Solution Review: while Loops

In this review, we provide a detailed analysis of the solution to this problem.

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Solution: Using while Loop

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testVariable <- 1
while (testVariable <= 20) { # Ccheck whether testVariable is less than or equal to 20
  if(testVariable %% 3 == 0 && testVariable %% 5 == 0)
  {
    cat(testVariable, "foo bar ")
  } else 
  {
    if(testVariable %% 3 == 0)
    {
      cat(testVariable, "foo ")
    }
    if(testVariable %% 5 == 0)
    {
      cat(testVariable, "bar ")
    }
  }
  testVariable = testVariable + 1 # increment testVariable to move on to the next number to check
}

Explanation

One simple approach is to copy paste the condition block (line number 4 to 17 in the above code) 2020 times. However, this method is redundant and lengthy.

We can write a while loop that runs 2020 times and for each iteration checks the condition. For this, we make a variable testVariable whose value is 11. Then, in each iteration, we check whether testVariable is less than or equal to 2020, and also increment at the end of each iteration.