LeetCode 1201: Ugly Number III Solution

Master LeetCode problem 1201 (Ugly Number III), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1201. Ugly Number III

Medium

Math Binary Search Combinatorics Number Theory

Problem Explanation

Explanation:

To solve this problem, we can use the concept of least common multiple (LCM) and binary search. We need to find the nth ugly number that is divisible by a, b, or c. We can take advantage of the fact that the LCM of a, b, and c will give us the period at which the next ugly number is generated.

We first find the LCM of a, b, and c.
Then, using binary search, we find the nth ugly number within the range of 1 to 2 * 10^9.
In the binary search process, we calculate the number of ugly numbers that are less than or equal to the mid value.
Based on this count, we adjust the search space to find the nth ugly number. :

Solution Code

class Solution {
    public int nthUglyNumber(int n, int a, int b, int c) {
        long lcmAB = lcm(a, b);
        long lcmBC = lcm(b, c);
        long lcmAC = lcm(a, c);
        long lcmABC = lcm(lcmAB, c);
        
        long left = 1, right = Long.MAX_VALUE;
        
        while (left < right) {
            long mid = left + (right - left) / 2;
            long count = mid / a + mid / b + mid / c - mid / lcmAB - mid / lcmBC - mid / lcmAC + mid / lcmABC;
            
            if (count < n) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        
        return (int)left;
    }
    
    private long lcm(long a, long b) {
        return a * b / gcd(a, b);
    }
    
    private long gcd(long a, long b) {
        while (b != 0) {
            long temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }
}

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