Posts published in “Math”

花花酱 LeetCode 592. Fraction Addition and Subtraction

By zxi on August 6, 2018

Problem

Given a string representing an expression of fraction addition and subtraction, you need to return the calculation result in string format. The final result should be irreducible fraction. If your final result is an integer, say 2, you need to change it to the format of fraction that has denominator 1. So in this case, 2 should be converted to 2/1.

Example 1:

Input:"-1/2+1/2"
Output: "0/1"

Example 2:

Input:"-1/2+1/2+1/3"
Output: "1/3"

Example 3:

Input:"1/3-1/2"
Output: "-1/6"

Example 4:

Input:"5/3+1/3"
Output: "2/1"

Note:

The input string only contains '0' to '9', '/', '+' and '-'. So does the output.
Each fraction (input and output) has format ±numerator/denominator. If the first input fraction or the output is positive, then '+' will be omitted.
The input only contains valid irreducible fractions, where the numerator and denominator of each fraction will always be in the range [1,10]. If the denominator is 1, it means this fraction is actually an integer in a fraction format defined above.
The number of given fractions will be in the range [1,10].
The numerator and denominator of the final result are guaranteed to be valid and in the range of 32-bit int.

Solution: Math

a/b+c/d = (a*d + b * c) / (b * d)

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

string fractionAddition(string expression) {

int a = 0;

int b = 1;

int c;

int d;

char slash;

istringstream in(expression);

while (in >> c >> slash >> d) {

a = a * d + b * c;

b *= d;

int gcd = abs(__gcd(a, b));

a /= gcd;

b /= gcd;

}

return to_string(a) + "/" + to_string(b);

}

};

C++/class

// Author: Huahua

// Running time: 0 ms

class Fraction {

public:

Fraction(int a, int b): a_(a), b_(b) {}

Fraction& operator+=(const Fraction& f) {

a_ = a_ * f.b_ + b_ * f.a_;

b_ = b_ * f.b_;

int d = abs(__gcd(a_, b_));

a_ /= d;

b_ /= d;

return *this;

}

string toString() {

return to_string(a_) + "/" + to_string(b_);

}

private:

int a_; // hold the sign, e.g. can be +-.

int b_; // always > 0.

};

class Solution {

public:

string fractionAddition(string expression) {

Fraction f(0, 1);

int a;

int b;

char op;

istringstream in(expression);

while (in >> a >> op >> b)

f += Fraction(a, b);

return f.toString();

}

};

花花酱 LeetCode 878. Nth Magical Number

By zxi on July 28, 2018

Problem

A positive integer is magical if it is divisible by either A or B.

Return the N-th magical number. Since the answer may be very large, return it modulo 10^9 + 7.

Example 1:

Input: N = 1, A = 2, B = 3
Output: 2

Example 2:

Input: N = 4, A = 2, B = 3
Output: 6

Example 3:

Input: N = 5, A = 2, B = 4
Output: 10

Example 4:

Input: N = 3, A = 6, B = 4
Output: 8

Note:

1 <= N <= 10^9
2 <= A <= 40000
2 <= B <= 40000

Solution: Math + Binary Search

Let n denote the number of numbers <= X that are divisible by either A or B.

n = X / A + X / B – X / lcm(A, B) = X / A + X / B – X / (A * B / gcd(A, B))

Binary search for the smallest X such that n >= N

Time complexity: O(log(1e9*4e5)

Space complexity: O(1)

// Author: Huahua

// Running time: 0 ms

class Solution {

public:

int nthMagicalNumber(int N, int A, int B) {

const int kMod = 1000000007;

long l = 2;

long r = static_cast<long>(1e9 * 4e5);

int d = gcd(A, B);

while (l < r) {

long m = (r - l) / 2 + l;

if (m / A + m / B - m / (A * B / d) < N)

l = m + 1;

else

r = m;

}

return l % kMod;

}

private:

int gcd(int a, int b) {

if (a % b == 0) return b;

return gcd(b, a % b);

}

};

花花酱 LeetCode 880. Random Pick with Weight

By zxi on July 28, 2018

Problem

Given an array w of positive integers, where w[i] describes the weight of index i, write a function pickIndex which randomly picks an index in proportion to its weight.

Note:

1 <= w.length <= 10000
1 <= w[i] <= 10^5
pickIndex will be called at most 10000 times.

Example 1:

Input: 
["Solution","pickIndex"]
[[[1]],[]]
Output: [null,0]

Example 2:

Input: 
["Solution","pickIndex","pickIndex","pickIndex","pickIndex","pickIndex"]
[[[1,3]],[],[],[],[],[]]
Output: [null,0,1,1,1,0]

Explanation of Input Syntax:

The input is two lists: the subroutines called and their arguments. Solution‘s constructor has one argument, the array w. pickIndex has no arguments. Arguments are always wrapped with a list, even if there aren’t any.

Solution: Binary Search

Convert PDF to CDF
Uniformly sample a value s in [1, sum(weights)].
Use binary search to find first index such that PDF[index] >= s.

Time complexity: Init O(n), query O(logn)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 184 ms

class Solution {

public:

Solution(vector<int> w) {

sums_ = std::move(w);

for (int i = 1; i < sums_.size(); ++i)

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

}

int pickIndex() {

int s = rand() / static_cast<double>(RAND_MAX) * sums_.back() + 1;

return lower_bound(sums_.begin(), sums_.end(), s) - sums_.begin();

// or

// int l = 0;

// int r = sums_.size();

// while (l < r) {

// int m = (r - l) / 2 + l;

// if (sums_[m] < s)

// l = m + 1;

// else

// r = m;

// }

// return l;

}

private:

vector<int> sums_;

};

Problem

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string.

Example 1:

Input: num1 = "2", num2 = "3"
Output: "6"

Example 2:

Input: num1 = "123", num2 = "456"
Output: "56088"

Note:

The length of both num1 and num2 is < 110.
Both num1 and num2 contain only digits 0-9.
Both num1 and num2 do not contain any leading zero, except the number 0 itself.
You must not use any built-in BigInteger library or convert the inputs to integer directly.

Solution: Simulation

Simulate multiplication one digit at a time.

Time complexity: O(l1*l2)

Space complexity: O(l1 + l2)

C++

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

string multiply(string num1, string num2) {

const int l1 = num1.length();

const int l2 = num2.length();

string ans(l1 + l2, '0');

for (int i = l1 - 1; i >= 0; --i)

for (int j = l2 - 1; j >= 0; --j) {

int sum = (ans[i + j + 1] - '0') + (num1[i] - '0') * (num2[j] - '0');

ans[i + j + 1] = (sum % 10) + '0';

ans[i + j] += sum / 10;

}

for (int i = 0; i < ans.length(); ++i)

if (ans[i] != '0' || i == ans.length() - 1) return ans.substr(i);

return "";

}

};

花花酱 LeetCode 357. Count Numbers with Unique Digits

By zxi on July 16, 2018

Problem

Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10ⁿ.

Example:
Given n = 2, return 91. (The answer should be the total numbers in the range of 0 ≤ x < 100, excluding [11,22,33,44,55,66,77,88,99])

Solution: Math

f(0) = 1 (0)

f(1) = 10 (0 – 9)

f(2) = 9 * 9 (1-9 * (0 ~ 9 exclude the one from first digit))

f(3) = 9 * 9 * 8

f(4) = 9 * 9 * 8 * 7

…

f(x) = 0 if x >= 10

ans = sum(f[1] ~ f[n])

Time complexity: O(1)

Space complexity: O(1)

// Author: Huahua

// Running time: 0 ms

class Solution {

public:

int countNumbersWithUniqueDigits(int n) {

if (n == 0) return 1;

vector<int> f(11);

f[1] = 10;

f[2] = 9 * 9;

for (int i = 3; i <= 10; ++i)

f[i] = f[i - 1] * (10 - i + 1);

int ans = 0;

for (int i = 0; i <= min(10, n); ++i)

ans += f[i];

return ans;

}

};

Posts published in “Math”

花花酱 LeetCode 592. Fraction Addition and Subtraction

Problem

Solution: Math

C++

C++/class

花花酱 LeetCode 878. Nth Magical Number

Problem

Solution: Math + Binary Search

花花酱 LeetCode 880. Random Pick with Weight

Problem

Solution: Binary Search

Related Problems

花花酱 LeetCode 43. Multiply Strings

Problem

Solution: Simulation

花花酱 LeetCode 357. Count Numbers with Unique Digits

Problem

Solution: Math