1201.Ugly Number III

Tags: Medium Math Binary Search

Links: https://leetcode.com/problems/ugly-number-iii/

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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]

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class Solution {
public:
    int nthUglyNumber(int n, long long a, long long b, long long c) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        long long x = a * b / GCD(a, b);
        long long y = a * c / GCD(a, c);
        long long z = b * c / GCD(b, c);
        long long t = x * c / GCD(x, c);

        long long left = 0, right = INT_MAX;
        while (left < right) {
            long long mid = left + ((right - left) >> 1);
            long long cnt = mid / a + mid / b + mid / c - mid / x - mid / y - mid / z + mid / t;
            if (cnt < n) left = mid + 1;
            else right = mid;
        }

        return left;
    }

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