# 1201. Ugly Number III

## Description

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 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 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.


Constraints:

• 1 <= n, a, b, c <= 109
• 1 <= a * b * c <= 1018
• It is guaranteed that the result will be in range [1, 2 * 109].

## Solutions

Solution 1: Binary Search + Inclusion-Exclusion Principle

We can transform the problem into: find the smallest positive integer $x$ such that the number of ugly numbers less than or equal to $x$ is exactly $n$.

For a positive integer $x$, there are $\left\lfloor \frac{x}{a} \right\rfloor$ numbers divisible by $a$, $\left\lfloor \frac{x}{b} \right\rfloor$ numbers divisible by $b$, $\left\lfloor \frac{x}{c} \right\rfloor$ numbers divisible by $c$, $\left\lfloor \frac{x}{lcm(a, b)} \right\rfloor$ numbers divisible by both $a$ and $b$, $\left\lfloor \frac{x}{lcm(a, c)} \right\rfloor$ numbers divisible by both $a$ and $c$, $\left\lfloor \frac{x}{lcm(b, c)} \right\rfloor$ numbers divisible by both $b$ and $c$, and $\left\lfloor \frac{x}{lcm(a, b, c)} \right\rfloor$ numbers divisible by $a$, $b$, and $c$ at the same time. According to the inclusion-exclusion principle, the number of ugly numbers less than or equal to $x$ is:

$\left\lfloor \frac{x}{a} \right\rfloor + \left\lfloor \frac{x}{b} \right\rfloor + \left\lfloor \frac{x}{c} \right\rfloor - \left\lfloor \frac{x}{lcm(a, b)} \right\rfloor - \left\lfloor \frac{x}{lcm(a, c)} \right\rfloor - \left\lfloor \frac{x}{lcm(b, c)} \right\rfloor + \left\lfloor \frac{x}{lcm(a, b, c)} \right\rfloor$

We can use binary search to find the smallest positive integer $x$ such that the number of ugly numbers less than or equal to $x$ is exactly $n$.

Define the left boundary of binary search as $l=1$ and the right boundary as $r=2 \times 10^9$, where $2 \times 10^9$ is the maximum value given by the problem. In each step of binary search, we find the middle number $mid$. If the number of ugly numbers less than or equal to $mid$ is greater than or equal to $n$, it means that the smallest positive integer $x$ falls in the interval $[l,mid]$, otherwise it falls in the interval $[mid+1,r]$. During the binary search process, we need to continuously update the number of ugly numbers less than or equal to $mid$ until we find the smallest positive integer $x$.

The time complexity is $O(\log m)$, where $m = 2 \times 10^9$. The space complexity is $O(1)$.

• class Solution {
public int nthUglyNumber(int n, int a, int b, int c) {
long ab = lcm(a, b);
long bc = lcm(b, c);
long ac = lcm(a, c);
long abc = lcm(ab, c);
long l = 1, r = 2000000000;
while (l < r) {
long mid = (l + r) >> 1;
if (mid / a + mid / b + mid / c - mid / ab - mid / bc - mid / ac + mid / abc >= n) {
r = mid;
} else {
l = mid + 1;
}
}
return (int) l;
}

private long gcd(long a, long b) {
return b == 0 ? a : gcd(b, a % b);
}

private long lcm(long a, long b) {
return a * b / gcd(a, b);
}
}

• class Solution {
public:
int nthUglyNumber(int n, int a, int b, int c) {
long long ab = lcm(a, b);
long long bc = lcm(b, c);
long long ac = lcm(a, c);
long long abc = lcm(ab, c);
long long l = 1, r = 2000000000;
while (l < r) {
long long mid = (l + r) >> 1;
if (mid / a + mid / b + mid / c - mid / ab - mid / bc - mid / ac + mid / abc >= n) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}

long long lcm(long long a, long long b) {
return a * b / gcd(a, b);
}

long long gcd(long long a, long long b) {
return b == 0 ? a : gcd(b, a % b);
}
};

• class Solution:
def nthUglyNumber(self, n: int, a: int, b: int, c: int) -> int:
ab = lcm(a, b)
bc = lcm(b, c)
ac = lcm(a, c)
abc = lcm(a, b, c)
l, r = 1, 2 * 10**9
while l < r:
mid = (l + r) >> 1
if (
mid // a
+ mid // b
+ mid // c
- mid // ab
- mid // bc
- mid // ac
+ mid // abc
>= n
):
r = mid
else:
l = mid + 1
return l


• func nthUglyNumber(n int, a int, b int, c int) int {
ab, bc, ac := lcm(a, b), lcm(b, c), lcm(a, c)
abc := lcm(ab, c)
var l, r int = 1, 2e9
for l < r {
mid := (l + r) >> 1
if mid/a+mid/b+mid/c-mid/ab-mid/bc-mid/ac+mid/abc >= n {
r = mid
} else {
l = mid + 1
}
}
return l
}

func gcd(a, b int) int {
if b == 0 {
return a
}
return gcd(b, a%b)
}

func lcm(a, b int) int {
return a * b / gcd(a, b)
}

• function nthUglyNumber(n: number, a: number, b: number, c: number): number {
const ab = lcm(BigInt(a), BigInt(b));
const bc = lcm(BigInt(b), BigInt(c));
const ac = lcm(BigInt(a), BigInt(c));
const abc = lcm(BigInt(a), bc);
let l = 1n;
let r = BigInt(2e9);
while (l < r) {
const mid = (l + r) >> 1n;
const count =
mid / BigInt(a) +
mid / BigInt(b) +
mid / BigInt(c) -
mid / ab -
mid / bc -
mid / ac +
mid / abc;
if (count >= BigInt(n)) {
r = mid;
} else {
l = mid + 1n;
}
}
return Number(l);
}

function gcd(a: bigint, b: bigint): bigint {
return b === 0n ? a : gcd(b, a % b);
}

function lcm(a: bigint, b: bigint): bigint {
return (a * b) / gcd(a, b);
}