Ugly Number III in C++

Suppose we have to write a program to find the n-th ugly number. The Ugly numbers are positive integers which are divisible by a or b or c. So for example, if n = 3 and a = 2, b = 3 and c = 5, then the output will be 4, as the ugly numbers are [2,3,4,5,6,8,9,10], the third one is 4.

To solve this, we will follow these steps −

• make a method called ok(), this will take x, a, b, c, this will act like below −

• return (x/a) + (x/b) + (x/c) – (x/lcm(a,b)) - (x/lcm(b, c)) - (x/lcm(b,c)) - (x/lcm(a,c)) + (x/lcm(a, lcm(b,c)))

• From the main method, do following −

• low := 1, high := 2 * (10^9)

• while low < high −

  • mid := low + (high - low) / 2

  • x := ok(mid, a, b, c)

  • if x >= n, then high := mid, otherwise low := mid + 1

• return high

Example (C++)

Let us see the following implementation to get a better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
class Solution {
public:
   lli gcd(lli a, lli b){
      return b == 0? a: gcd(b, a % b);
   }
   lli lcm(lli a, lli b){
      return a * b / gcd(a, b);
   }
   lli ok(lli x, lli a, lli b, lli c){
      return (x / a) + (x / b) + (x / c) - (x / lcm(a, b)) - (x / lcm(b, c)) - (x / lcm(a, c)) + (x / lcm(a, lcm(b, c)));
   }
   int nthUglyNumber(int n, int a, int b, int c) {
      int low = 1;
      int high = 2 * (int) 1e9;
      while(low < high){
         int mid = low + (high - low) / 2;
         int x = ok(mid, a, b, c);
         if(x>= n){
            high = mid;
         }
         else low = mid + 1;
      }
      return high;
   }
};
main(){
   Solution ob;
   cout << (ob.nthUglyNumber(3,2,3,5));
}

Input

3
2
3
5

Output

4