Posts tagged as “greedy”

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 860. Lemonade Change

By zxi on June 30, 2018

Problem

At a lemonade stand, each lemonade costs $5.

Customers are standing in a queue to buy from you, and order one at a time (in the order specified by bills).

Each customer will only buy one lemonade and pay with either a $5, $10, or $20 bill. You must provide the correct change to each customer, so that the net transaction is that the customer pays $5.

Note that you don’t have any change in hand at first.

Return true if and only if you can provide every customer with correct change.

Example 1:

Input: [5,5,5,10,20]
Output: true
Explanation: 
From the first 3 customers, we collect three $5 bills in order.
From the fourth customer, we collect a $10 bill and give back a $5.
From the fifth customer, we give a $10 bill and a $5 bill.
Since all customers got correct change, we output true.

Example 2:

Input: [5,5,10]
Output: true

Example 3:

Input: [10,10]
Output: false

Example 4:

Input: [5,5,10,10,20]
Output: false
Explanation: 
From the first two customers in order, we collect two $5 bills.
For the next two customers in order, we collect a $10 bill and give back a $5 bill.
For the last customer, we can't give change of $15 back because we only have two $10 bills.
Since not every customer received correct change, the answer is false.

Solution: Simulation + Greedy

Always use 10 bill first.

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua

// Running time: 18 ms

class Solution {

public:

bool lemonadeChange(vector<int>& bills) {

int fives = 0;

int tens = 0;

for (const int bill : bills) {

if (bill == 5) {

++fives;

} else if (bill == 10) {

if (!fives) return false;

++tens;

--fives;

} else if (bill == 20) {

if (tens && fives) {

--tens;

--fives;

} else if (fives >= 3) {

fives -= 3;

} else {

return false;

}

return true;

}

};

花花酱 LeetCode 452. Minimum Number of Arrows to Burst Balloons

By zxi on June 23, 2018

Problem

There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it’s horizontal, y-coordinates don’t matter and hence the x-coordinates of start and end of the diameter suffice. Start is always smaller than end. There will be at most 10⁴ balloons.

An arrow can be shot up exactly vertically from different points along the x-axis. A balloon with x_start and x_endbursts by an arrow shot at x if x_start ≤ x ≤ x_end. There is no limit to the number of arrows that can be shot. An arrow once shot keeps travelling up infinitely. The problem is to find the minimum number of arrows that must be shot to burst all balloons.

Example:

Input:
[[10,16], [2,8], [1,6], [7,12]]

Output:
2

Explanation:
One way is to shoot one arrow for example at x = 6 (bursting the balloons [2,8] and [1,6]) and another arrow at x = 11 (bursting the other two balloons).

Solution: Sweep Line

Time complexity: O(nlogn)

Space complexity: O(1)

C++

// Author: Huahua

// Running time: 82 ms

class Solution {

public:

int findMinArrowShots(vector<pair<int, int>>& points) {

if (points.empty()) return 0;

sort(points.begin(), points.end(),

[&](const pair<int,int>& a, const pair<int,int>& b){

return a.second < b.second;

});

int right = points.front().second;

int ans = 1;

for (const auto& point : points) {

if (point.first > right) {

right = point.second;

++ans;

}

return ans;

}

};

Problem

题目大意：给你一些牌，问你能否分组，要求每组w张连续的牌。

https://leetcode.com/problems/hand-of-straights/description/

Alice has a hand of cards, given as an array of integers.

Now she wants to rearrange the cards into groups so that each group is size W, and consists of W consecutive cards.

Return true if and only if she can.

Example 1:

Input: hand = [1,2,3,6,2,3,4,7,8], W = 3
Output: true
Explanation: Alice's hand can be rearranged as [1,2,3],[2,3,4],[6,7,8].

Example 2:

Input: hand = [1,2,3,4,5], W = 4
Output: false
Explanation: Alice's hand can't be rearranged into groups of 4.

Note:

1 <= hand.length <= 10000
0 <= hand[i] <= 10^9
1 <= W <= hand.length

Solution: Greedy

Time complexity: O(nlogn)

Space complexity: O(n)

// Author: Huahua

// Running time: 65 ms

class Solution {

public:

bool isNStraightHand(vector<int>& hand, int W) {

map<int, int> counts;

for (int h : hand) ++counts[h];

while (!counts.empty()) {

auto it = counts.begin();

int start = it->first;

for (int j = 0; j < W; ++j) {

if (it == counts.end()

|| it->first != start + j) return false;

auto prev = it++;

if (--(prev->second) == 0)

counts.erase(prev);

}

return true;

}

};

Posts tagged as “greedy”

花花酱 LeetCode 502. IPO

Problem

Solution: Greedy

花花酱 LeetCode 135. Candy

Problem

Solution: Greedy

花花酱 LeetCode 860. Lemonade Change

Problem

Solution: Simulation + Greedy

花花酱 LeetCode 452. Minimum Number of Arrows to Burst Balloons

Problem

Solution: Sweep Line

Related Problems

花花酱 LeetCode 846. Hand of Straights

Problem

Solution: Greedy