Posts published in “Difficulty”

花花酱 LeetCode 312. Burst Balloons

By zxi on November 27, 2017

Problem:

Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by array nums. You are asked to burst all the balloons. If the you burst balloon i you will get nums[left] * nums[i] * nums[right] coins. Here left and right are adjacent indices of i. After the burst, the left and rightthen becomes adjacent.

Find the maximum coins you can collect by bursting the balloons wisely.

Note:
(1) You may imagine nums[-1] = nums[n] = 1. They are not real therefore you can not burst them.
(2) 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100

Example:

Given [3, 1, 5, 8]

Return 167

1 2	nums = [3,1,5,8] --> [3,5,8] --> [3,8] --> [8] --> [] coins = 315 + 358 + 138 + 181 = 167

Idea

Solution1:

C++ / Recursion with memoization

// Author: Huahua

// Runtime: 16 ms

class Solution {

public:

int maxCoins(vector<int>& nums) {

int n = nums.size();

nums.insert(nums.begin(), 1);

nums.push_back(1);

// c[i][j] = maxCoins(nums[i:j+1])

c_ = vector<vector<int>>(n+2, vector<int>(n+2, 0));

return maxCoins(nums, 1, n);

}

private:

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

if(i>j) return 0;

if(c_[i][j]>0) return c_[i][j];

if(i==j) return nums[i-1]*nums[i]*nums[i+1];

int ans=0;

for(int k=i;k<=j;++k)

ans=max(ans, maxCoins(nums,i,k-1) + nums[i-1]*nums[k]*nums[j+1] + maxCoins(nums, k+1, j));

c_[i][j] = ans;

return c_[i][j];

}

vector<vector<int>> c_;

};

Java

// Author: Huahua

class Solution {

private int[][] memo;

private int[] vals;

public int maxCoins(int[] nums) {

final int n = nums.length;

vals = new int[n + 2];

for (int i = 0; i < n; ++i) vals[i + 1] = nums[i];

vals[0] = vals[n + 1] = 1;

memo = new int[n + 2][n + 2];

return maxCoins(1, n);

}

private int maxCoins(int i, int j) {

if (i > j) return 0;

if (i == j) return vals[i - 1] * vals[i] * vals[i + 1];

if (memo[i][j] > 0) return memo[i][j];

int ans = 0;

for (int k = i; k <= j; ++k)

ans = Math.max(ans, maxCoins(i, k - 1) + vals[i - 1] * vals[k] * vals[j + 1] + maxCoins(k + 1, j));

memo[i][j] = ans;

return memo[i][j];

}

Solution2:

C++ / DP

// Author: Huahua

// Runtime: 9 ms

class Solution {

public:

int maxCoins(vector<int>& nums) {

int n = nums.size();

nums.insert(nums.begin(), 1);

nums.push_back(1);

// c[i][j] = maxCoins(nums[i:j+1])

vector<vector<int>> c(n+2, vector<int>(n+2, 0));

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

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

int j=i+l-1;

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

c[i][j] = max(c[i][j], c[i][k-1] + nums[i-1]*nums[k]*nums[j+1] + c[k+1][j]);

}

return c[1][n];

}

};

Java / DP

Java

// Author: Huahua

class Solution {

public int maxCoins(int[] nums) {

final int n = nums.length;

int[] vals = new int[n + 2];

for (int i = 0; i < n; ++i) vals[i + 1] = nums[i];

vals[0] = vals[n + 1] = 1;

int[][] dp = new int[n + 2][n + 2];

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

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

int j = i + l - 1;

int best = 0;

for (int k = i; k <= j; ++k)

best = Math.max(best,

dp[i][k - 1] + vals[i - 1] * vals[k] * vals[j + 1] + dp[k + 1][j]);

dp[i][j] = best;

}

return dp[1][n];

}

花花酱 LeetCode 1. Two Sum

By zxi on November 23, 2017

Problem:

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Idea:

Brute force / Hashtable

Solution1:

Brute force / C++

Time complexity: O(n^2)

Space complexity: O(1)

// Author: Huahua

// Runtime: 166 ms

class Solution {

public:

vector<int> twoSum(vector<int>& nums, int target) {

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

for (int j = i + 1;j < nums.size(); ++j) {

int sum = nums[i] + nums[j];

if (sum == target) {

return {i, j};

}

return {};

}

};

Solution 2:

Hashtable / C++

Time complexity: O(n)

Space complexity: O(n)

// Author: Huahua

// Runtime: 6 ms

class Solution {

public:

vector<int> twoSum(vector<int>& nums, int target) {

unordered_map<int, int> indies;

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

indies[nums[i]] = i;

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

int left = target - nums[i];

if (indies.count(left) && indies[left] != i) {

return {i, indies[left]};

}

return {};

}

};

花花酱 LeetCode 731. My Calendar II – 花花酱

By zxi on November 19, 2017

Problem:

Implement a MyCalendarTwo class to store your events. A new event can be added if adding the event will not cause a triple booking.

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 triple booking happens when three events have some non-empty intersection (ie., there is some time that is common to all 3 events.)

For each call to the method MyCalendar.book, return true if the event can be added to the calendar successfully without causing a triple booking. Otherwise, return false and do not add the event to the calendar.

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

Example 1:

MyCalendar();

MyCalendar.book(10, 20); // returns true

MyCalendar.book(50, 60); // returns true

MyCalendar.book(10, 40); // returns true

MyCalendar.book(5, 15); // returns false

MyCalendar.book(5, 10); // returns true

MyCalendar.book(25, 55); // returns true

Explanation<b>:</b>

The first two events can be booked. The third event can be double booked.

The fourth event (5, 15) can't be booked, because it would result in a triple booking.

The fifth event (5, 10) can be booked, as it does not use time 10 which is already double booked.

The sixth event (25, 55) can be booked, as the time in [25, 40) will be double booked with the third event;

the time [40, 50) will be single booked, and the time [50, 55) will be double booked with the second event.

Note:

The number of calls to MyCalendar.book per test case will be at most 1000.
In calls to MyCalendar.book(start, end), start and end are integers in the range [0, 10^9].

Idea:

Brute Force

Solution1:

Brute Force

Time Complexity: O(n^2)

Space Complexity: O(n)

// Author: Huahua

// Runtime: 82 ms

class MyCalendarTwo {

public:

MyCalendarTwo() {}

bool book(int start, int end) {

for (const auto& kv : overlaps_)

if (max(start, kv.first) < min(end, kv.second)) return false;

for (const auto& kv : booked_) {

const int ss = max(start, kv.first);

const int ee = min(end, kv.second);

if (ss < ee) overlaps_.emplace_back(ss, ee);

}

booked_.emplace_back(start, end);

return true;

}

private:

vector<pair<int, int>> booked_;

vector<pair<int, int>> overlaps_;

};

Java

// Author: Huahua

// Runtime: 156 ms

class MyCalendarTwo {

private List<int[]> booked_;

private List<int[]> overlaps_;

public MyCalendarTwo() {

booked_ = new ArrayList<>();

overlaps_ = new ArrayList<>();

}

public boolean book(int start, int end) {

for (int[] range : overlaps_)

if (Math.max(range[0], start) < Math.min(range[1], end))

return false;

for (int[] range : booked_) {

int ss = Math.max(range[0], start);

int ee = Math.min(range[1], end);

if (ss < ee) overlaps_.add(new int[]{ss, ee});

}

booked_.add(new int[]{start, end});

return true;

}

Python

"""

Author: Huahua

Runtime: 638 ms

"""

class MyCalendarTwo:

def __init__(self):

self.booked_ = []

self.overlaps_ = []

def book(self, start, end):

for s, e in self.overlaps_:

if start < e and end > s: return False

for s, e in self.booked_:

if start < e and end > s:

self.overlaps_.append([max(start, s), min(end, e)])

self.booked_.append([start, end])

return True

Solution 2:

Counting

Time Complexity: O(n^2logn)

Space Complexity: O(n)

// Author: Huahua

// Runtime: 212 ms

class MyCalendarTwo {

public:

MyCalendarTwo() {}

bool book(int start, int end) {

++delta_[start];

--delta_[end];

int count = 0;

for (const auto& kv : delta_) {

count += kv.second;

if (count == 3) {

--delta_[start];

++delta_[end];

return false;

}

if (kv.first > end) break;

}

return true;

}

private:

map<int, int> delta_;

};

Related Problems:

花花酱 LeetCode 729. My Calendar I

By zxi on November 19, 2017

Problem:

Implement a MyCalendar class to store your events. A new event can be added if adding the event will not cause a double booking.

Your class will have the 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 double booking happens when two events have some non-empty intersection (ie., there is some time that is common to both events.)

For each call to the method MyCalendar.book, return true if the event can be added to the calendar successfully without causing a double booking. Otherwise, return false and do not add the event to the calendar.

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

Example 1:

MyCalendar();

MyCalendar.book(10, 20); // returns true

MyCalendar.book(15, 25); // returns false

MyCalendar.book(20, 30); // returns true

Explanation:

The first event can be booked. The second can't because time 15 is already booked by another event.

The third event can be booked, as the first event takes every time less than 20, but not including 20.

Note:

The number of calls to MyCalendar.book per test case will be at most 1000.
In calls to MyCalendar.book(start, end), start and end are integers in the range [0, 10^9].

Idea:

Binary Search

Solution1:

Brute Force: O(n^2)

C++

// Author: Huahua

// Runtime: 99 ms

class MyCalendar {

public:

MyCalendar() {}

bool book(int start, int end) {

for (const auto& event : booked_) {

int s = event.first;

int e = event.second;

if (max(s, start) < min(e, end)) return false;

}

booked_.emplace_back(start, end);

return true;

}

private:

vector<pair<int, int>> booked_;

};

Solution 2:

Binary Search O(nlogn)

C++

// Author: Huahua

// Runtime: 82 ms

class MyCalendar {

public:

MyCalendar() {}

bool book(int start, int end) {

auto it = booked_.lower_bound(start);

if (it != booked_.cend() && it->first < end)

return false;

if (it != booked_.cbegin() && (--it)->second > start)

return false;

booked_[start] = end;

return true;

}

private:

map<int, int> booked_;

};

Java

// Author: Huahua

// Runtime: 170 ms

class MyCalendar {

TreeMap<Integer, Integer> booked_;

public MyCalendar() {

booked_ = new TreeMap<>();

}

public boolean book(int start, int end) {

Integer lb = booked_.floorKey(start);

if (lb != null && booked_.get(lb) > start) return false;

Integer ub = booked_.ceilingKey(start);

if (ub != null && ub < end) return false;

booked_.put(start, end);

return true;

}

Related Problems:

花花酱 LeetCode 728. Self Dividing Numbers

By zxi on November 19, 2017

Problem:

A self-dividing number is a number that is divisible by every digit it contains.

For example, 128 is a self-dividing number because 128 % 1 == 0, 128 % 2 == 0, and 128 % 8 == 0.

Also, a self-dividing number is not allowed to contain the digit zero.

Given a lower and upper number bound, output a list of every possible self dividing number, including the bounds if possible.

Example 1:

Input:

left = 1, right = 22

Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22]

Note:

The boundaries of each input argument are 1 <= left <= right <= 10000.

Idea:

Brute Force

Time Complexity: O(n)

Space Complexity: O(1)

Solution:

C++

// Author: Huahua

// Runtime: 3 ms

class Solution {

public:

vector<int> selfDividingNumbers(int left, int right) {

vector<int> ans;

for (int n = left; n <= right; ++n) {

int t = n;

bool valid = true;

while (t && valid) {

const int r = t % 10;

if (r == 0 || n % r)

valid = false;

t /= 10;

}

if (valid) ans.push_back(n);

}

return ans;

}

};

String

// Author: Huahua

// Runtime: 6 ms

class Solution {

public:

vector<int> selfDividingNumbers(int left, int right) {

vector<int> ans;

for (int n = left; n <= right; ++n) {

const string t = std::to_string(n);

bool valid = true;

for (const char c : t) {

if (c == '0' || n % (c - '0')) {

valid = false;

break;

}

if (valid) ans.push_back(n);

}

return ans;

}

};

Related Problems:

[解题报告] LeetCode 611. Valid Triangle Number