見習村01 - Multiples of 3 or 5

01 - Multiples of 3 or 5

Don’t say so much, just coding…

Instruction

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Finish the solution so that it returns the sum of all the multiples of 3 or 5 below the number passed in.

Note: If the number is a multiple of both 3 and 5, only count it once. Also, if a number is negative, return 0(for languages that do have them)

Ruby

Init

1
2
3
def solution(number)
  # put your solution here
end

Sample Testing

1
2
3
4
5
6
7
8
9
def test(actual, expected)
  Test.assert_equals(actual, expected)
end

Test.describe("example tests") do
  test(solution(10), 23)
  test(solution(20), 78)
  test(solution(200), 9168)
end

Javascript

Init

1
2
3
function solution(number){
  // put your solution here
}

Sample Testing

1
2
3
4
5
6
7
8
function test(n, expected) {
  let actual = solution(n)
  Test.assertEquals(actual, expected, `Expected ${expected}, got ${actual}`)
}

Test.describe("basic tests", function(){
  test(10,23)
})

Thinking

想法(1): 從題目概述 natural numbers below 10 中開始模擬會出現的數字
想法(2): 計算兩者的加總，可以知道在算 3 or 5 的倍數
想法(3): 處理級距，需要從 1 ~ 參數 然後再來選取符合項目，並加總

https://ithelp.ithome.com.tw/upload/images/20200916/20120826ve4OpCsRKr.jpg
圖片來源：Unsplash Jesus Kiteque

Hint & Reference

Solution

Ruby

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
# Solution 1
def solution(number)
  [*1..number-1].select{ |n| n % 3 == 0 || n % 5 == 0}.reduce(&:+)
end

# Solution 2
def solution(number)
  [*1...number].select{ |n| n % 3 == 0 || n % 5 == 0}.reduce(&:+)
end

# Solution 3
def solution(number)
  (1...number).select{ |n| n % 3 == 0 || n % 5 == 0}.inject(&:+)
end

# Solution 4
def solution(number)
  (1...number).select{ |n| (n % 5).zero? || (n % 3).zero?}.inject(:+)
end

Javascript

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
// Solution 1
function solution(number){
  let sum = 0;

  for(i = 1; i < number; i++){
      if( i % 3 == 0 || i % 5 == 0){
        sum += i;
      }
  }
  return sum;
}

// Solution 2
function solution(number){
  return number < 3 ? 0
    : [...Array(number).keys()]
      .map(int => (int % 3 === 0 || int % 5 === 0) ? int : 0 )
      .reduce((a, b) => a + b )
}