Lintcode: Backpack

Given n items with size A[i], an integer m denotes the size of a backpack. How full you can fill this backpack? 

Note
You can not divide any item into small pieces.

Example
If we have 4 items with size [2, 3, 5, 7], the backpack size is 11, we can select 2, 3 and 5, so that the max size we can fill this backpack is 10. If the backpack size is 12. we can select [2, 3, 7] so that we can fulfill the backpack.

You function should return the max size we can fill in the given backpack.

DP.

boolean d[i][j]: For the first i items, can we fill a backpack of size j? true or false.

d[i][j] = d[i-1][j] || (j>=A[i-1] && d[i-1][j-A[i-1]]).

d[0][0] = true;

We can use 1D array to perform the DP.

d[j] = d[j] || d[j-A[i-1]].

NOTE: for 1D array, the j must be decreased from m to 0 rather increasing from 0 to m!

2D code:

 1 public class Solution {
 2     /**
 3      * @param m: An integer m denotes the size of a backpack
 4      * @param A: Given n items with size A[i]
 5      * @return: The maximum size
 6      */
 7     public int backPack(int m, int[] A) {
 8         // write your code here
 9         boolean[][] res = new boolean[A.length+1][m+1];
10         res[0][0] = true;
11         for (int i=1; i<=A.length; i++) {
12             for (int j=0; j<=m; j++) {
13                 res[i][j] = res[i-1][j] || (j-A[i-1]>=0 && res[i-1][j-A[i-1]]);
14             }
15         }
16         for (int j=m; j>=0; j--) {
17             if (res[A.length][j]) return j;
18         }
19         return 0;
20     }
21 }

1D code:

 1 public class Solution {
 2     /**
 3      * @param m: An integer m denotes the size of a backpack
 4      * @param A: Given n items with size A[i]
 5      * @return: The maximum size
 6      */
 7     public int backPack(int m, int[] A) {
 8         if (A.length==0) return 0;
 9         
10         int len = A.length;
11         boolean[] size = new boolean[m+1];
12         Arrays.fill(size,false);
13         size[0] = true;
14         for (int i=1;i<=len;i++)
15             for (int j=m;j>=0;j--){
16                 if (j-A[i-1]>=0 && size[j-A[i-1]])
17                     size[j] = size[j-A[i-1]];
18             }
19 
20         for (int i=m; i>=0;i--)
21             if (size[i]) return i;
22 
23         return 0;
24     }
25 }