1 Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get. 2 3 For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36 (10 = 3 + 3 + 4). 4 5 Note: You may assume that n is not less than 2 and not larger than 58.
O(N^2)解法： DP
dp[i] represent the maximum product of breaking up integer i
1 public class Solution { 2 public int integerBreak(int n) { 3 int[] dp = new int[n+1]; 4 dp[1] = 1; 5 for (int i=2; i<=n; i++) { 6 for (int j=1; j<=i-1; j++) { 7 dp[i] = Math.max(dp[i], Math.max(j, dp[j]) * Math.max(i-j, dp[i-j])); 8 } 9 } 10 return dp[n]; 11 } 12 }
O(N)解法：
the best factor is 3. we keep breaking n into 3's until n gets smaller than 10, then solve the problem by brute-force.
1 public class Solution { 2 public int integerBreak(int n) { 3 if (n == 2) return 1; 4 if (n == 3) return 2; 5 if (n == 4) return 4; 6 if (n == 5) return 6; 7 return 3*Math.max(n-3, integerBreak(n-3)); 8 } 9 }
Why 3 is the best factor?
I saw many solutions were referring to factors of 2 and 3. But why these two magic numbers? Why other factors do not work?
Let's study the math behind it.
For convenience, say n is sufficiently large and can be broken into any smaller real positive numbers. We now try to calculate which real number generates the largest product.
Assume we break n into (n / x) x's, then the product will be xn/x, and we want to maximize it.
Taking its derivative gives us n * x^(n/x-2) * (1 - ln(x)).
The derivative is positive when 0 < x < e, and equal to 0 when x = e, then becomes negative when x > e,
which indicates that the product increases as x increases, then reaches its maximum when x = e, then starts dropping.
This reveals the fact that if n is sufficiently large and we are allowed to break n into real numbers,
the best idea is to break it into nearly all e's.
On the other hand, if n is sufficiently large and we can only break n into integers, we should choose integers that are closer to e.
The only potential candidates are 2 and 3 since 2 < e < 3, but we will generally prefer 3 to 2. Why?
Of course, one can prove it based on the formula above, but there is a more natural way shown as follows.
6 = 2 + 2 + 2 = 3 + 3. But 2 * 2 * 2 < 3 * 3.
Therefore, if there are three 2's in the decomposition, we can replace them by two 3's to gain a larger product.
All the analysis above assumes n is significantly large. When n is small (say n <= 10), it may contain flaws.
For instance, when n = 4, we have 2 * 2 > 3 * 1.