Backpack II

Description

There are n items and a backpack with size m. Given array A representing the size of each item and array V representing the value of each item.

What's the maximum value can you put into the backpack?

1. A[i], V[i], n, m are all integers.
2. You can not split an item.
3. The sum size of the items you want to put into backpack can not exceed m.
4. Each item can only be picked up once

Example

Example 1:

Input: m = 10, A = [2, 3, 5, 7], V = [1, 5, 2, 4]
Output: 9
Explanation: Put A[1] and A[3] into backpack, getting the maximum value V[1] + V[3] = 9 

Example 2:

Input: m = 10, A = [2, 3, 8], V = [2, 5, 8]

Output: 10
Explanation: Put A[0] and A[2] into backpack, getting the maximum value V[0] + V[2] = 10 

Challenge

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

 

Ideas:

The classic 01 knapsack problem, dynamic programming resource allocation.

Set F [i] [j] denotes the i-th former items into size backpack j, the sum of the maximum value that can be acquired. Decision is the i-th item is not loaded into the backpack, the state transition equation is f[i][j] = max(f[i - 1][j], f[i - 1][j - A[i]] + V[i])

Optimized spatial arrays may be used to scroll to O (m).

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