Huahua's Tech Road

花花酱 LeetCode 1361. Validate Binary Tree Nodes

By zxi on February 23, 2020

You have n binary tree nodes numbered from 0 to n - 1 where node i has two children leftChild[i] and rightChild[i], return true if and only if all the given nodes form exactly one valid binary tree.

If node i has no left child then leftChild[i] will equal -1, similarly for the right child.

Note that the nodes have no values and that we only use the node numbers in this problem.

Example 1:

Input: n = 4, leftChild = [1,-1,3,-1], rightChild = [2,-1,-1,-1]
Output: true

Example 2:

Input: n = 4, leftChild = [1,-1,3,-1], rightChild = [2,3,-1,-1]
Output: false

Example 3:

Input: n = 2, leftChild = [1,0], rightChild = [-1,-1]
Output: false

Example 4:

Input: n = 6, leftChild = [1,-1,-1,4,-1,-1], rightChild = [2,-1,-1,5,-1,-1]
Output: false

Constraints:

1 <= n <= 10^4
leftChild.length == rightChild.length == n
-1 <= leftChild[i], rightChild[i] <= n - 1

Solution: Count in-degrees for each node

in degree must <= 1 and there must be exact one node that has 0 in-degree.

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

C++

// Author: Huahua

class Solution {

public:

bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {

vector<int> in(n);

for (int l : leftChild) if (l >= 0) ++in[l];

for (int r : rightChild) if (r >= 0) ++in[r];

int zero = 0;

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

if (in[i] > 1) return false;

if (in[i] == 0) ++zero;

}

// # of nodes without parent must be 1.

return zero == 1;

}

};

花花酱 LeetCode 1360. Number of Days Between Two Dates

By zxi on February 23, 2020

Write a program to count the number of days between two dates.

The two dates are given as strings, their format is YYYY-MM-DD as shown in the examples.

Example 1:

Input: date1 = "2019-06-29", date2 = "2019-06-30"
Output: 1

Example 2:

Input: date1 = "2020-01-15", date2 = "2019-12-31"
Output: 15

Constraints:

The given dates are valid dates between the years 1971 and 2100.

Solution: Convert to days since epoch

Time complexity: O(1)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int daysBetweenDates(string date1, string date2) {

array<int, 12> m{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

auto isLeap = [](int year) {

return year % 4 == 0 && (year % 100 != 0 || year % 400 == 0);

};

auto daysFromEpoch = [&](const string& date) {

int year = stoi(date.substr(0, 4));

int month = stoi(date.substr(5, 2));

int days = stoi(date.substr(8, 2));

for (int i = 1970; i < year; ++i)

days += 365 + isLeap(i);

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

days += m[i - 1];

days += month > 2 && isLeap(year);

return days;

};

return abs(daysFromEpoch(date1) - daysFromEpoch(date2));

}

};

花花酱 LeetCode 1359. Count All Valid Pickup and Delivery Options

By zxi on February 23, 2020

Given n orders, each order consist in pickup and delivery services.

Count all valid pickup/delivery possible sequences such that delivery(i) is always after of pickup(i).

Since the answer may be too large, return it modulo 10^9 + 7.

Example 1:

Input: n = 1
Output: 1
Explanation: Unique order (P1, D1), Delivery 1 always is after of Pickup 1.

Example 2:

Input: n = 2
Output: 6
Explanation: All possible orders: 
(P1,P2,D1,D2), (P1,P2,D2,D1), (P1,D1,P2,D2), (P2,P1,D1,D2), (P2,P1,D2,D1) and (P2,D2,P1,D1).
This is an invalid order (P1,D2,P2,D1) because Pickup 2 is after of Delivery 2.

Example 3:

Input: n = 3
Output: 90

Constraints:

1 <= n <= 500

Solution: Combination

Let dp[i] denote the number of valid sequence of i nodes.

For i-1 nodes, the sequence length is 2(i-1).
For the i-th nodes,
If we put Pi at index = 0, then we can put Di at 1, 2, …, 2i – 2 => 2i-1 options.
If we put Pi at index = 1, then we can put Di at 2,3,…, 2i – 2 => 2i – 2 options.
…
If we put Pi at index = 2i-1, then we can put Di at 2i – 1=> 1 option.
There are total (2i – 1 + 1) / 2 * (2i – 1) = i * (2*i – 1) options

dp[i] = dp[i – 1] * i * (2*i – 1)

dp[i] = 2n! / 2^n

C++

// Author: Huahua

class Solution {

public:

int countOrders(int n) {

constexpr int kMod = 1e9 + 7;

long ans = 1;

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

ans = ans * i * (2 * i - 1) % kMod;

return ans;

}

};

花花酱 1358. Number of Substrings Containing All Three Characters

By zxi on February 22, 2020

Given a string s consisting only of characters a, b and c.

Return the number of substrings containing at least one occurrence of all these characters a, b and c.

Example 1:

Input: s = "abcabc"
Output: 10
Explanation: The substrings containing at least one occurrence of the characters a, b and c are "abc", "abca", "abcab", "abcabc", "bca", "bcab", "bcabc", "cab", "cabc" and "abc" (again).

Example 2:

Input: s = "aaacb"
Output: 3
Explanation: The substrings containing at least one occurrence of the characters a, b and c are "aaacb", "aacb" and "acb".

Example 3:

Input: s = "abc"
Output: 1

Constraints:

3 <= s.length <= 5 x 10^4
s only consists of a, b or c characters.

Solution

Record the last index of each character.

At each index i, we can choose any index j that j <= min(last_a, last_b, last_c) as the starting point, and there will be min(last_a, last_b, last_c) + 1 valid substrings.

e.g. aabbabcc…
last_a = 4
last_b = 5
last_c = 7
min(last_a, last_b, last_c) = 4
aabba | bcc
We can choose any char with index <= 4 as string point, there are 5 of them:
aabbabcc
abbabcc
bbabcc
babcc
abcc

Time complexity: O(n)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int numberOfSubstrings(string s) {

vector<int> l{-1,-1,-1};

int ans = 0;

for (size_t i = 0; i < s.length(); ++i) {

l[s[i] - 'a'] = i;

ans += 1 + *min_element(begin(l), end(l));

}

return ans;

}

};

Python3

# Author: Huahua

class Solution:

def numberOfSubstrings(self, s: str) -> int:

l = {'a':-1, 'b':-1, 'c':-1}

ans = 0

for i, ch in enumerate(s):

l[ch] = i

ans += min(l.values()) + 1

return ans

花花酱 LeetCode 1357. Apply Discount Every n Orders

By zxi on February 22, 2020

There is a sale in a supermarket, there will be a discount every n customer.
There are some products in the supermarket where the id of the i-th product is products[i] and the price per unit of this product is prices[i].
The system will count the number of customers and when the n-th customer arrive he/she will have a discount on the bill. (i.e if the cost is x the new cost is x - (discount * x) / 100). Then the system will start counting customers again.
The customer orders a certain amount of each product where product[i] is the id of the i-th product the customer ordered and amount[i] is the number of units the customer ordered of that product.

Implement the Cashier class:

Cashier(int n, int discount, int[] products, int[] prices) Initializes the object with n, the discount, the products and their prices.
double getBill(int[] product, int[] amount) returns the value of the bill and apply the discount if needed. Answers within 10^-5 of the actual value will be accepted as correct.

Example 1:

Input
["Cashier","getBill","getBill","getBill","getBill","getBill","getBill","getBill"]
[[3,50,[1,2,3,4,5,6,7],[100,200,300,400,300,200,100]],[[1,2],[1,2]],[[3,7],[10,10]],[[1,2,3,4,5,6,7],[1,1,1,1,1,1,1]],[[4],[10]],[[7,3],[10,10]],[[7,5,3,1,6,4,2],[10,10,10,9,9,9,7]],[[2,3,5],[5,3,2]]]
Output
[null,500.0,4000.0,800.0,4000.0,4000.0,7350.0,2500.0]
Explanation
Cashier cashier = new Cashier(3,50,[1,2,3,4,5,6,7],[100,200,300,400,300,200,100]);
cashier.getBill([1,2],[1,2]);                        // return 500.0, bill = 1 * 100 + 2 * 200 = 500.
cashier.getBill([3,7],[10,10]);                      // return 4000.0
cashier.getBill([1,2,3,4,5,6,7],[1,1,1,1,1,1,1]);    // return 800.0, The bill was 1600.0 but as this is the third customer, he has a discount of 50% which means his bill is only 1600 - 1600 * (50 / 100) = 800.
cashier.getBill([4],[10]);                           // return 4000.0
cashier.getBill([7,3],[10,10]);                      // return 4000.0
cashier.getBill([7,5,3,1,6,4,2],[10,10,10,9,9,9,7]); // return 7350.0, Bill was 14700.0 but as the system counted three more customers, he will have a 50% discount and the bill becomes 7350.0
cashier.getBill([2,3,5],[5,3,2]);                    // return 2500.0

Constraints:

1 <= n <= 10^4
0 <= discount <= 100
1 <= products.length <= 200
1 <= products[i] <= 200
There are not repeated elements in the array products.
prices.length == products.length
1 <= prices[i] <= 1000
1 <= product.length <= products.length
product[i] exists in products.
amount.length == product.length
1 <= amount[i] <= 1000
At most 1000 calls will be made to getBill.
Answers within 10^-5 of the actual value will be accepted as correct.

Solution: Simulation

Time complexity: O(|Q|)
Space complexity: O(|P|)

C++

// Author: Huahua

class Cashier {

public:

Cashier(int n, int discount,

vector<int>& products,

vector<int>& prices): n_(n), c_(0), discount_(discount) {

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

prices_[products[i]] = prices[i];

}

double getBill(vector<int> product, vector<int> amount) {

++c_;

double bill = 0.0;

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

bill += prices_[product[i]] * amount[i];

if (c_ % n_ == 0)

bill *= 1.0 - discount_ / 100.0;

return bill;

}

private:

int n_;

int c_;

int discount_;

array<int, 201> prices_;

};