Description

Given n items with size Ai and value Vi, and a backpack with size _m. What's the maximum value can you put into the backpack?

Notice

You cannot divide item into small pieces and the total size of items you choose should smaller or equal to m.

Example

Given 4 items with size [2, 3, 5, 7]and value [1, 5, 2, 4], and a backpack with size 10. The maximum value is 9.

Challenge

O(n x m) memory is acceptable, can you do it in O(m) memory?

1. Solution

这一题和backpack I非常相似，唯一不同的是状态不是体积而是价值。参考backpack I即可。

public class Solution {
    /**
     * @param m: An integer m denotes the size of a backpack
     * @param A & V: Given n items with size A[i] and value V[i]
     * @return: The maximum value
     */
    public int backPackII(int m, int[] A, int V[]) {
        // write your code here
        int[][] dp = new int[A.length+1][m+1];
        int max = 0;
        for (int i = 1; i <= A.length; i++) {
            for (int j = 1; j <= m; j++) {
                dp[i][j] = dp[i-1][j];
                if (A[i-1] <= j) {
                    dp[i][j] = Math.max(dp[i][j], dp[i-1][j-A[i-1]] + V[i-1]);
                }
                max = Math.max(max, dp[i][j]);
            }
        }
        return dp[A.length][m];
    }
}

2. Optimized Solution

public class Solution {
    /**
     * @param m: An integer m denotes the size of a backpack
     * @param A & V: Given n items with size A[i] and value V[i]
     * @return: The maximum value
     */
    public int backPackII(int m, int[] A, int V[]) {
        // write your code here
        int[] dp = new int[m+1];
        for (int i = 0; i < A.length; i++) {
            for (int j = m; j > 0; j--) {
                if (A[i] <= j) {
                    dp[j] = Math.max(dp[j], dp[j-A[i]] + V[i]);
                }
            }
        }
        return dp[m];
    }
}