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();

}

};

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);

}

};

Posts tagged as “gcd”

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Problem

Solution: Math

C++

C++/class

花花酱 LeetCode 878. Nth Magical Number

Problem

Solution: Math + Binary Search