Ugly Number II LeetCode Solution

Problem – Ugly Number II LeetCode Solution

An ugly number is a positive integer whose prime factors are limited to 23, and 5.

Given an integer n, return the nth ugly number.

Example 1:

Input: n = 10
Output: 12
Explanation: [1, 2, 3, 4, 5, 6, 8, 9, 10, 12] is the sequence of the first 10 ugly numbers.

Example 2:

Input: n = 1
Output: 1
Explanation: 1 has no prime factors, therefore all of its prime factors are limited to 2, 3, and 5.

Constraints:

  • 1 <= n <= 1690

Ugly Number II LeetCode Solution in C++

class Solution {
public:
    int nthUglyNumber(int n) {
        if(n <= 0) return false; // get rid of corner cases 
        if(n == 1) return true; // base case
        int t2 = 0, t3 = 0, t5 = 0; //pointers for 2, 3, 5
        vector<int> k(n);
        k[0] = 1;
        for(int i  = 1; i < n ; i ++)
        {
            k[i] = min(k[t2]*2,min(k[t3]*3,k[t5]*5));
            if(k[i] == k[t2]*2) t2++; 
            if(k[i] == k[t3]*3) t3++;
            if(k[i] == k[t5]*5) t5++;
        }
        return k[n-1];
    }
};

Ugly Number II LeetCode Solution in Java

public class Solution {
    public int nthUglyNumber(int n) {
        int[] ugly = new int[n];
        ugly[0] = 1;
        int index2 = 0, index3 = 0, index5 = 0;
        int factor2 = 2, factor3 = 3, factor5 = 5;
        for(int i=1;i<n;i++){
            int min = Math.min(Math.min(factor2,factor3),factor5);
            ugly[i] = min;
            if(factor2 == min)
                factor2 = 2*ugly[++index2];
            if(factor3 == min)
                factor3 = 3*ugly[++index3];
            if(factor5 == min)
                factor5 = 5*ugly[++index5];
        }
        return ugly[n-1];
    }
}

Ugly Number II LeetCode Solution in Python

def nthUglyNumber(self, n):
    ugly = [1]
    i2, i3, i5 = 0, 0, 0
    while n > 1:
        u2, u3, u5 = 2 * ugly[i2], 3 * ugly[i3], 5 * ugly[i5]
        umin = min((u2, u3, u5))
        if umin == u2:
            i2 += 1
        if umin == u3:
            i3 += 1
        if umin == u5:
            i5 += 1
        ugly.append(umin)
        n -= 1
    return ugly[-1]
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