Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Program to find nth ugly number in C++
Suppose we have a number n; we have to find the nth ugly number. As we know that the ugly numbers are those numbers, whose prime factors are only 2, 3 and 5. So if we want to find 10th ugly number, the output will be 12, as the first few ugly numbers are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 and so on.
To solve this, we will follow these steps:
- Define an array v of size(n + 1)
- if n is same as 1, then:
- return 1
- two := 2, three := 3, five := 5
- twoIdx := 2, threeIdx := 2, fiveIdx := 2
- for initialize i := 2, when i <= n, update (increase i by 1), do:
- curr := minimum of two, three and five
- v[i] := curr
- if curr is same as two, then:
- two := v[twoIdx] * 2;
- (increase twoIdx by 1)
- if curr is same as three, then:
- three := v[threeIdx] * 3
- (increase threeIdx by 1)
- if curr is same as five, then:
- five := v[fiveIdx] * 5
- (increase fiveIdx by 1)
- return v[n]
Let us see the following implementation to get better understanding:
Example
#include
using namespace std;
class Solution {
public:
int nthUglyNumber(int n) {
vector v(n + 1);
if(n == 1){
return 1;
}
int two = 2, three = 3, five = 5;
int twoIdx = 2;
int threeIdx = 2;
int fiveIdx = 2;
for(int i = 2; i <= n; i++){
int curr = min({two, three, five});
v[i] = curr;
if(curr == two){
two = v[twoIdx] * 2;;
twoIdx++;
}
if(curr == three){
three = v[threeIdx] * 3;
threeIdx++;
}
if(curr == five){
five = v[fiveIdx] * 5;
fiveIdx++;
}
}
return v[n];
}
};
main(){
Solution ob;
cout << (ob.nthUglyNumber(15));
}Input
15
Output
24
Updated on: 26-Nov-2020
205 Views
Advertisements