Posts tagged as “combination”

Knapsack Problem 背包问题

By zxi on November 10, 2018

Videos

上期节目中我们对动态规划做了一个总结，这期节目我们来聊聊背包问题。

背包问题是一个NP-complete的组合优化问题，Search的方法需要O(2^N)时间才能获得最优解。而使用动态规划，我们可以在伪多项式（pseudo-polynomial time）时间内获得最优解。

0-1 Knapsack Problem 0-1背包问题

Problem

Given N items, w[i] is the weight of the i-th item and v[i] is value of the i-th item. Given a knapsack with capacity W. Maximize the total value. Each item can be use 0 or 1 time.

0-1背包问题的通常定义是：一共有N件物品，第i件物品的重量为w[i]，价值为v[i]。在总重量不超过背包承载上限W的情况下，能够获得的最大价值是多少？每件物品可以使用0次或者1次。

例子：

重量 w = [1, 1, 2, 2]

价值 v = [1, 3, 4, 5]

背包承重 W = 4

最大价值为9，可以选第1,2,4件物品，也可以选第3，4件物品；总重量为4，总价值为9。

动态规划的状态转移方程为：

1	dp[i][j] = max(dp[i-1][j], dp[i][j-w[i]] + v[i])

Solutions

Search

def knapsack01DFS(w, v, W):

def dfs(s, cur_w, cur_v, ans):

ans[0] = max(ans[0], cur_v)

if s == N: return

for i in range(s, N):

if cur_w + w[i] <= W:

dfs(i + 1, cur_w + w[i], cur_v + v[i], ans)

ans = [0]

dfs(0, 0, 0, ans)

return ans[0]

DP2

def knapsack01(w, v, W):

dp = [[0] * (W + 1) for _ in range(N+1)]

for i in range(1, N + 1):

dp[i] = dp[i-1].clone()

for j in range(w[i], W + 1):

dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - w[i - 1]] + v[i - 1])

return max(dp[N])

DP1/tmp

def knapsack01R(w, v, W):

dp = [0] * (W + 1)

for i in range(0, N):

tmp = list(dp)

for j in range(w[i], W + 1):

tmp[j] = max(tmp[j], dp[j - w[i]] + v[i])

dp = tmp

return max(dp)

DP1/push

def knapsack01R2(w, v, W):

dp = [0] * (W + 1)

for i in range(0, N):

for j in range(W, w[i] - 1, -1):

dp[j] = max(dp[j], dp[j - w[i]] + v[i])

return max(dp)

DP1/pull

def knapsack01R1(w, v, W):

dp = [0] * (W + 1)

for i in range(0, N):

for j in range(W - w[i], -1, -1):

dp[j + w[i]] = max(dp[j + w[i]], dp[j] + v[i])

return max(dp)

Unbounded Knapsack Problem 完全背包

完全背包、多重背包是常见的变形。和01背包的区别在于，完全背包每件物品可以使用无限多次，而多重背包每件物品最多可以使用n[i]次。两个问题都可以转换成01背包问题进行求解。

但是Naive的转换会大大增加时间复杂度：

完全背包：“复制”第i件物品到一共有 W/w[i] 件

多重背包：“复制”第i件物品到一共有 n[i] 件

然后直接调用01背包进行求解。

时间复杂度：

完全背包 O(Σ(W/w[i])*W)

多重背包 O(Σn[i]*W)

不难看出时间复杂度 = O(物品数量*背包承重)

背包承重是给定的，要降低运行时候，只有减少物品数量。但怎样才能减少总的物品数量呢？

这就涉及到二进制思想：任何一个正整数都可以用 (1, 2, 4, …, 2^K)的组合来表示。例如14 = 2 + 4 + 8。
原本需要放入14件相同的物品，现在只需要放入3件（重量和价值是原物品的2倍，4倍，8倍）。大幅降低了总的物品数量从而降低运行时间。

完全背包：对于第i件物品，我们只需要创建k = log(W/w[i])件虚拟物品即可。

每件虚拟物品的重量和价值为：1*(w[i], v[i]), 2*(w[i], v[i]), …, 2^k*(w[i], v[i])。

多重背包：对于第i件物品，我们只需要创建k + 1件虚拟物品即可，其中k = log(n[i])。

每件虚拟物品的重量和价值为：1*(w[i], v[i]), 2*(w[i], v[i]), …, 2^(k-1)*(w[i], v[i]), 以及 (n[i] – 2^k – 1) * (w[i], v[i])。

例如：n[i] = 14, k = 3, 虚拟物品的倍数为 1, 2, 4 和 7，这4个数组合可以组成1 ~ 14中的任何一个数，并且不会>14，即不超过n[i]。

二进制转换后直接调用01背包即可

时间复杂度：

完全背包 O(Σlog(W/w[i])*W)

多重背包 O(Σlog(n[i])*W)

空间复杂度 O(W)

其实完全背包和多重背包都可以在 O(NW)时间内完成，前者在视频中有讲到，后者属于超纲内容，以后有机会再和大家深入分享。

Bounded Knapsack Problem 多重背包

花花酱 LeetCode 923. 3Sum With Multiplicity

By zxi on October 13, 2018

Problem

Given an integer array A, and an integer target, return the number of tuples i, j, k such that i < j < k and A[i] + A[j] + A[k] == target.

As the answer can be very large, return it modulo 10^9 + 7.

Example 1:

Input: A = [1,1,2,2,3,3,4,4,5,5], target = 8
Output: 20
Explanation: 
Enumerating by the values (A[i], A[j], A[k]):
(1, 2, 5) occurs 8 times;
(1, 3, 4) occurs 8 times;
(2, 2, 4) occurs 2 times;
(2, 3, 3) occurs 2 times.

Example 2:

Input: A = [1,1,2,2,2,2], target = 5
Output: 12
Explanation: 
A[i] = 1, A[j] = A[k] = 2 occurs 12 times:
We choose one 1 from [1,1] in 2 ways,
and two 2s from [2,2,2,2] in 6 ways.

Note:

3 <= A.length <= 3000
0 <= A[i] <= 100
0 <= target <= 300

Solution: Math / Combination

Time complexity: O(n + |target|^2)

Space complexity: O(|target|)

C++

// Author: Huahua, runtime: 4 ms

class Solution {

public:

int threeSumMulti(vector<int>& A, int target) {

constexpr int kMaxN = 100;

constexpr int kMod = 1e9 + 7;

vector<long> c(kMaxN + 1, 0);

for (int a : A) ++c[a];

long ans = 0;

for (int i = 0; i <= target; ++i) {

for (int j = i; j <= target; ++j) {

const int k = target - i - j;

if (k < 0 || k >= c.size() || k < j) continue;

if (!c[i] || !c[j] || !c[k]) continue;

if (i == j && j == k)

ans += (c[i] - 2) * (c[i] - 1) * c[i] / 6;

else if (i == j && j != k)

ans += c[i] * (c[i] - 1) / 2 * c[k];

else if (i != j && j == k)

ans += c[i] * (c[j] - 1) * c[j] / 2;

else

ans += c[i] * c[j] * c[k];

}

return ans % kMod;

}

};

花花酱 LeetCode 920. Number of Music Playlists

By zxi on October 6, 2018

Problem

Your music player contains N different songs and she wants to listen to L (not necessarily different) songs during your trip. You create a playlist so that:

Every song is played at least once
A song can only be played again only if K other songs have been played

Return the number of possible playlists. As the answer can be very large, return it modulo 10^9 + 7.

Example 1:

Input: N = 3, L = 3, K = 1
Output: 6
Explanation: There are 6 possible playlists. [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1].

Example 2:

Input: N = 2, L = 3, K = 0
Output: 6
Explanation: There are 6 possible playlists. [1, 1, 2], [1, 2, 1], [2, 1, 1], [2, 2, 1], [2, 1, 2], [1, 2, 2]

Example 3:

Input: N = 2, L = 3, K = 1
Output: 2
Explanation: There are 2 possible playlists. [1, 2, 1], [2, 1, 2]

Note:

0 <= K < N <= L <= 100

Solution: DP

dp[i][j] := # of playlists of length i using j different songs.

dp[i][j] = dp[i – 1][j – 1] * (N – (j – 1)) + // Adding a new song. j – 1 used, choose any one from (N – (j – 1)) unused.
dp[i -1][j] * max(j – K, 0) // Reuse an existing song.

Time complexity: O(LN)

Space complexity: O(LN) -> O(N)

C++/O(LN)

// Author: Huahua

class Solution {

public:

int numMusicPlaylists(int N, int L, int K) {

constexpr long kMod = 1e9 + 7;

vector<vector<long>> dp(L + 1, vector<long>(N + 1, 0));

dp[0][0] = 1;

for (int i = 1; i <= L; ++i)

for (int j = 1; j <= min(i, N); ++j)

dp[i][j] = (dp[i - 1][j - 1] * (N - (j - 1)) +

dp[i - 1][j] * max(j - K, 0)) % kMod;

return dp[L][N];

}

};

C++/O(N)

// Author: Huahua, 0 ms

class Solution {

public:

int numMusicPlaylists(int N, int L, int K) {

constexpr long kMod = 1e9 + 7;

vector<long> dp(N + 1, 0);

for (int i = 1; i <= L; ++i) {

dp[0] = i == 1;

for (int j = min(i, N); j >= 1; --j)

dp[j] = (dp[j - 1] * (N - (j - 1)) +

dp[j] * max(j - K, 0)) % kMod;

}

return dp[N];

}

};

花花酱 LeetCode 902. Numbers At Most N Given Digit Set

By zxi on September 13, 2018

Problem

We have a sorted set of digits D, a non-empty subset of {'1','2','3','4','5','6','7','8','9'}. (Note that '0' is not included.)

Now, we write numbers using these digits, using each digit as many times as we want. For example, if D = {'1','3','5'}, we may write numbers such as '13', '551', '1351315'.

Return the number of positive integers that can be written (using the digits of D) that are less than or equal to N.

Example 1:

Input: D = ["1","3","5","7"], N = 100
Output: 20
Explanation: 
The 20 numbers that can be written are:
1, 3, 5, 7, 11, 13, 15, 17, 31, 33, 35, 37, 51, 53, 55, 57, 71, 73, 75, 77.

Example 2:

Input: D = ["1","4","9"], N = 1000000000
Output: 29523
Explanation: 
We can write 3 one digit numbers, 9 two digit numbers, 27 three digit numbers,
81 four digit numbers, 243 five digit numbers, 729 six digit numbers,
2187 seven digit numbers, 6561 eight digit numbers, and 19683 nine digit numbers.
In total, this is 29523 integers that can be written using the digits of D.

Note:

D is a subset of digits '1'-'9' in sorted order.
1 <= N <= 10^9

Solution -1: DFS (TLE)

Time complexity: O(|D|^log10(N))

Space complexity: O(n)

// Author: Huahua

class Solution {

public:

int atMostNGivenDigitSet(vector<string>& D, int N) {

int ans = 0;

dfs(D, 0, N, ans);

return ans;

}

private:

void dfs(const vector<string>& D, int cur, int N, int& ans) {

if (cur > N) return;

if (cur > 0 && cur <= N) ++ans;

for (const string& d : D)

dfs(D, cur * 10 + d[0] - '0', N, ans);

}

};

Solution 1: Math

Time complexity: O(log10(N))

Space complexity: O(1)

Suppose N has n digits.

less than n digits

We can use all the numbers from D to construct numbers of with length 1,2,…,n-1 which are guaranteed to be less than N.

e.g. n = 52125, D = [1, 2, 5]

format X: e.g. 1, 2, 5 counts = |D| ^ 1

format XX: e.g. 11,12,15,21,22,25,51,52,55, counts = |D|^2

format XXX: counts = |D|^3

format XXXX: counts = |D|^4

exact n digits

if all numbers in D != N[0], counts = |d < N[0] | d in D| * |D|^(n-1), and we are done.

e.g. N = 34567, D = [1,2,8]

we can make:

X |3|^1
XX |3| ^ 2
XXX |3| ^ 3
XXXX |3| ^ 3
1XXXX, |3|^4
2XXXX, |3|^4
we can’t do 8XXXX

Total = (3^1 + 3^2 + 3^3 + 3^4) + 2 * |3|^ 4 = 120 + 162 = 282

N = 52525, D = [1,2,5]

However, if d = N[i], we need to check the next digit…

X |3|^1
XX |3| ^ 2
XXX |3| ^ 3
XXXX |3| ^ 3
1XXXX, |3|^4
2XXXX, |3|^4
5????
- 51XXX |3|^3
- 52???
  - 521XX |3|^2
  - 522XX |3|^2
  - 525??
    - 5251X |3|^1
    - 5252?
      - 52521 |3|^0
      - 52522 |3|^0
      - 52525 +1

total = (120) + 2 * |3|^4 + |3|^3 + 2*|3|^2 + |3|^1 + 2 * |3|^0 + 1 = 120 + 213 = 333

if every digit of N is from D, then we also have a valid solution, thus need to + 1.

C++

class Solution {

public:

int atMostNGivenDigitSet(vector<string>& D, int N) {

const string s = to_string(N);

const int n = s.length();

int ans = 0;

for (int i = 1; i < n; ++i)

ans += pow(D.size(), i);

for (int i = 0; i < n; ++i) {

bool prefix = false;

for (const string& d : D) {

if (d[0] < s[i]) {

ans += pow(D.size(), n - i - 1);

} else if (d[0] == s[i]) {

prefix = true;

break;

}

if (!prefix) return ans;

}

// plus one for solution N itself, all the digits are from D.

return ans + 1;

}

};

Java

// Author: Huahua, 4 ms

class Solution {

public int atMostNGivenDigitSet(String[] D, int N) {

char[] s = String.valueOf(N).toCharArray();

int n = s.length;

int ans = 0;

for (int i = 1; i < n; ++i)

ans += (int)Math.pow(D.length, i);

for (int i = 0; i < n; ++i) {

boolean prefix = false;

for (String d : D) {

if (d.charAt(0) < s[i]) {

ans += Math.pow(D.length, n - i - 1);

} else if (d.charAt(0) == s[i]) {

prefix = true;

break;

}

if (!prefix) return ans;

}

return ans + 1;

}

Python3

# Author: Huahua, 36 ms

class Solution:

def atMostNGivenDigitSet(self, D, N):

s = str(N)

n = len(s)

ans = sum(len(D) ** i for i in range(1, n))

for i, c in enumerate(s):

ans += len(D) ** (n - i - 1) * sum(d < c for d in D)

if c not in D: return ans

return ans + 1

花花酱 LeetCode 22. Generate Parentheses

By zxi on July 26, 2018

Problem

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

[
  "((()))",
  "(()())",
  "(())()",
  "()(())",
  "()()()"
]

Solution: DFS

Time complexity: O(2^n)

Space complexity: O(k + n)

C++

// Author: Huahua

// Running time: 0 ms

class Solution {

public:

vector<string> generateParenthesis(int n) {

vector<string> ans;

string cur;

if (n > 0) dfs(n, n, cur, ans);

return ans;

}

private:

void dfs(int l, int r, string& s, vector<string>& ans) {

if (l + r == 0) {

ans.push_back(s);

return;

}

if (r < l) return;

if (l > 0) {

dfs(l - 1, r, s += "(", ans);

s.pop_back();

}

if (r > 0) {

dfs(l, r - 1, s += ")", ans);

s.pop_back();

}

};

Posts tagged as “combination”

Knapsack Problem 背包问题

Videos

0-1 Knapsack Problem 0-1背包问题

Problem

Solutions

Search

DP2

DP1/tmp

DP1/push

DP1/pull

Unbounded Knapsack Problem 完全背包

Bounded Knapsack Problem 多重背包

花花酱 LeetCode 923. 3Sum With Multiplicity

Problem

Solution: Math / Combination

C++

花花酱 LeetCode 920. Number of Music Playlists

Problem

Solution: DP

C++/O(LN)

C++/O(N)

花花酱 LeetCode 902. Numbers At Most N Given Digit Set

Problem

Solution -1: DFS (TLE)

Solution 1: Math

less than n digits

exact n digits

C++

Java

Python3

花花酱 LeetCode 22. Generate Parentheses

Problem

Solution: DFS

Related Problems