Description
Given n items with size Ai, an integer m denotes the size of a backpack. How full you can fill this backpack?
You can not divide any item into small pieces.
Example
Example 1:
Input: [3,4,8,5], backpack size=10
Output: 9
Example 2:
Input: [2,3,5,7], backpack size=12
Output: 12
Challenge
O(n x m) time and O(m) memory.
O(n x m) memory is also acceptable if you do not know how to optimize memory.
Thinking: f [i] [j] denotes the i-th former article is selected from a number of items into the capacity j backpack can filled.
public class Solution { /** * @param m: An integer m denotes the size of a backpack * @param A: Given n items with size A[i] * @return: The maximum size */ public int backPack(int m, int[] A) { boolean f[][] = new boolean[A.length + 1][m + 1]; for (int i = 0; i <= A.length; i++) { for (int j = 0; j <= m; j++) { f[i][j] = false; } } f[0][0] = true; for (int i = 1; i <= A.length; i++) { for (int j = 0; j <= m; j++) { f[i][j] = f[i - 1][j]; if (j >= A[i-1] && f[i-1][j - A[i-1]]) { f[i][j] = true; } } // for j } // for i for (int i = m; i >= 0; i--) { if (f[A.length][i]) { return i; } } return 0; } }
The complexity of O (m) space Solution
public class Solution { /** * @param m: An integer m denotes the size of a backpack * @param A: Given n items with size A[i] * @return: The maximum size */ public int backPack(int m, int[] A) { int f[] = new int[m + 1]; for (int i = 0; i < A.length; i++) { for (int j = m; j >= A[i]; j--) { f[j] = Math.max(f[j], f[j - A[i]] + A[i]); } } return f[m]; } }