1201. Ugly Number III #

Problem #

Write a program to find the n-th ugly number.

Ugly numbers are positive integers which are divisible by a or b or c.

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 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 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 5th is 10.

Example 4:

Input: n = 1000000000, a = 2, b = 217983653, c = 336916467
Output: 1999999984

Constraints:

• 1 <= n, a, b, c <= 10^9
• 1 <= a * b * c <= 10^18
• It’s guaranteed that the result will be in range [1, 2 * 10^9]

Problem Summary #

Please help design a program to find the nth ugly number. Ugly numbers are positive integers that can be divided by a or b or c.

Notes:

• 1 <= n, a, b, c <= 10^9
• 1 <= a * b * c <= 10^18
• The result of this problem is within the range [1, 2 * 10^9]

Solution Approach #

• Given 4 numbers, a, b, c, and n. The task is to output the nth number that is divisible by a, or divisible by b, or divisible by c.
• This problem limits the range of the answer to [1, 2 * 10^9], so we can directly use binary search to solve it. Gradually narrow down to the low value until finding the minimum low that satisfies the condition, which is the final answer.
• The key to this problem is how to determine the rank of a number. If a number is divisible by a, divisible by b, and divisible by c, then it should be the num/a + num/b + num/c - num/ab - num/bc - num/ac + num/abcth number. This is the Venn diagram principle. Note that when calculating ab, bc, ac, and abc, you also need to divide by their respective greatest common divisor gcd().

Code #

package leetcode

func nthUglyNumber(n int, a int, b int, c int) int {
	low, high := int64(0), int64(2*1e9)
	for low < high {
		mid := low + (high-low)>>1
		if calNthCount(mid, int64(a), int64(b), int64(c)) < int64(n) {
			low = mid + 1
		} else {
			high = mid
		}
	}
	return int(low)
}

func calNthCount(num, a, b, c int64) int64 {
	ab, bc, ac := a*b/gcd(a, b), b*c/gcd(b, c), a*c/gcd(a, c)
	abc := a * bc / gcd(a, bc)
	return num/a + num/b + num/c - num/ab - num/bc - num/ac + num/abc
}

func gcd(a, b int64) int64 {
	for b != 0 {
		a, b = b, a%b
	}
	return a
}