LeetCode 878: Nth Magical Number Solution
Master LeetCode problem 878 (Nth Magical Number), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
878. Nth Magical Number
Problem Explanation
Explanation
To solve this problem, we can use the concept of binary search. We can find the least common multiple (LCM) of a and b and then use binary search to find the nth magical number within a range of [1, n * min(a, b)]. We can calculate the LCM using the formula LCM(a, b) = a * b / GCD(a, b) where GCD is the greatest common divisor. Once we have the LCM, we can find how many magical numbers are there in each interval of length LCM. Using binary search, we can efficiently find the nth magical number.
Solution Code
class Solution {
public int nthMagicalNumber(int n, int a, int b) {
long mod = 1000000007;
long lcm = a / gcd(a, b) * b;
long low = 1, high = (long)1e18;
while (low < high) {
long mid = low + (high - low) / 2;
long count = mid / a + mid / b - mid / lcm;
if (count < n) {
low = mid + 1;
} else {
high = mid;
}
}
return (int)(low % mod);
}
private long gcd(long a, long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
}Try It Yourself
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