花花酱 LeetCode 808. Soup Servings

By zxi on August 1, 2018

Problem

There are two types of soup: type A and type B. Initially we have N ml of each type of soup. There are four kinds of operations:

Serve 100 ml of soup A and 0 ml of soup B
Serve 75 ml of soup A and 25 ml of soup B
Serve 50 ml of soup A and 50 ml of soup B
Serve 25 ml of soup A and 75 ml of soup B

When we serve some soup, we give it to someone and we no longer have it. Each turn, we will choose from the four operations with equal probability 0.25. If the remaining volume of soup is not enough to complete the operation, we will serve as much as we can. We stop once we no longer have some quantity of both types of soup.

Note that we do not have the operation where all 100 ml’s of soup B are used first.

Return the probability that soup A will be empty first, plus half the probability that A and B become empty at the same time.

Example:
Input: N = 50
Output: 0.625
Explanation: 
If we choose the first two operations, A will become empty first. For the third operation, A and B will become empty at the same time. For the fourth operation, B will become empty first. So the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.25 * (1 + 1 + 0.5 + 0) = 0.625.

Notes:

0 <= N <= 10^9.
Answers within 10^-6 of the true value will be accepted as correct.

Solution 1: Recursion with Memorization

Time complexity: O(N^2) N ~ 5000 / 25 = 200

Space complexity: O(N^2)

C++

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

double soupServings(int N) {

if (N >= 5000) return 1.0;

return prob(N, N);

}

private:

unordered_map<int, unordered_map<int, double>> m_;

double prob(int A, int B) {

static int ops[][2] = {{100, 0}, {75, 25}, {50, 50}, {25, 75}};

if (A <= 0 && B <= 0) return 0.5;

if (A <= 0) return 1.0;

if (B <= 0) return 0.0;

if (m_.count(A) && m_[A].count(B)) return m_[A][B];

m_[A][B] = 0.0;

for (int i = 0; i < 4; ++i)

m_[A][B] += 0.25 * prob(A - ops[i][0], B - ops[i][1]);

return m_[A][B];

}

};

花花酱 LeetCode 881. Random Flip Matrix

By zxi on July 31, 2018

Problem

You are given the number of rows n_rows and number of columns n_cols of a 2D binary matrix where all values are initially 0. Write a function flip which chooses a 0 value uniformly at random, changes it to 1, and then returns the position [row.id, col.id] of that value. Also, write a function reset which sets all values back to 0. Try to minimize the number of calls to system’s Math.random() and optimize the time and space complexity.

Note:

1 <= n_rows, n_cols <= 10000
0 <= row.id < n_rows and 0 <= col.id < n_cols
flip will not be called when the matrix has no 0 values left.
the total number of calls to flip and reset will not exceed 1000.

Example 1:

Input: 
["Solution","flip","flip","flip","flip"]
[[2,3],[],[],[],[]]
Output: [null,[0,1],[1,2],[1,0],[1,1]]

Example 2:

Input: 
["Solution","flip","flip","reset","flip"]
[[1,2],[],[],[],[]]
Output: [null,[0,0],[0,1],null,[0,0]]

Explanation of Input Syntax:

The input is two lists: the subroutines called and their arguments. Solution‘s constructor has two arguments, n_rows and n_cols. flip and reset have no arguments. Arguments are always wrapped with a list, even if there aren’t any.

Solution 1: Hashtable + Resample

Time complexity: O(|flip|) = O(1000) = O(1)

Space complexity: O(|flip|) = O(1000) = O(1)

// Author: Huahua

// Running time: 12 ms

class Solution {

public:

Solution(int n_rows, int n_cols):

rows_(n_rows), cols_(n_cols), n_(rows_ * cols_) {}

vector<int> flip() {

int index = rand() % n_;

while (m_.count(index))

index = rand() % n_;

m_.insert(index);

return {index / cols_, index % cols_};

}

void reset() {

m_.clear();

n_ = rows_ * cols_;

}

private:

const int rows_;

const int cols_;

int n_;

unordered_set<int> m_;

};

Solution 2: Fisher–Yates shuffle

Generate a random shuffle of 0 to n – 1, one number at a time.

Time complexity: flip: O(1)

Space complexity: O(|flip|) = O(1000) = O(1)

C++

// Author: Huahua

// Running time: 12 ms

class Solution {

public:

Solution(int n_rows, int n_cols):

rows_(n_rows), cols_(n_cols), n_(rows_ * cols_) {}

vector<int> flip() {

int s = rand() % (n_--);

int index = s;

if (m_.count(s))

index = m_[s];

m_[s] = m_.count(n_) ? m_[n_] : n_;

return {index / cols_, index % cols_};

}

void reset() {

m_.clear();

n_ = rows_ * cols_;

}

private:

const int rows_;

const int cols_;

int n_;

unordered_map<int, int> m_;

};

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By zxi on July 31, 2018

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花花酱 LeetCode 855. Exam Room

By zxi on July 29, 2018

Problem

In an exam room, there are N seats in a single row, numbered 0, 1, 2, ..., N-1.

When a student enters the room, they must sit in the seat that maximizes the distance to the closest person. If there are multiple such seats, they sit in the seat with the lowest number. (Also, if no one is in the room, then the student sits at seat number 0.)

Return a class ExamRoom(int N) that exposes two functions: ExamRoom.seat() returning an int representing what seat the student sat in, and ExamRoom.leave(int p) representing that the student in seat number p now leaves the room. It is guaranteed that any calls to ExamRoom.leave(p) have a student sitting in seat p.

Example 1:

Input: ["ExamRoom","seat","seat","seat","seat","leave","seat"], [[10],[],[],[],[],[4],[]]
Output: [null,0,9,4,2,null,5]
Explanation:
ExamRoom(10) -> null
seat() -> 0, no one is in the room, then the student sits at seat number 0.
seat() -> 9, the student sits at the last seat number 9.
seat() -> 4, the student sits at the last seat number 4.
seat() -> 2, the student sits at the last seat number 2.
leave(4) -> null
seat() -> 5, the student sits at the last seat number 5.

Note:

1 <= N <= 10^9
ExamRoom.seat() and ExamRoom.leave() will be called at most 10^4 times across all test cases.
Calls to ExamRoom.leave(p) are guaranteed to have a student currently sitting in seat number p.

Solution: BST

Use a BST (ordered set) to track the current seatings.

Time Complexity:

init: O(1)

seat: O(P)

leave: O(logP)

Space complexity: O(P)

// Author: Huahua

// Running time: 24 ms

class ExamRoom {

public:

ExamRoom(int N): N_(N) {}

int seat() {

int p = 0;

int max_dist = s_.empty() ? 0 : *s_.begin();

auto left = s_.begin();

auto right = left;

while (left != s_.end()) {

++right;

int l = *left;

int r = right != s_.end() ? *right : (2 * (N_ - 1) - *left);

int d = (r - l) / 2;

if (d > max_dist) {

max_dist = d;

p = l + d;

}

left = right;

}

s_.insert(p);

return p;

}

void leave(int p) {

s_.erase(p);

}

private:

const int N_;

set<int> s_;

};

花花酱 LeetCode 879. Profitable Schemes

By zxi on July 28, 2018

Problem

There are G people in a gang, and a list of various crimes they could commit.

The i-th crime generates a profit[i] and requires group[i] gang members to participate.

If a gang member participates in one crime, that member can’t participate in another crime.

Let’s call a profitable scheme any subset of these crimes that generates at least P profit, and the total number of gang members participating in that subset of crimes is at most G.

How many schemes can be chosen? Since the answer may be very large, return it modulo 10^9 + 7.

Example 1:

Input: G = 5, P = 3, group = [2,2], profit = [2,3]
Output: 2
Explanation: 
To make a profit of at least 3, the gang could either commit crimes 0 and 1, or just crime 1.
In total, there are 2 schemes.

Example 2:

Input: G = 10, P = 5, group = [2,3,5], profit = [6,7,8]
Output: 7
Explanation: 
To make a profit of at least 5, the gang could commit any crimes, as long as they commit one.
There are 7 possible schemes: (0), (1), (2), (0,1), (0,2), (1,2), and (0,1,2).

Note:

1 <= G <= 100
0 <= P <= 100
1 <= group[i] <= 100
0 <= profit[i] <= 100
1 <= group.length = profit.length <= 100

Solution: DP

Time complexity: O(KPG)

Space complexity: O(KPG)

C++

// Author: Huahua

// Running time: 64 ms

class Solution {

public:

int profitableSchemes(int G, int P, vector<int>& group, vector<int>& profit) {

const int kMod = 1000000007;

const int K = group.size();

// dp[k][i][j]:= # of schemes of making profit i with j people by commiting first k crimes.

vector<vector<vector<int>>> dp(K + 1, vector<vector<int>>(P + 1, vector<int>(G + 1, 0)));

dp[0][0][0] = 1;

for (int k = 1; k <= K; ++k) {

const int p = profit[k - 1];

const int g = group[k - 1];

for (int i = 0; i <= P; ++i)

for (int j = 0; j <= G; ++j)

dp[k][i][j] = (dp[k - 1][i][j] + (j < g ? 0 : dp[k - 1][max(0, i - p)][j - g])) % kMod;

}

return accumulate(begin(dp[K][P]), end(dp[K][P]), 0LL) % kMod;

}

};

Space complexity: O(PG)

v1: Dimension reduction by copying.

// Author: Huahua

// Running time: 24 ms

class Solution {

public:

int profitableSchemes(int G, int P, vector<int>& group, vector<int>& profit) {

const int kMod = 1000000007;

const int K = group.size();

// dp[i][j]:= # of schemes of making profit i with j people.

vector<vector<int>> dp(P + 1, vector<int>(G + 1, 0));

dp[0][0] = 1;

for (int k = 1; k <= K; ++k) {

auto tmp = dp;

const int p = profit[k - 1];

const int g = group[k - 1];

for (int i = 0; i <= P; ++i)

for (int j = 0; j <= G; ++j)

tmp[i][j] = (tmp[i][j] + (j < g ? 0 : dp[max(0, i - p)][j - g])) % kMod;

dp.swap(tmp);

}

return accumulate(begin(dp[P]), end(dp[P]), 0LL) % kMod;

}

};

v2: Dimension reduction by using rolling array.

// Author: Huahua

// Running time: 16 ms

class Solution {

public:

int profitableSchemes(int G, int P, vector<int>& group, vector<int>& profit) {

const int kMod = 1000000007;

// dp[i][j]:= # of schemes of making profit i with j people.

vector<vector<int>> dp(P + 1, vector<int>(G + 1, 0));

dp[0][0] = 1;

const int K = group.size();

for (int k = 0; k < K; ++k) {

const int p = profit[k];

const int g = group[k];

for (int i = P; i >= 0; --i) {

const int ip = min(P, i + p);

for (int j = G - g; j >= 0; --j)

dp[ip][j + g] = (dp[ip][j + g] + dp[i][j]) % kMod;

}

return accumulate(begin(dp[P]), end(dp[P]), 0LL) % kMod;

}

};

Huahua's Tech Road

花花酱 LeetCode 808. Soup Servings

Problem

Solution 1: Recursion with Memorization

花花酱 LeetCode 881. Random Flip Matrix

Problem

Solution 1: Hashtable + Resample

Solution 2: Fisher–Yates shuffle

Upgraded to HTTPS

花花酱 LeetCode 855. Exam Room

Problem

Solution: BST

花花酱 LeetCode 879. Profitable Schemes

Problem

Solution: DP