FizzBuzz

FizzBuzz

Task: Run 1 through 100, then:

1. For multiples of 3, print "Fizz"
2. For multiples of 5, print "Buzz"
3. For multiples of both “3” and “5”, print "FizzBuzz".

Otherwise, print the number.

To get this done, we need a for Loop and if else, for a simple solution for this. The Modulus function - % will be quite handy also.

# Run 1 through 100
x <- c(1:100)

for (i in x){
  # For multiples of both "3" and "5"
  if ((i %% 3 == 0) & (i %% 5 == 0)){
    print("FizzBuzz")
  }
  # For multiples of 3
  else if(i %% 3 == 0){
    print("Fizz")
    }
  # For multiples of 5
  else if(i %% 5 == 0){
    print("Buzz")
  }
  # This outputs same number when the conditions are not applicable
  else print(i)
}

## [1] 1
## [1] 2
## [1] "Fizz"
## [1] 4
## [1] "Buzz"
## [1] "Fizz"
## [1] 7
## [1] 8
## [1] "Fizz"
## [1] "Buzz"
## [1] 11
## [1] "Fizz"
## [1] 13
## [1] 14
## [1] "FizzBuzz"
## [1] 16
## [1] 17
## [1] "Fizz"
## [1] 19
## [1] "Buzz"
## [1] "Fizz"
## [1] 22
## [1] 23
## [1] "Fizz"
## [1] "Buzz"
## [1] 26
## [1] "Fizz"
## [1] 28
## [1] 29
## [1] "FizzBuzz"
## [1] 31
## [1] 32
## [1] "Fizz"
## [1] 34
## [1] "Buzz"
## [1] "Fizz"
## [1] 37
## [1] 38
## [1] "Fizz"
## [1] "Buzz"
## [1] 41
## [1] "Fizz"
## [1] 43
## [1] 44
## [1] "FizzBuzz"
## [1] 46
## [1] 47
## [1] "Fizz"
## [1] 49
## [1] "Buzz"
## [1] "Fizz"
## [1] 52
## [1] 53
## [1] "Fizz"
## [1] "Buzz"
## [1] 56
## [1] "Fizz"
## [1] 58
## [1] 59
## [1] "FizzBuzz"
## [1] 61
## [1] 62
## [1] "Fizz"
## [1] 64
## [1] "Buzz"
## [1] "Fizz"
## [1] 67
## [1] 68
## [1] "Fizz"
## [1] "Buzz"
## [1] 71
## [1] "Fizz"
## [1] 73
## [1] 74
## [1] "FizzBuzz"
## [1] 76
## [1] 77
## [1] "Fizz"
## [1] 79
## [1] "Buzz"
## [1] "Fizz"
## [1] 82
## [1] 83
## [1] "Fizz"
## [1] "Buzz"
## [1] 86
## [1] "Fizz"
## [1] 88
## [1] 89
## [1] "FizzBuzz"
## [1] 91
## [1] 92
## [1] "Fizz"
## [1] 94
## [1] "Buzz"
## [1] "Fizz"
## [1] 97
## [1] 98
## [1] "Fizz"
## [1] "Buzz"

This code can still be done more efficiently.