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, 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 <= 10^91 <= a * b * c <= 10^18- It’s guaranteed that the result will be in range
[1, 2 * 10^9]
Solution:
Binary search.
Key ob: N/a is the number of numbers divisible by a and <= N.
Number of ugly <= n is
n/a+n/b+n/c-n/(lcm(a,b))-n/(lcm(b,c))-n/(lcm(a,c))+n/(lcm(a,b,c))
Code:
class Solution {
public:
long long gcd(long long a, long long b){
if(a < b) swap(a,b);
if(a % b == 0) return b;
else return gcd(b, a % b);
}
long long lcm(long long a, long long b){
return a*b/gcd(a,b);
}
long long check(long long m, long long n, long long a, long long b, long long c){
long long ab = lcm(a,b);
long long bc = lcm(b,c);
long long ac = lcm(a,c);
long long abc = lcm(lcm(a,b),c);
if(m/a+m/b+m/c-m/ab-m/ac-m/bc+m/abc >= n){
return true;
}
return false;
}
int nthUglyNumber(int n, int a, int b, int c) {
long long l = 1;
long long r = 2e9+1;
while(l <= r){
long long m = (l+r)/2;
if(check(m,n,a,b,c)){
r = m-1;
}else{
l = m+1;
}
}
return (int)l;
}
};