878. Nth Magical Number

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. 1 <= N <= 10^9
2. 2 <= A <= 40000
3. 2 <= B <= 40000

 

Approach #1: Binary Serach + Brute Froce. [Time limit exceeded]

class Solution {
public:
    int nthMagicalNumber(int N, int A, int B) {
        long l = 1, r = N * min(A, B);
        while (l <= r) {
            long m = l + (r - l) / 2;
            int num = 0;
            for (int i = 1; i <= m; ++i) {
                if (i % A == 0 || i % B == 0)
                    num++;
            }
            if (num >= N) r = m - 1;
            else l = m + 1;
        }
        return l;
    }
};

　　

Approach #2: Binary Serach + LCM.

class Solution {
public:
    int nthMagicalNumber(int N, int A, int B) {
        int MOD = 1e9 + 7;
        int L = A / gcd(A, B) * B;
        
        long l = 0, r = (long) 1e15;
        while (l < r) {
            long m = l + (r - l) / 2;
            long num = m / A + m / B - m / L;       // not int type
            if (num < N) l = m + 1;
            else r = m;
        }
        return (int) (l % MOD);
    }
private:
    int gcd(int x, int y) {
        if (x == 0) return y;
        return gcd(y % x, x);
    }
};

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Nth Magical Number.

 

 Approach #3: Mathemacital.

class Solution {
public:
    int nthMagicalNumber(int N, int A, int B) {
        int MOD = 1e9 + 7;
        int L = A / gcd(A, B) * B;
        int M = L / A + L / B - 1;
        int q = N / M, r = N % M;

        long ans = (long) q * L % MOD;
        if (r == 0)
            return (int) ans;

        int heads[2] = {A, B};
        for (int i = 0; i < r - 1; ++i) {
            if (heads[0] <= heads[1])
                heads[0] += A;
            else
                heads[1] += B;
        }

        ans += min(heads[0], heads[1]);
        return (int) (ans % MOD);
    }

    int gcd(int x, int y) {
        if (x == 0) return y;
        return gcd(y % x, x);
    }
};