1201. Ugly Number III
An ugly number is a positive integer that is divisible by a, b, or c.
Given four integers n, a, b, and c, return the n t h n^{th} nth ugly number.
Example 1:
Input: n = 3, a = 2, b = 3, c = 5
Output: 4
Explanation: The ugly numbers are 2, 3, 4, 5, 6, 8, 9, 10... The 3 r d 3^{rd} 3rd is 4.
Example 2:
Input: n = 4, a = 2, b = 3, c = 4
Output: 6
Explanation: The ugly numbers are 2, 3, 4, 6, 8, 9, 10, 12... The 4 t h 4^{th} 4th is 6.
Example 3:
Input: n = 5, a = 2, b = 11, c = 13
Output: 10
Explanation: The ugly numbers are 2, 4, 6, 8, 10, 11, 12, 13... The 5 t h 5^{th} 5th is 10.
Constraints:
- 1 < = n , a , b , c < = 10 9 1 <= n, a, b, c <= 10^9 1<=n,a,b,c<=109
- 1 < = a ∗ b ∗ c < = 10 18 1 <= a * b * c <= 10^{18} 1<=a∗b∗c<=1018
- It is guaranteed that the result will be in range 1 , 2 ∗ 10 9 1, 2 \* 10\^9 1,2∗109.
From: LeetCode
Link: 1201. Ugly Number III
Solution:
Ideas:
binary search the answer.
For a number x, count how many numbers <= x are divisible by a, b, or c using inclusion-exclusion.
Code:
c
long long gcd(long long x, long long y) {
while (y) {
long long t = x % y;
x = y;
y = t;
}
return x;
}
long long lcm(long long x, long long y) {
return x / gcd(x, y) * y;
}
int nthUglyNumber(int n, int a, int b, int c) {
long long A = a, B = b, C = c;
long long ab = lcm(A, B);
long long ac = lcm(A, C);
long long bc = lcm(B, C);
long long abc = lcm(ab, C);
long long left = 1, right = 2000000000LL;
while (left < right) {
long long mid = left + (right - left) / 2;
long long count = mid / A + mid / B + mid / C
- mid / ab - mid / ac - mid / bc
+ mid / abc;
if (count >= n) {
right = mid;
} else {
left = mid + 1;
}
}
return (int)left;
}