Posts published in “Greedy”

花花酱 LeetCode 455. Assign Cookies

By zxi on August 15, 2018

Problem

Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie. Each child i has a greed factor g_i, which is the minimum size of a cookie that the child will be content with; and each cookie j has a size s_j. If s_j >= g_i, we can assign the cookie j to the child i, and the child i will be content. Your goal is to maximize the number of your content children and output the maximum number.

Note:
You may assume the greed factor is always positive.
You cannot assign more than one cookie to one child.

Example 1:

Input: [1,2,3], [1,1]

Output: 1

Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3. 
And even though you have 2 cookies, since their size is both 1, you could only make the child whose greed factor is 1 content.
You need to output 1.

Example 2:

Input: [1,2], [1,2,3]

Output: 2

Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2. 
You have 3 cookies and their sizes are big enough to gratify all of the children, 
You need to output 2.

Solution: Greedy + Two Pointers

Time complexity: O(mlogm + nlogn)

Space complexity: O(1)

// Author: Huahua

// Running time: 28 ms

class Solution {

public:

int findContentChildren(vector<int>& g, vector<int>& s) {

sort(begin(g), end(g));

sort(begin(s), end(s));

int ans = 0;

int j = 0;

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

while (j < s.size() && s[j] < g[i]) {

++j;

}

if (j < s.size()) {

++ans;

++j;

continue;

}

return ans;

}

};

花花酱 LeetCode 885. Boats to Save People

By zxi on August 4, 2018

Problem

The i-th person has weight people[i], and each boat can carry a maximum weight of limit.

Each boat carries at most 2 people at the same time, provided the sum of the weight of those people is at most limit.

Return the minimum number of boats to carry every given person. (It is guaranteed each person can be carried by a boat.)

Example 1:

Input: people = [1,2], limit = 3
Output: 1
Explanation: 1 boat (1, 2)

Example 2:

Input: people = [3,2,2,1], limit = 3
Output: 3
Explanation: 3 boats (1, 2), (2) and (3)

Example 3:

Input: people = [3,5,3,4], limit = 5
Output: 4
Explanation: 4 boats (3), (3), (4), (5)

Note:

1 <= people.length <= 50000
1 <= people[i] <= limit <= 30000

Solution: Greedy + Two Pointers

Time complexity: O(nlogn)

Space complexity: O(1)

Put one heaviest guy and put the lightest guy if not full.

// Author: Huahua

// Running time: 80 ms

class Solution {

public:

int numRescueBoats(vector<int>& people, int limit) {

sort(people.rbegin(), people.rend());

int i = 0;

int j = people.size() - 1;

int ans = 0;

while (i <= j) {

++ans;

if (i == j) break;

if (people[i++] + people[j] <= limit) --j;

}

return ans;

}

};

花花酱 LeetCode 502. IPO

By zxi on August 4, 2018

Problem

Suppose LeetCode will start its IPO soon. In order to sell a good price of its shares to Venture Capital, LeetCode would like to work on some projects to increase its capital before the IPO. Since it has limited resources, it can only finish at most k distinct projects before the IPO. Help LeetCode design the best way to maximize its total capital after finishing at most k distinct projects.

You are given several projects. For each project i, it has a pure profit P_i and a minimum capital of C_i is needed to start the corresponding project. Initially, you have W capital. When you finish a project, you will obtain its pure profit and the profit will be added to your total capital.

To sum up, pick a list of at most k distinct projects from given projects to maximize your final capital, and output your final maximized capital.

Example 1:

Input: k=2, W=0, Profits=[1,2,3], Capital=[0,1,1].

Output: 4

Explanation: Since your initial capital is 0, you can only start the project indexed 0.
             After finishing it you will obtain profit 1 and your capital becomes 1.
             With capital 1, you can either start the project indexed 1 or the project indexed 2.
             Since you can choose at most 2 projects, you need to finish the project indexed 2 to get the maximum capital.
             Therefore, output the final maximized capital, which is 0 + 1 + 3 = 4.

Note:

You may assume all numbers in the input are non-negative integers.
The length of Profits array and Capital array will not exceed 50,000.
The answer is guaranteed to fit in a 32-bit signed integer.

Solution: Greedy

For each round, find the most profitable job whose capital requirement <= W.

Finish that job and increase W.

Brute force (TLE)

Time complexity: O(kn)

Space complexity: O(1)

C++

class Solution {

public:

int findMaximizedCapital(int k, int W, vector<int>& Profits, vector<int>& Capital) {

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

int best_j = -1;

int best_profit = 0;

for (int j = 0; j < Profits.size(); ++j) {

if (Capital[j] <= W && Profits[j] > best_profit) {

best_j = j;

best_profit = Profits[j];

}

if (best_profit == 0) break;

W += Profits[best_j];

Profits[best_j] = INT_MIN; // Used.

}

return W;

}

};

Use priority queue and multiset to track doable and undoable projects at given W.

Time complexity: O(nlogn)

Space complexity: O(n)

// Author: Huahua

// Running time: 28 ms (beats 96.4%)

class Solution {

public:

int findMaximizedCapital(int k, int W, vector<int>& Profits, vector<int>& Capital) {

priority_queue<int> doable; // sorted by profit high to low.

multiset<pair<int, int>> undoable; // {capital, profit}

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

if (Profits[i] <= 0) continue;

if (Capital[i] <= W)

doable.push(Profits[i]);

else

undoable.emplace(Capital[i], Profits[i]);

}

auto it = undoable.cbegin();

while (!doable.empty() && k--) {

W += doable.top(); doable.pop();

while (it != undoable.cend() && it->first <= W)

doable.push(it++->second);

}

return W;

}

};

Or use an array and sort by capital

// Author: Huahua

// Running time: 28 ms (beats 96.4%)

class Solution {

public:

int findMaximizedCapital(int k, int W, vector<int>& Profits, vector<int>& Capital) {

priority_queue<int> doable; // sorted by profit high to low.

vector<pair<int, int>> undoable; // {capital, profit}

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

if (Profits[i] <= 0) continue;

if (Capital[i] <= W)

doable.push(Profits[i]);

else

undoable.emplace_back(Capital[i], Profits[i]);

}

sort(begin(undoable), end(undoable));

auto it = undoable.cbegin();

while (!doable.empty() && k--) {

W += doable.top(); doable.pop();

while (it != undoable.cend() && it->first <= W)

doable.push(it++->second);

}

return W;

}

};

花花酱 LeetCode 135. Candy

By zxi on July 27, 2018

Problem

There are N children standing in a line. Each child is assigned a rating value.

You are giving candies to these children subjected to the following requirements:

Each child must have at least one candy.
Children with a higher rating get more candies than their neighbors.

What is the minimum candies you must give?

Example 1:

Input: [1,0,2]
Output: 5
Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.

Example 2:

Input: [1,2,2]
Output: 4
Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
             The third child gets 1 candy because it satisfies the above two conditions.

Solution: Greedy

First pass: left to right, the right one will have one more candy than the left one if taller.

Second pass: right to left, the left one will be at least one more candy than the right one if taller.

Time Complexity: O(n)

Space Complexity: O(n)

C++

class Solution {

public:

int candy(vector<int>& ratings) {

const int n = ratings.size();

vector<int> ans(n, 1);

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

if (ratings[i] > ratings[i - 1]) ans[i] = ans[i - 1] + 1;

for (int i = n - 2; i >= 0; --i)

if (ratings[i] > ratings[i + 1]) ans[i] = max(ans[i], ans[i + 1] + 1);

return accumulate(begin(ans), end(ans), 0);

}

};

花花酱 LeetCode 870. Advantage Shuffle

By zxi on July 14, 2018

Problem

Given two arrays A and B of equal size, the advantage of A with respect to B is the number of indices i for which A[i] > B[i].

Return any permutation of A that maximizes its advantage with respect to B.

Example 1:

Input: A = [2,7,11,15], B = [1,10,4,11]
Output: [2,11,7,15]

Example 2:

Input: A = [12,24,8,32], B = [13,25,32,11]
Output: [24,32,8,12]

Note:

1 <= A.length = B.length <= 10000
0 <= A[i] <= 10^9
0 <= B[i] <= 10^9

Solution: Greedy 田忌赛马

Use the smallest unused number A[j] in A such that A[j] > B[i], if not possible, use the smallest number in A.

Time complexity: O(nlogn)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 124 ms

class Solution {

public:

vector<int> advantageCount(vector<int>& A, vector<int>& B) {

multiset<int> s(begin(A), end(A));

vector<int> ans;

for (int b : B) {

auto it = s.upper_bound(b);

if (it == s.end()) it = s.begin();

ans.push_back(*it);

s.erase(it);

}

return ans;

}

};