Given n items with size A[i], an integer m denotes the size of a backpack. How full you can fill this backpack? NoteYou can not divide any item into small pieces.ExampleIf 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 }