Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/2413.html

2413. Smallest Even Multiple

  • Difficulty: Easy.
  • Related Topics: .
  • Similar Questions: Greatest Common Divisor of Strings, Three Divisors, Find Greatest Common Divisor of Array.

Problem

Given a positive integer n, return the smallest positive integer that is a multiple of **both 2 and n.   **Example 1:

Input: n = 5
Output: 10
Explanation: The smallest multiple of both 5 and 2 is 10.

Example 2:

Input: n = 6
Output: 6
Explanation: The smallest multiple of both 6 and 2 is 6. Note that a number is a multiple of itself.

  Constraints:

  • 1 <= n <= 150

Solution (Java, C++, Python)

  • class Solution {
        public int smallestEvenMultiple(int n) {
            int k=2*n;
            for(int i=2*n;i>=n;i--)
            {
                if(i%n==0&&i%2==0)
                    k=i;
            }
            return k;
        }
    }
    
    ############
    
    class Solution {
        public int smallestEvenMultiple(int n) {
            return n % 2 == 0 ? n : n * 2;
        }
    }
    
  • class Solution:
        def smallestEvenMultiple(self, n: int) -> int:
            return n if n % 2 == 0 else n * 2
    
    ############
    
    # 2413. Smallest Even Multiple
    # https://leetcode.com/problems/smallest-even-multiple/
    
    class Solution:
        def smallestEvenMultiple(self, n: int) -> int:
            for x in range(1, n * 2 + 1):
                if x % 2 == 0 and x % n == 0:
                    return x
            
            return -1
    
    
  • class Solution {
    public:
        int smallestEvenMultiple(int n) {
            return n % 2 == 0 ? n : n * 2;
        }
    };
    
  • func smallestEvenMultiple(n int) int {
    	if n%2 == 0 {
    		return n
    	}
    	return n * 2
    }
    
  • function smallestEvenMultiple(n: number): number {
        return n % 2 === 0 ? n : n * 2;
    }
    
    
  • impl Solution {
        pub fn smallest_even_multiple(n: i32) -> i32 {
            if n % 2 == 0 {
                return n;
            }
            n * 2
        }
    }
    
    

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

All Problems

All Solutions