add 343. Integer Break

Author: selfboot

Math/343_IntegerBreak.cpp

@@ -0,0 +1,47 @@
+/*
+ * @Author: [email protected]
+ * @Last Modified time: 2016-07-04 15:01:50
+ */
+
+class Solution {
+public:
+    /*
+    Magic factor 2 and 3.
+
+    Break the numbers into magic factor only 2 and 3 if number >= 4,
+    Then we will get the max product.
+
+    If we break a number N into two factors x, N-x, product is p=x(N-x).
+    To get the maximum of p,  x=N/2 when N is even, x=(N-1)/2 when N is odd.
+    If x can be break again and the product is bigger than x, then break recursively.
+
+    Now the question is, for a given number N, when to stop break.  It's clearly that:
+    (N/2)*(N/2) < N (N is even),     then N < 4,  N = 2
+    (N-1)/2 *(N+1)/2 < N (N id odd), then N < 5,  N = 3, N = 1
+
+    Thus, the factors of the perfect product should only be 2 or 3.
+
+    According to:
+    https://discuss.leetcode.com/topic/45341/an-simple-explanation-of-the-math-part-and-a-o-n-solution
+    */
+    int integerBreak(int n) {
+        if(n==2 || n==3){
+            return n-1;
+        }
+        int three_cnt = n / 3;
+        int two_cnt = (n - n/3 * 3) / 2;
+
+        if((n-n/3*3) == 1){
+            three_cnt -= 1;
+            two_cnt = 2;
+        }
+        return pow(3, three_cnt) * pow(2, two_cnt);
+    }
+};
+
+/*
+2
+7
+10
+102
+*/

Math/343_IntegerBreak.py

@@ -0,0 +1,44 @@
+#! /usr/bin/env python
+# -*- coding: utf-8 -*-
+# @Author: [email protected]
+# @Last Modified time: 2016-07-04 15:01:41
+
+
+class Solution(object):
+    """ Magic factor 2 and 3.
+
+    Break the numbers into magic factor only 2 and 3 if number >= 4,
+    Then we will get the max product.
+
+    If we break a number N into two factors x, N-x, product is p=x(N-x).
+    To get the maximum of p,  x=N/2 when N is even, x=(N-1)/2 when N is odd.
+    If x can be break again and the product is bigger than x, then break recursively.
+
+    Now the question is, for a given number N, when to stop break.  It's clearly that:
+    (N/2)*(N/2) < N (N is even),     then N < 4,  N = 2
+    (N-1)/2 *(N+1)/2 < N (N id odd), then N < 5,  N = 3, N = 1
+
+    Thus, the factors of the perfect product should only be 2 or 3.
+
+    According to:
+    https://discuss.leetcode.com/topic/45341/an-simple-explanation-of-the-math-part-and-a-o-n-solution
+    """
+    def integerBreak(self, n):
+        assert(n >= 2)
+        if n == 2 or n == 3:
+            return n - 1
+        three_cnt = n / 3
+        two_cnt = (n - three_cnt * 3) / 2
+
+        # We should minus one 3 and add two 2,  number may be 10, 13
+        if n - three_cnt * 3 == 1:
+            two_cnt = 2
+            three_cnt -= 1
+        return 3 ** three_cnt * (2 ** two_cnt)
+
+"""
+2
+7
+10
+102
+"""

README.md

@@ -274,6 +274,7 @@
 * 326. [Power of Three](Math/326_PowerOfThree.py)
 * 319. [Bulb Switcher](Math/319_BulbSwitcher.py)
 * 335. [Self Crossing](Math/335_SelfCrossing.py)
+* 343. [Integer Break](Math/343_IntegerBreak.py)
 
 # [Bit Manipulation](BitManipulation/)