Given n items with size A[i] (i from 0 to n - 1), an integer m denotes the size of a backpack. How full you can fill this backpack?
Put this mathematically, you are trying to:
max(sum(j[i] * A[i]))
subject to: j[i] = {0, 1} and sum(j[i] * A[i]) <= m
where 0 =< i < n
j[i] indicates that whether you put item i in the backpack.
Backtracking
If we start from the first item, we have two choices, put it or do not put it in the bag. So it is a decision binary tree of depth n where each level corresponding to a item. The complexity is O(2^n) .
If the remaining capacity is enough (bigger than the current size of item), otherwise we can choose skipping current item.
class Solution {
public:
/**
* @param m: An integer m denotes the size of a backpack
* @param A: Given n items with size A[i]
* @return: The maximum size
*/
int backPack(int m, vector<int> &A) {
pack(0, m, A);
return curMax;
}
void pack(int n, int weight, vector<int> &A) {
int sz = A.size();
int m = 0;
for (int i = n; i < sz; i ++) {
pack(n + 1, weight, A); // do not put it
if (weight >= A[i]) { // we can put it
m += A[i];
if (m > curMax) {
curMax = m;
}
weight -= A[i];
pack(n + 1, weight, A); // put it
}
}
}
private:
int curMax = 0;
};
However, this recursion backtracking is too slow because of the large search space especially if n is large.
Dynamic Programming
If, we use dp[i][j] to represent that if we can use first i items (maximum, could use less) to pack at most j weight. Thus, the following stands:
dp[i][j] = dp[i - 1][j] || dp[i - 1][j - A[i - 1]];
This can be interpreted as:
- by achieving dp[i][j] we automatically achieve dp[i + 1][j]
- by achieving dp[i][j] we automatically achieve dp[i + 1][j + A[i + 1]]
The DP solution allows you to cache intermediate results so you don't have to calculate them over and over again.
class Solution {
public:
/**
* @param m: An integer m denotes the size of a backpack
* @param A: Given n items with size A[i]
* @return: The maximum size
*/
int backPack(int m, vector<int> &A) {
int size = A.size();
vector< vector<bool> > dp(size + 1, vector<bool>(m + 1, false));
for (int i = 0; i < size + 1; ++ i) {
dp[i][0] = true;
}
for(int i = 1; i < A.size() + 1; i++) {
for(int j = 1; j < m + 1; j++) {
if(j < A[i - 1]) { // insufficient capacity
dp[i][j] = dp[i - 1][j];
} else {
dp[i][j] = dp[i - 1][j] || dp[i - 1][j - A[i - 1]];
}
}
}
for (int i = m; i >= 0; i--) { // reverse checking the maximum weight
if (dp[size][i]) {
return i;
}
}
return 0;
}
};
The O(nm) solution (with O(mn) space) fills the DP array and in the end, we need to check if we can fulfil the backpack from m downards to zero (get the maximum)
Memory Optimisation
We can see that the dp[i] is fully dependent on the dp[i-1], thus, we can optimise the memory consumption from O(mn) to O(m).
class Solution {
public:
/**
* @param m: An integer m denotes the size of a backpack
* @param A: Given n items with size A[i]
* @return: The maximum size
*/
int backPack(int m, vector<int> &A) {
int size = A.size();
vector<bool> dp(m + 1, false);
dp[0] = true;
for (int i = 1; i < A.size() + 1; i++) {
for (int j = m; j >= 1; --j) {
if(j >= A[i - 1]) {
dp[j] = dp[j] || dp[j - A[i - 1]];
}
}
}
for (int i = m; i >= 0; i--) {
if (dp[i]) {
return i;
}
}
return 0;
}
};
The runtime complexity is still O(mn) where you need to reverse the inner loop from m downwards to 1.
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0/1背包问题是非常经典的算法问题。要求是：给定 n 个物件，每个物件的重量（或体积）是 A[i] 那么请问一个最多装重（或体积） m 的背包最多可以装下多少这些物件。
如果用数学表示这个问题：
max(sum(j[i] * A[i]))
subject to: j[i] = {0, 1} and sum(j[i] * A[i]) <= m
where 0 =< i < n
j[i] 就是控制是否装下第 i 个 物件 的变量（0或者1）
回溯算法
把整个搜索空间从第一个物件开始当成树根，那么这是一棵二叉树，每层的节点就能对应到每个物品，两个分叉分别代表是否把当前节点（物品）装入背包中。
不难想像，时间复杂度为 O(2^n)。如果当前背包并不能装下当前物件，那么我们需要继续尝试下一个物品，直到所有物品尝试完并回溯。
class Solution {
public:
/**
* @param m: An integer m denotes the size of a backpack
* @param A: Given n items with size A[i]
* @return: The maximum size
*/
int backPack(int m, vector<int> &A) {
pack(0, m, A);
return curMax;
}
void pack(int n, int weight, vector<int> &A) {
int sz = A.size();
int m = 0;
for (int i = n; i < sz; i ++) {
pack(n + 1, weight, A); // 不放
if (weight >= A[i]) { // 可以放
m += A[i];
if (m > curMax) {
curMax = m;
}
weight -= A[i];
pack(n + 1, weight, A); // 放
}
}
}
private:
int curMax = 0;
};
这个算法的问题就是慢。特别是物品数量多的时候，整个搜索空间就变得巨大。
动态规化
如果我们用 dp[i][j] 来表示是否可以用最多 前 i 件物品来装最多重量（或者体积） 为 j 。那么：
dp[i][j] = dp[i - 1][j] || dp[i - 1][j - A[i - 1]];
我们可以反过来理解：
- 如果dp[i][j] 那么下一件物品我们不装的话： dp[i + 1][j]
- 如果dp[i][j] 那么装下下一件物品： dp[i + 1][j + A[i + 1]]
动态规化的优点就是可以把中间结果给缓存起来，这样就不会像递规回溯一样有些结果（中间节点）会重复计算。
class Solution {
public:
/**
* @param m: An integer m denotes the size of a backpack
* @param A: Given n items with size A[i]
* @return: The maximum size
*/
int backPack(int m, vector<int> &A) {
int size = A.size();
vector< vector<bool> > dp(size + 1, vector<bool>(m + 1, false));
for (int i = 0; i < size + 1; ++ i) {
dp[i][0] = true;
}
for(int i = 1; i < A.size() + 1; i++) {
for(int j = 1; j < m + 1; j++) {
if(j < A[i - 1]) { // 装不下了
dp[i][j] = dp[i - 1][j];
} else {
dp[i][j] = dp[i - 1][j] || dp[i - 1][j - A[i - 1]];
}
}
}
for (int i = m; i >= 0; i--) { // 反过来从大到小检查能装下的最大重量
if (dp[size][i]) {
return i;
}
}
return 0;
}
};
时间复杂度和空间复杂度都是 O(nm)。最后面我们只要反过来检查 dp[A.size()][j] 为 true 的第一个下标 j 就是当前背包能装下的最大重量。
内存优化
上面的DP公式我们不难发现， 在计算 dp[i] 的时候只依赖于 dp[i-1], 这样我们就可以只用一维来存取 dp[i-1] 的情况。这样，空间复杂度从 O(mn) 就降到了 O(m).
class Solution {
public:
/**
* @param m: An integer m denotes the size of a backpack
* @param A: Given n items with size A[i]
* @return: The maximum size
*/
int backPack(int m, vector<int> &A) {
int size = A.size();
vector<bool> dp(m + 1, false);
dp[0] = true;
for (int i = 1; i < A.size() + 1; i++) {
for (int j = m; j >= 1; --j) { // 内循环需要从 m 到 1 的检查
if(j >= A[i - 1]) {
dp[j] = dp[j] || dp[j - A[i - 1]];
}
}
}
for (int i = m; i >= 0; i--) {
if (dp[i]) {
return i;
}
}
return 0;
}
};
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