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 $O(1)$.

Python Code
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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

Java Code
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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);
    }
}

C++ Code
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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);
    }
};

Go Code
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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)
}

TypeScript Code
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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);
}