Given n items with size Ai, an integer m denotes the size of a backpack. How full you can fill this backpack?
Notice
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, 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.
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.
Tags
LintCode Copyright Dynamic Programming Backpack
Related Problems
Medium Backpack VI 30 %
Medium Backpack V 38 %
Medium Backpack IV 36 %
Hard Backpack III 50 %
Medium Backpack II 37 %
Analysis
背包问题序列I
要点: 单次选择+最大体积
常用是DP,但也可以用递归+记忆化搜索来做。
这个问题没有给每个物品的价值,也没有问最多能获得多少价值,而是问最多能装多满。
实际上在这里,如果我们把每个物品的大小当作每个物品的价值,就可以完美解决这个问题。
Solution
Preferred Solution
2D - DP
状态:
dp[i][j] - 代表在前 i 件物品中选择若干件,这若干件物品的体积和不超过 j 的情况下,所能获得的最大容量
public class Solution {
public int backPack(int m, int[] A) {
int[] dp = new int[m+1];
for (int i = 0; i < A.length; i++) {
for (int j = m; j > 0; j--) {
if (j >= A[i]) {
dp[j] = Math.max(dp[j], dp[j - A[i]] + A[i]);
}
}
}
return dp[m];
}
}
1D DP
j = m, m-1, ..., A[i]
public int backPack(int m, int[] A) {
int[] dp = new int[m + 1];
for (int i = 0; i < A.length; i++) {
for (int j = m; j >= A[i]; j--) {
dp[j] = Math.max(dp[j], dp[j - A[i]] + A[i]);
}
}
return dp[m];
}
Another DP Approach (NOT Preferred)
2D with O(n * m) time and O(n * m) space.
动态规划4要素
State:
f[i][S] “前i”个物品,取出一些能否组成和为S
Function:
f[i][S] = f[i-1][S - a[i]] or f[i-1][S]
Initialize:
f[i][0] = true; f[0][1..target] = false
Answer:
检查所有的f[n][j]
O(n * S) , 滚动数组优化
注意这里A[i - 1]的数组下标,因为新建的是i = 1, ..., A.length
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[][] dp = new boolean[A.length + 1][m + 1];
dp[0][0] = true;
for (int i = 1; i <= A.length; i++) {
for (int j = 0; j <= m; j++) {
dp[i][j] = dp[i - 1][j] || (j - A[i - 1] >= 0 && dp[i - 1][j - A[i - 1]]);
}
}
for (int j = m; j >= 0; j--) {
if (dp[A.length][j]) {
return j;
}
}
return 0;
}
}
1D version with O(n * m) time and O(m) memory.
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) {
if (A.length == 0) return 0;
int n = A.length;
boolean[] dp = new boolean[m + 1];
Arrays.fill(dp, false);
dp[0] = true;
for (int i = 1; i <= n; i++) {
for (int j = m; j >= 0; j--) {
if (j - A[i - 1] >= 0 && dp[j - A[i - 1]]) {
dp[j] = dp[j - A[i - 1]];
}
}
}
for (int i = m; i >= 0;i--) {
if (dp[i]) return i;
}
return 0;
}
}
int max = Integer.MAX_VALUE;
public int backPack(int m, int[] A) {
Arrays.sort(A);
boolean[] visited = new boolean[m + 1];
int[] memo = new int[m + 1];
dfs( m , A , 0 ,visited , memo );
return m - max ;
}
private int dfs( int target , int[] A , int startIndex , boolean[] visited , int[] memo){
if(visited[target]){
return memo[target];
}
int res = target;
for(int i = startIndex ; i < A.length ; i++){
if(target - A[i] >= 0){
res = dfs( target - A[i] , A , i + 1 ,visited ,memo);
}
else{
break;
}
}
visited[target] = true;
memo[target] = res;
max = Math.min(max ,res);
return res;
}