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[LeetCode] Integer Break

leonardofed / 994人閱讀

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.

For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36 (10 = 3 + 3 + 4).

Note: you may assume that n is not less than 2.

Hint

There is a simple O(n) solution to this problem.
You may check the breaking results of n ranging from 7 to 10 to discover the regularities.

Solution

1. beat 40.83% O(log(n))

public class Solution {
    public int integerBreak(int n) {
        if (n < 4) return n-1;
        else if(n%3 == 0)
            return (int)Math.pow(3, n/3);
        else if(n%3 == 1)
            return 4*(int)Math.pow(3, (n-4)/3);
        else 
            return 2*(int)Math.pow(3, n/3);
    }
}

2. beat 40.83% O(n)

public class Solution {
    public int integerBreak(int n) {
        if (n<4) return n-1;
        int product = 1;
        while(n>4){
            product*=3;
            n-=3;
        }
        return product*n;
    }
}

3. DP beat 40.83% O(n)

public class Solution {
    public int integerBreak(int n) {
        if (n < 4) return n-1;
        int[] dp = new int[n+1];
        dp[2] = 2;
        dp[3] = 3;
        for (int i = 4; i <= n; i++) {
            dp[i] = Math.max(3*dp[i-3], 2*dp[i-2]);
        }
        return dp[n];
    }
}

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