Return true in an incorrect manner

Question

Return true in an incorrect manner

Coderbyte Challenge 7

Need help with a function that checks if every letter in the string is bounded by '+' signs. The code provided seems to be returning incorrect results - it should return false for the input string below as 'k' is missing a right-hand side '+'. Any assistance would be appreciated.

str = '+f+++l+k=+'

alpha = 'abcdefghijklmnopqrstuvwxyz'
function SimpleSymbols(str) {
  str = str.toLowerCase()
  for (i=0 ; i<str.length ; i++)  {

    if (alpha.indexOf(str.charAt(i)) > 0 && i === 0)  {
        return false 
    }
    else if (alpha.indexOf(str.charAt(str.length -1)) > 0) {
        return false 
    }
     else  if (alpha.indexOf(str.charAt(i)) > 0 && 
        (str.charAt(i-1) !== '+' || str.charAt(i+1) != '+')) {
         return false
    }
     else {return true}    
   }       
}

javascript algorithm

Answer 1

Answer №1

If you break the for loop at the first iteration, your function will stop prematurely. Instead, only break the loop on the branches where the condition is false. After completing the for loop without encountering any false conditions, return true.

Answer 2

If you break the for loop at the first iteration, your function will stop prematurely. Instead, only break the loop on the branches where the condition is false. After completing the for loop without encountering any false conditions, return true.

Answer 3

Answer №2

Here is another approach:

You can implement the following logic:

Find the count of alphabets and the count of alphabets enclosed in a + sign on both sides. If the two counts are equal, then return true, otherwise return false

function CheckSymbols(str) {
   return str.match(/[a-z]/gi).length == str.match(/\+[a-z]\+/g).length
}

Answer 4

Here is another approach:

You can implement the following logic:

Find the count of alphabets and the count of alphabets enclosed in a + sign on both sides. If the two counts are equal, then return true, otherwise return false

function CheckSymbols(str) {
   return str.match(/[a-z]/gi).length == str.match(/\+[a-z]\+/g).length
}

Answer 5

Answer №3

What are your thoughts on this straightforward reasoning?

str = '+f+++l+k++'

function SimpleSymbols(str) {
  str = str.toLowerCase()
  for (i=0 ; i<str.length ; i++)  {
      if (str[i] === '+') {
          continue;
      }

      if (i === 0 || i === str.length - 1) {
          return false;
      }

      if (str[i - 1] === '+' && str[i].match(/[a-z]/) && str[i + 1] === '+') {
          i += 1;
          continue;
      } else {return false;}
  } 

  return true;
}

Fiddle

Answer 6

What are your thoughts on this straightforward reasoning?

str = '+f+++l+k++'

function SimpleSymbols(str) {
  str = str.toLowerCase()
  for (i=0 ; i<str.length ; i++)  {
      if (str[i] === '+') {
          continue;
      }

      if (i === 0 || i === str.length - 1) {
          return false;
      }

      if (str[i - 1] === '+' && str[i].match(/[a-z]/) && str[i + 1] === '+') {
          i += 1;
          continue;
      } else {return false;}
  } 

  return true;
}

Fiddle

Answer 7

Answer №4

While the Regex solution is concise, opting for a version using string.split can greatly enhance readability:

function CheckSymbols(input){
     if(input.length === 0) return false;
     if(input[0] !== '+' || input[input.length-1] !== '+') return false;
     var sections = input.split('+');
     for(var j=0; j < sections.length; j++){
          if(sections.length > 1) return false;
     }
     return true;
}

Answer 8

While the Regex solution is concise, opting for a version using string.split can greatly enhance readability:

function CheckSymbols(input){
     if(input.length === 0) return false;
     if(input[0] !== '+' || input[input.length-1] !== '+') return false;
     var sections = input.split('+');
     for(var j=0; j < sections.length; j++){
          if(sections.length > 1) return false;
     }
     return true;
}

Return true in an incorrect manner

Answer №1

Answer №2

Answer №3

Answer №4

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