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;

}

};

花花酱 LeetCode 878. Nth Magical Number

By zxi on July 28, 2018

Problem

A positive integer is magical if it is divisible by either A or B.

Return the N-th magical number. Since the answer may be very large, return it modulo 10^9 + 7.

Example 1:

Input: N = 1, A = 2, B = 3
Output: 2

Example 2:

Input: N = 4, A = 2, B = 3
Output: 6

Example 3:

Input: N = 5, A = 2, B = 4
Output: 10

Example 4:

Input: N = 3, A = 6, B = 4
Output: 8

Note:

1 <= N <= 10^9
2 <= A <= 40000
2 <= B <= 40000

Solution: Math + Binary Search

Let n denote the number of numbers <= X that are divisible by either A or B.

n = X / A + X / B – X / lcm(A, B) = X / A + X / B – X / (A * B / gcd(A, B))

Binary search for the smallest X such that n >= N

Time complexity: O(log(1e9*4e5)

Space complexity: O(1)

// Author: Huahua

// Running time: 0 ms

class Solution {

public:

int nthMagicalNumber(int N, int A, int B) {

const int kMod = 1000000007;

long l = 2;

long r = static_cast<long>(1e9 * 4e5);

int d = gcd(A, B);

while (l < r) {

long m = (r - l) / 2 + l;

if (m / A + m / B - m / (A * B / d) < N)

l = m + 1;

else

r = m;

}

return l % kMod;

}

private:

int gcd(int a, int b) {

if (a % b == 0) return b;

return gcd(b, a % b);

}

};

花花酱 LeetCode 877. Stone Game

By zxi on July 28, 2018

Problem

Alex and Lee play a game with piles of stones. There are an even number of piles arranged in a row, and each pile has a positive integer number of stones piles[i].

The objective of the game is to end with the most stones. The total number of stones is odd, so there are no ties.

Alex and Lee take turns, with Alex starting first. Each turn, a player takes the entire pile of stones from either the beginning or the end of the row. This continues until there are no more piles left, at which point the person with the most stones wins.

Assuming Alex and Lee play optimally, return True if and only if Alex wins the game.

Example 1:

Input: [5,3,4,5]
Output: true
Explanation: 
Alex starts first, and can only take the first 5 or the last 5.
Say he takes the first 5, so that the row becomes [3, 4, 5].
If Lee takes 3, then the board is [4, 5], and Alex takes 5 to win with 10 points.
If Lee takes the last 5, then the board is [3, 4], and Alex takes 4 to win with 9 points.
This demonstrated that taking the first 5 was a winning move for Alex, so we return true.

Note:

2 <= piles.length <= 500
piles.length is even.
1 <= piles[i] <= 500
sum(piles) is odd.

Solution 1: min-max (TLE)

Time complexity: O(2^n)

Space complexity: O(n)

// Author: Huahua

// Running time: TLE 26/46 passed.

class Solution {

public:

bool stoneGame(vector<int>& piles) {

return score(piles, 0, piles.size() - 1) > 0;

}

private:

int score(const vector<int>& piles, int l, int r) {

if (l == r) return piles[l];

return max(piles[l] - score(piles, l + 1, r),

piles[r] - score(piles, l, r - 1));

}

};

Solution 2: min-max + memorization

Time complexity: O(n^2)

Space complexity: O(n^2)

// Author: Huahua

// Running time: 12 ms

class Solution {

public:

bool stoneGame(vector<int>& piles) {

const int n = piles.size();

m_ = vector<vector<int>>(n, vector<int>(n, INT_MIN));

return score(piles, 0, n - 1) > 0;

}

private:

vector<vector<int>> m_;

int score(const vector<int>& piles, int l, int r) {

if (l == r) return piles[l];

if (m_[l][r] == INT_MIN)

m_[l][r] = max(piles[l] - score(piles, l + 1, r),

piles[r] - score(piles, l, r - 1));

return m_[l][r];

}

};

Solution 3: min-max + DP

Time complexity: O(n^2)

Space complexity: O(n^2)

// Author: Huahua

// Running time: 8 ms

class Solution {

public:

bool stoneGame(vector<int>& piles) {

const int n = piles.size();

// dp[i][j] := max(your_stones - op_stones) for piles[i] ~ piles[j]

vector<vector<int>> dp(n, vector<int>(n, 0));

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

dp[i][i] = piles[i];

for (int l = 2; l <= n; ++l)

for (int i = 0; i < n - l + 1; ++i) {

int j = i + l - 1;

dp[i][j] = max(piles[i] - dp[i + 1][j], piles[j] - dp[i][j - 1]);

}

return dp[0][n - 1] > 0;

}

};

O(n) Space

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

bool stoneGame(vector<int>& piles) {

const int n = piles.size();

// dp[i] := max(your_stones - op_stones) for piles[i] to piles[i + l - 1]

vector<int> dp(piles);

for (int l = 2; l <= n; ++l)

for (int i = 0; i < n - l + 1; ++i)

dp[i] = max(piles[i] - dp[i + 1], piles[i + l - 1] - dp[i]);

return dp[0] > 0;

}

};

Problem

Given a list of non-overlapping axis-aligned rectangles rects, write a function pick which randomly and uniformily picks an integer point in the space covered by the rectangles.

Note:

An integer point is a point that has integer coordinates.
A point on the perimeter of a rectangle is included in the space covered by the rectangles.
ith rectangle = rects[i] = [x1,y1,x2,y2], where [x1, y1] are the integer coordinates of the bottom-left corner, and [x2, y2] are the integer coordinates of the top-right corner.
length and width of each rectangle does not exceed 2000.
1 <= rects.length <= 100
pick return a point as an array of integer coordinates [p_x, p_y]
pick is called at most 10000 times.

Example 1:

Input: 
["Solution","pick","pick","pick"]
[[[[1,1,5,5]]],[],[],[]]
Output: 
[null,[4,1],[4,1],[3,3]]

Example 2:

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

Explanation of Input Syntax:

The input is two lists: the subroutines called and their arguments. Solution‘s constructor has one argument, the array of rectangles rects. pick has no arguments. Arguments are always wrapped with a list, even if there aren’t any.

Solution: Binary Search

Same as LeetCode 880. Random Pick with Weight

Use area of the rectangles as weights.

Time complexity: Init: O(n) Pick: O(logn)

Space complexity: O(n)

// Author: Huahua

// Running time: 92 ms

class Solution {

public:

Solution(vector<vector<int>> rects): rects_(std::move(rects)) {

sums_ = vector<int>(rects_.size());

for (int i = 0; i < rects_.size(); ++i) {

sums_[i] = (rects_[i][2] - rects_[i][0] + 1) * (rects_[i][3] - rects_[i][1] + 1);

if (i > 0) sums_[i] += sums_[i - 1];

}

vector<int> pick() {

const int s = nextInt(sums_.back()) + 1;

int index = lower_bound(sums_.begin(), sums_.end(), s) - sums_.begin();

const auto& rect = rects_[index];

return {rect[0] + nextInt(rect[2] - rect[0] + 1),

rect[1] + nextInt(rect[3] - rect[1] + 1)};

}

private:

vector<int> sums_;

vector<vector<int>> rects_;

// Returns a random int in [0, n - 1]

int nextInt(int n) {

return rand() / static_cast<double>(RAND_MAX) * n;

}

};

花花酱 LeetCode 879. Profitable Schemes

Problem

Solution: DP

花花酱 LeetCode 878. Nth Magical Number

Problem

Solution: Math + Binary Search

花花酱 LeetCode 877. Stone Game

Problem

Solution 1: min-max (TLE)

Solution 2: min-max + memorization

Solution 3: min-max + DP

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Solution: Binary Search

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Huahua's Tech Road

花花酱 LeetCode 879. Profitable Schemes

Problem

Solution: DP

花花酱 LeetCode 878. Nth Magical Number

Problem

Solution: Math + Binary Search

花花酱 LeetCode 877. Stone Game

Problem

Solution 1: min-max (TLE)

Solution 2: min-max + memorization

Solution 3: min-max + DP

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花花酱 LeetCode 882. Random Point in Non-overlapping Rectangles

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