic/92854/java-solution-result-array
+     * and https://leetcode.com/articles/minimum-factorization/#approach-3-using-factorizationaccepted*/
+    public int smallestFactorization(int a) {
+        //case 1: a < 10
+        if (a < 10) return a;
+
+        //case 2: start with 9 and try every possible digit
+        List<Integer> resultArray = new ArrayList<>();
+        for (int i = 9; i > 1; i--) {
+            //if current digit divides a, then store all occurences of current digit in res
+            while (a % i == 0) {
+                a = a/i;
+                resultArray.add(i);
+            }
+        }
+
+        //if a could not be broken in form of digits, return 0
+        if (a != 0) return 0;
+
+        //get the result from the result array in reverse order
+        long result = 0;
+        for (int i = resultArray.size()-1; i >=0; i--) {
+            result = result*10 + resultArray.get(i);
+            if (result > Integer.MAX_VALUE) return 0;
+        }
+        return (int) result;
+    }
+
+}
