Math Archives - Page 14 of 23 - Huahua's Tech Road

花花酱 LeetCode 1185. Day of the Week

By zxi on September 7, 2019

Given a date, return the corresponding day of the week for that date.

The input is given as three integers representing the day, month and year respectively.

Return the answer as one of the following values {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}.

Example 1:

Input: day = 31, month = 8, year = 2019
Output: "Saturday"

Example 2:

Input: day = 18, month = 7, year = 1999
Output: "Sunday"

Example 3:

Input: day = 15, month = 8, year = 1993
Output: "Sunday"

Constraints:

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

Solution: LeapYear

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

C++

// Author: Huahua

inline bool leapYear(int y) {

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

}

class Solution {

public:

string dayOfTheWeek(int day, int month, int year) {

vector<string> names = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"};

vector<int> days = {31, 28 + leapYear(year), 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

int sum = 0;

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

sum += 365 + leapYear(i);

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

sum += days[i - 1];

sum += day;

return names[(sum + 3) % 7];

}

};

花花酱 LeetCode 1175. Prime Arrangements

By zxi on August 31, 2019

Return the number of permutations of 1 to n so that prime numbers are at prime indices (1-indexed.)

(Recall that an integer is prime if and only if it is greater than 1, and cannot be written as a product of two positive integers both smaller than it.)

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

Example 1:

Input: n = 5
Output: 12
Explanation: For example [1,2,5,4,3] is a valid permutation, but [5,2,3,4,1] is not because the prime number 5 is at index 1.

Example 2:

Input: n = 100
Output: 682289015

Constraints:

1 <= n <= 100

Solution: Permutation

Count the number of primes in range [1, n], assuming there are p primes and n – p non-primes, we can permute each group separately.
ans = p! * (n – p)!

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

C++

// Author: Huahua

class Solution {

public:

int numPrimeArrangements(int n) {

const int kMod = 1e9 + 7;

int p = 0;

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

p += isPrime(i);

long ans = 1;

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

ans = (ans * i) % kMod;

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

ans = (ans * i) % kMod;

return ans;

}

private:

bool isPrime(int x) {

if (x < 2) return false;

if (x == 2) return true;

for (int i = 2; i <= sqrt(x); ++i)

if (x % i == 0) return false;

return true;

}

};

花花酱 LeetCode 1109. Corporate Flight Bookings

By zxi on July 11, 2019

There are n flights, and they are labeled from 1 to n.

We have a list of flight bookings. The i-th booking bookings[i] = [i, j, k] means that we booked kseats from flights labeled i to j inclusive.

Return an array answer of length n, representing the number of seats booked on each flight in order of their label.

Example 1:

Input: bookings = [[1,2,10],[2,3,20],[2,5,25]], n = 5
Output: [10,55,45,25,25]

Constraints:

1 <= bookings.length <= 20000
1 <= bookings[i][0] <= bookings[i][1] <= n <= 20000
1 <= bookings[i][2] <= 10000

Solution: Marking start and end

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

C++

// Author: Huahua

class Solution {

public:

vector<int> corpFlightBookings(vector<vector<int>>& bookings, int n) {

vector<int> ans(n + 1);

for (const auto& b : bookings) {

ans[b[0] - 1] += b[2];

ans[b[1]] -= b[2];

}

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

ans[i] += ans[i - 1];

ans.pop_back();

return ans;

}

};

花花酱 LeetCode 50. Pow(x, n)

By zxi on April 24, 2019

Implement pow(x, n), which calculates x raised to the power n (xⁿ).

Example 1:

Input: 2.00000, 10
Output: 1024.00000

Example 2:

Input: 2.10000, 3
Output: 9.26100

Example 3:

Input: 2.00000, -2
Output: 0.25000
Explanation: 2^-2 = 1/2² = 1/4 = 0.25

Note:

-100.0 < x < 100.0
n is a 32-bit signed integer, within the range [−2³¹, 2³¹− 1]

Solution: Recursion

square x and cut n in half.
if n is negative, compute 1.0 / pow(x, |n|)

pow(x, n) := pow(x * x, n / 2) * (x if n % 2 else 1)
pow(x, 0) := 1

Example:
pow(x, 5) = pow(x^2, 2) * x 
          = pow(x^4, 1) * x 
          = pow(x^8, 0) * x^4 * x
          = 1 * x^4 * x = x^5

Time complexity: O(logn)
Space complexity: O(logn)

C++

// Author: Huahua

class Solution {

public:

double myPow(double x, int n) {

return n >= 0 ? myPowImp(x, n) : 1.0 / myPowImp(x, n);

}

private:

double myPowImp(double x, int n) {

if (n == 0) return 1;

return myPowImp(x * x, n / 2) * (n % 2 ? x : 1);

}

};

花花酱 LeetCode 67. Add Binary

By zxi on April 10, 2019

Given two binary strings, return their sum (also a binary string).

The input strings are both non-empty and contains only characters 1 or 0.

Example 1:

Input: a = "11", b = "1"
Output: "100"

Example 2:

Input: a = "1010", b = "1011"
Output: "10101"

Solution: Big integer

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

C++

// Author: Huahua

class Solution {

public:

string addBinary(string a, string b) {

int carry = 0;

int i = a.length() - 1;

int j = b.length() - 1;

string ans;

while (i >= 0 || j >= 0) {

int s = (i >= 0 ? a[i--] - '0' : 0) +

(j >= 0 ? b[j--] - '0' : 0) +

carry;

carry = s >> 1;

ans += '0' + (s & 1);

}

if (carry) ans += '1';

return {rbegin(ans), rend(ans)};

}

};

Posts published in “Math”

花花酱 LeetCode 1185. Day of the Week

C++

花花酱 LeetCode 1175. Prime Arrangements

Solution: Permutation

C++

花花酱 LeetCode 1109. Corporate Flight Bookings

Solution: Marking start and end

C++

花花酱 LeetCode 50. Pow(x, n)

Solution: Recursion

C++

花花酱 LeetCode 67. Add Binary

C++