Posts published in December 2017

花花酱 LeetCode 732. My Calendar III

By zxi on December 6, 2017

Problem:

link: https://leetcode.com/problems/my-calendar-iii/description/

Implement a MyCalendarThree class to store your events. A new event can always be added.

Your class will have one method, book(int start, int end). Formally, this represents a booking on the half open interval [start, end), the range of real numbers x such that start <= x < end.

A K-booking happens when K events have some non-empty intersection (ie., there is some time that is common to all K events.)

For each call to the method MyCalendar.book, return an integer K representing the largest integer such that there exists a K-booking in the calendar.

Your class will be called like this: MyCalendarThree cal = new MyCalendarThree();MyCalendarThree.book(start, end)

Example 1:

MyCalendarThree();
MyCalendarThree.book(10, 20); // returns 1
MyCalendarThree.book(50, 60); // returns 1
MyCalendarThree.book(10, 40); // returns 2
MyCalendarThree.book(5, 15); // returns 3
MyCalendarThree.book(5, 10); // returns 3
MyCalendarThree.book(25, 55); // returns 3
Explanation: 
The first two events can be booked and are disjoint, so the maximum K-booking is a 1-booking.
The third event [10, 40) intersects the first event, and the maximum K-booking is a 2-booking.
The remaining events cause the maximum K-booking to be only a 3-booking.
Note that the last event locally causes a 2-booking, but the answer is still 3 because
eg. [10, 20), [10, 40), and [5, 15) are still triple booked.

Note:

The number of calls to MyCalendarThree.book per test case will be at most 400.
In calls to MyCalendarThree.book(start, end), start and end are integers in the range [0, 10^9].

Idea:

Similar to LeetCode 731 My Calendar II Use an ordered / tree map to track the number of event at current time.

For a new book event, increase the number of events at start, decrease the number of events at end.

Scan the timeline to find the maximum number of events.

Solution 1: Count Boundaries

Time complexity: O(n^2)

Space complexity: O(n)

C++

// Author: Huahua

// Runtime: 116 ms

class MyCalendarThree {

public:

MyCalendarThree() {}

int book(int start, int end) {

++deltas_[start];

--deltas_[end];

int ans = 0;

int curr = 0;

for (const auto& kv : deltas_)

ans = max(ans, curr += kv.second);

return ans;

}

private:

map<int, int> deltas_;

};

Solution 2

C++

// Author: Huahua

// Runtime: 66 ms

class MyCalendarThree {

public:

MyCalendarThree() {

counts_[INT_MIN] = 0;

counts_[INT_MAX] = 0;

max_count_ = 0;

}

int book(int start, int end) {

auto l = prev(counts_.upper_bound(start)); // l->first < start

auto r = counts_.lower_bound(end); // r->first >= end

for (auto curr = l, next = curr; curr != r; curr = next) {

++next;

if (next->first > end)

counts_[end] = curr->second;

if (curr->first <= start && next->first > start)

max_count_ = max(max_count_, counts_[start] = curr->second + 1);

else

max_count_ = max(max_count_, ++curr->second);

}

return max_count_;

}

private:

map<int, int> counts_;

int max_count_;

};

Solution 3: Segment Tree

C++

// Author: Huahua

// Runtime: 63 ms (<95.65%)

class MyCalendarThree {

public:

MyCalendarThree(): max_count_(0) {

root_.reset(new Node(0, 100000000, 0));

}

int book(int start, int end) {

Add(start, end, root_.get());

return max_count_;

}

private:

struct Node {

Node(int l, int r, int count):l(l), m(-1), r(r), count(count){}

int l;

int m;

int r;

int count;

std::unique_ptr<Node> left;

std::unique_ptr<Node> right;

};

void Add(int start, int end, Node* root) {

if (root->m != -1) {

if (end <= root->m)

Add(start, end, root->left.get());

else if(start >= root->m)

Add(start, end, root->right.get());

else {

Add(start, root->m, root->left.get());

Add(root->m, end, root->right.get());

}

return;

}

if (start == root->l && end == root->r)

max_count_ = max(max_count_, ++root->count);

else if (start == root->l) {

root->m = end;

root->left.reset(new Node(start, root->m, root->count + 1));

root->right.reset(new Node(root->m, root->r, root->count));

max_count_ = max(max_count_, root->count + 1);

} else if(end == root->r) {

root->m = start;

root->left.reset(new Node(root->l, root->m, root->count));

root->right.reset(new Node(root->m, root->r, root->count + 1));

max_count_ = max(max_count_, root->count + 1);

} else {

root->m = start;

root->left.reset(new Node(root->l, root->m, root->count));

root->right.reset(new Node(root->m, root->r, root->count));

Add(start, end, root->right.get());

}

std::unique_ptr<Node> root_;

int max_count_;

};

Python3

"""

Author: Huahua

Runtime: 436 ms (<88.88%)

"""

class Node:

def __init__(self, l, r, count):

self.l = l

self.m = -1

self.r = r

self.count = count

self.left = None

self.right = None

class MyCalendarThree:

def __init__(self):

self.root = Node(0, 10**9, 0)

self.max = 0

def book(self, start, end):

self.add(start, end, self.root)

return self.max

def add(self, start, end, root):

if root.m != -1:

if end <= root.m: self.add(start, end, root.left)

elif start >= root.m: self.add(start, end, root.right)

else:

self.add(start, root.m, root.left)

self.add(root.m, end, root.right)

return

if start == root.l and end == root.r:

root.count += 1

self.max = max(self.max, root.count)

elif start == root.l:

root.m = end

root.left = Node(start, root.m, root.count + 1)

root.right = Node(root.m, root.r, root.count)

self.max = max(self.max, root.count + 1)

elif end == root.r:

root.m = start;

root.left = Node(root.l, root.m, root.count)

root.right = Node(root.m, root.r, root.count + 1)

self.max = max(self.max, root.count + 1)

else:

root.m = start

root.left = Node(root.l, root.m, root.count)

root.right = Node(root.m, root.r, root.count)

self.add(start, end, root.right)

Related Problems:

花花酱 LeetCode 740. Delete and Earn

By zxi on December 5, 2017

Problem:

Given an array nums of integers, you can perform operations on the array.

In each operation, you pick any nums[i] and delete it to earn nums[i] points. After, you must delete everyelement equal to nums[i] - 1 or nums[i] + 1.

You start with 0 points. Return the maximum number of points you can earn by applying such operations.

Example 1:

Input: nums = [3, 4, 2]

Output: 6

Explanation:

Delete 4 to earn 4 points, consequently 3 is also deleted.

Then, delete 2 to earn 2 points. 6 total points are earned.

Example 2:

Input: nums = [2, 2, 3, 3, 3, 4]

Output: 9

Explanation:

Delete 3 to earn 3 points, deleting both 2's and the 4.

Then, delete 3 again to earn 3 points, and 3 again to earn 3 points.

9 total points are earned.

Note:

The length of nums is at most 20000.
Each element nums[i] is an integer in the range [1, 10000].

Idea:

Reduce the problem to House Robber Problem

Key observations: If we take nums[i]

We can safely take all of its copies.
We can’t take any of copies of nums[i – 1] and nums[i + 1]

This problem is reduced to 198 House Robber.

Houses[i] has all the copies of num whose value is i.

[3 4 2] -> [0 2 3 4], rob([0 2 3 4]) = 6

[2, 2, 3, 3, 3, 4] -> [0 2*2 3*3 4], rob([0 2*2 3*3 4]) = 9

Time complexity: O(n+r) reduction + O(r) solving rob = O(n + r)

Space complexity: O(r)

r = max(nums) – min(nums) + 1

Time complexity: O(n + r)

Space complexity: O(r)

Solution:

// Author: Huahua

// Runtime: 6 ms

class Solution {

public:

int deleteAndEarn(vector<int>& nums) {

if (nums.empty()) return 0;

const auto range = minmax_element(nums.begin(), nums.end());

const int l = *(range.first);

const int r = *(range.second);

vector<int> points(r - l + 1, 0);

for (const int num : nums)

points[num - l] += num;

return rob(points);

}

private:

// From LeetCode 198. House Robber

int rob(const vector<int>& nums) {

int dp2 = 0;

int dp1 = 0;

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

int dp = max(dp2 + nums[i], dp1);

dp2 = dp1;

dp1 = dp;

}

return dp1;

}

};

Related Problem:

花花酱 LeetCode 198. House Robber

花花酱 LeetCode 198. House Robber

By zxi on December 5, 2017

Problem:

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

Idea:

Time complexity: O(n)

Space complexity: O(n) -> O(1)

Solution:

C++ / Recursion + Memorization

Time complexity: O(n)

Space complexity: O(n)

// Author: Huahua

// Runtime: 0 ms

class Solution {

public:

int rob(vector<int>& nums) {

const int n = nums.size();

m_ = vector<int>(n, -1);

return rob(nums, n - 1);

}

private:

int rob(const vector<int>& nums, int i) {

if (i < 0) return 0;

if (m_[i] >= 0) return m_[i];

return m_[i] = max(rob(nums, i - 2) + nums[i],

rob(nums, i - 1));

}

vector<int> m_;

};

C++ / DP

Time complexity: O(n)

Space complexity: O(n)

// Author: Huahua

// Runtime: 3 ms

class Solution {

public:

int rob(vector<int>& nums) {

if (nums.empty()) return 0;

vector<int> dp(nums.size(), 0);

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

dp[i] = max((i > 1 ? dp[i - 2] : 0) + nums[i],

(i > 0 ? dp[i - 1] : 0));

return dp.back();

}

};

C++ / O(1) Space

// Author: Huahua

// Runtime: 3 ms

class Solution {

public:

int rob(vector<int>& nums) {

if (nums.empty()) return 0;

int dp2 = 0;

int dp1 = 0;

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

int dp = max(dp2 + nums[i], dp1);

dp2 = dp1;

dp1 = dp;

}

return dp1;

}

};

花花酱 LeetCode 741. Cherry Pickup

By zxi on December 4, 2017

题目大意：给你樱桃田的地图（1: 樱桃, 0: 空, -1: 障碍物）。然你从左上角走到右下角（只能往右或往下），再从右下角走回左上角（只能往左或者往上）。问你最多能采到多少棵樱桃。

Problem:

In a N x N grid representing a field of cherries, each cell is one of three possible integers.

0 means the cell is empty, so you can pass through;
1 means the cell contains a cherry, that you can pick up and pass through;
-1 means the cell contains a thorn that blocks your way.

Your task is to collect maximum number of cherries possible by following the rules below:

Starting at the position (0, 0) and reaching (N-1, N-1) by moving right or down through valid path cells (cells with value 0 or 1);
After reaching (N-1, N-1), returning to (0, 0) by moving left or up through valid path cells;
When passing through a path cell containing a cherry, you pick it up and the cell becomes an empty cell (0);
If there is no valid path between (0, 0) and (N-1, N-1), then no cherries can be collected.

Example 1:

Input: grid =

[[0, 1, -1],

[1, 0, -1],

[1, 1, 1]]

Output: 5

Explanation:

The player started at (0, 0) and went down, down, right right to reach (2, 2).

4 cherries were picked up during this single trip, and the matrix becomes [[0,1,-1],[0,0,-1],[0,0,0]].

Then, the player went left, up, up, left to return home, picking up one more cherry.

The total number of cherries picked up is 5, and this is the maximum possible.

Note:

grid is an N by N 2D array, with 1 <= N <= 50.
Each grid[i][j] is an integer in the set {-1, 0, 1}.
It is guaranteed that grid[0][0] and grid[N-1][N-1] are not -1.

Idea:

Key observation: (0,0) to (n-1, n-1) to (0, 0) is the same as (n-1,n-1) to (0,0) twice

Two people starting from (n-1, n-1) and go to (0, 0).
They move one step (left or up) at a time simultaneously. And pick up the cherry within the grid (if there is one).
if they ended up at the same grid with a cherry. Only one of them can pick up it.

Solution: DP / Recursion with memoization.

x1, y1, x2 to represent a state y2 can be computed: y2 = x1 + y1 – x2

dp(x1, y1, x2) computes the max cherries if start from {(x1, y1), (x2, y2)} to (0, 0), which is a recursive function.

Since two people move independently, there are 4 subproblems: (left, left), (left, up), (up, left), (left, up). Finally, we have:

dp(x1, y1, x2)= g[y1][x1] + g[y2][x2] + max{dp(x1-1,y1,x2-1), dp(x1,y1-1,x2-1), dp(x1-1,y1,x2), dp(x1,y1-1,x2)}

Time complexity: O(n^3)

Space complexity: O(n^3)

Solution: DP

Time complexity: O(n^3)

Space complexity: O(n^3)

C ++

// Author: Huahua

// Runtime: 32 ms

class Solution {

public:

int cherryPickup(vector<vector<int>>& grid) {

const int n = grid.size();

grid_ = &grid;

m_ = vector<vector<vector<int>>>(

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

return max(0, dp(n - 1, n - 1, n - 1));

}

private:

// max cherries from (x1, y1) to (0, 0) + (x2, y2) to (0, 0)

int dp(int x1, int y1, int x2) {

const int y2 = x1 + y1 - x2;

if (x1 < 0 || y1 < 0 || x2 < 0 || y2 < 0) return -1;

if ((*grid_)[y1][x1] < 0 || (*grid_)[y2][x2] < 0) return -1;

if (x1 == 0 && y1 == 0) return (*grid_)[y1][x1];

if (m_[x1][y1][x2] != INT_MIN) return m_[x1][y1][x2];

int ans = max(max(dp(x1 - 1, y1, x2 - 1), dp(x1, y1 - 1, x2)),

max(dp(x1, y1 - 1, x2 - 1), dp(x1 - 1, y1, x2)));

if (ans < 0) return m_[x1][y1][x2] = -1;

ans += (*grid_)[y1][x1];

if (x1 != x2) ans += (*grid_)[y2][x2];

return m_[x1][y1][x2] = ans;

}

vector<vector<vector<int>>> m_;

vector<vector<int>>* grid_; // does not own the object

};

Java

// Author: Huahua

class Solution {

private int[][] grid;

private int[][][] memo;

public int cherryPickup(int[][] grid) {

this.grid = grid;

int m = grid.length;

int n = grid[0].length;

memo = new int[m][n][n];

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

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

Arrays.fill(memo[i][j], Integer.MIN_VALUE);

return Math.max(0, dp(n - 1, m - 1, n - 1));

}

private int dp(int x1, int y1, int x2) {

final int y2 = x1 + y1 - x2;

if (x1 < 0 || y1 < 0 || x2 < 0 || y2 < 0) return -1;

if (grid[y1][x1] < 0 || grid[y2][x2] < 0) return -1;

if (x1 == 0 && y1 == 0) return grid[y1][x1];

if (memo[y1][x1][x2] != Integer.MIN_VALUE) return memo[y1][x1][x2];

memo[y1][x1][x2] = Math.max(Math.max(dp(x1 - 1, y1, x2 - 1), dp(x1, y1 - 1, x2)),

Math.max(dp(x1, y1 - 1, x2 - 1), dp(x1 - 1, y1, x2)));

if (memo[y1][x1][x2] >= 0) {

memo[y1][x1][x2] += grid[y1][x1];

if (x1 != x2) memo[y1][x1][x2] += grid[y2][x2];

}

return memo[y1][x1][x2];

}

Related Problems:

花花酱 LeetCode 735. Asteroid Collision

By zxi on December 3, 2017

Problem:

We are given an array asteroids of integers representing asteroids in a row.

For each asteroid, the absolute value represents its size, and the sign represents its direction (positive meaning right, negative meaning left). Each asteroid moves at the same speed.

Find out the state of the asteroids after all collisions. If two asteroids meet, the smaller one will explode. If both are the same size, both will explode. Two asteroids moving in the same direction will never meet.

Example 1:

Input:

asteroids = [5, 10, -5]

Output: [5, 10]

Explanation:

The 10 and -5 collide resulting in 10. The 5 and 10 never collide.

Example 2:

Input:

asteroids = [8, -8]

Output: []

Explanation:

The 8 and -8 collide exploding each other.

Example 3:

Input:

asteroids = [10, 2, -5]

Output: [10]

Explanation:

The 2 and -5 collide resulting in -5. The 10 and -5 collide resulting in 10.

Example 4:

Input:

asteroids = [-2, -1, 1, 2]

Output: [-2, -1, 1, 2]

Explanation:

The -2 and -1 are moving left, while the 1 and 2 are moving right.

Asteroids moving the same direction never meet, so no asteroids will meet each other.

Note:

The length of asteroids will be at most 10000.
Each asteroid will be a non-zero integer in the range [-1000, 1000]..

Idea:

Simulation

Time complexity: O(n), Space complexity: O(n)

Solution:

C ++

// Author: Huahua

// Runtime: 22 ms

class Solution {

public:

vector<int> asteroidCollision(vector<int>& asteroids) {

vector<int> s;

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

const int size = asteroids[i];

if (size > 0) { // To right, OK

s.push_back(size);

} else {

// To left

if (s.empty() || s.back() < 0) // OK when all are negtives

s.push_back(size);

else if (abs(s.back()) <= abs(size)) {

// Top of the stack is going right.

// Its size is less than the current one.

// The current one still alive!

if (abs(s.back()) < abs(size)) --i;

s.pop_back(); // Destory the top one moving right

}

// s must look like [-s1, -s2, ... , si, sj, ...]

return s;

}

};