343. Integer Break #

Problem #

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.

Example 1:

Input: 2
Output: 1
Explanation: 2 = 1 + 1, 1 × 1 = 1.

Example 2:

Input: 10
Output: 36
Explanation: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36.

Note: You may assume that n is not less than 2 and not larger than 58.

Summary #

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.

Solution Approach #

• This is a DP problem. Split a number into the sum of multiple numbers, at least into the sum of 2 numbers, and find the maximum product of the decomposed numbers.
• The dynamic transition equation for this problem is dp[i] = max(dp[i], j * (i - j), j * dp[i-j]). A number is decomposed into two numbers, j and i - j, or decomposed into j and more decomposed numbers; more decomposed numbers is dp[i-j]. Since the index of dp[i-j] is less than i, dp[i-j] must have already been calculated when computing dp[i].

Code #

package leetcode

func integerBreak(n int) int {
	dp := make([]int, n+1)
	dp[0], dp[1] = 1, 1
	for i := 1; i <= n; i++ {
		for j := 1; j < i; j++ {
			// dp[i] = max(dp[i], j * (i - j), j*dp[i-j])
			dp[i] = max(dp[i], j*max(dp[i-j], i-j))
		}
	}
	return dp[n]
}