LeetCode 878: Nth Magical Number
LeetCode 878 Solution 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.
LeetCode 878 Solutions in Java, C++, Python
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);
}
}Interactive Code Editor for LeetCode 878
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