CODEWITHSUNDEEP
javascriptstring
Daniyal Javaid
Daniyal Javaid

Reputation: 1636

string comparison "=" != "+" , incorrect results

Task is to check whether string results true or false for following given condition:

If a letter a to z occurs it must have '+' before and after the particular letter This works fine expect the case when '=' comes before or after the letter. Why this condition is wrong for '=' ? This code returns true for these strings Thanks

function SimpleSymbols(str) { 
    var temp=str.split("")
    if(temp.length==1 || temp.length==2 )
    return false;
    for(i=0;i<temp.length;i++){
        if(temp[i]>='a' && temp[i]<='z'){
            if(i===0 || i===(temp.length-1)|| 
            (temp[i-1]!=='+' && temp[i+1]!=='+')){
                return false;
            }   
        }
    } 
  return true; 
}

Upvotes: 0

Views: 67

Answers (3)

Christoph B&#246;hme
Christoph B&#246;hme

Reputation: 4021

The condition (temp[i-1]!=='+' && temp[i+1]!=='+') is only true if both the character before and after the letter are not pluses. If one of them is a plus sign the condition is false.

You need to change the logical and to an or operator: (temp[i-1]!=='+' || temp[i+1]!=='+')

Background: De Morgan's Laws

The original condition is that a letter is surrounded by plus signs:

temp[i-1] === '+' && temp[i+1] === '+'

In your if clause you test that this condition is not matched. So, the original condition becomes:

!(temp[i-1] === '+' && temp[i+1] === '+')

To transform this into a condition using not-equals, you need to apply De Morgan's Laws which basically say that a logical and becomes an or and vice-versa if you factor in a negation. This makes the resulting condition:

temp[i-1] !== '+' || temp[i+1] !== '+'

Upvotes: 2

user3310334
user3310334

Reputation:

Hint: check out these strings too, they all return true even though they should return false.

  • "+a4"
  • "1a+"
  • "%a+"

Let's simplify your condition:

str[i-1] !== '+' && str[i+1] !== '+'

is the same as

!(str[i-1] === '+' || str[i+1] === '+')

This makes it easier to see what you were really checking.

We can see now that the condition str[i-1] !== '+' && str[i+1] !== '+' returns true only if neither of those two characters are a +.

You want it to return true if at least one is not a +. So you should use this instead:

str[i-1] !== '+' || str[i+1] !== '+'

I re-wrote your code with this correct condition here:

function SimpleSymbols(str) { 
    if (str.length == 1 || str.length == 2) return false;
    for (var i = 0; i < str.length; i++) {
        if (str[i] >= 'a' && str[i] <= 'z') {
            if (i === 0 || i === (str.length-1) || (str[i-1] !== '+' || str[i+1] !== '+')) {
                return false;
            }   
        }
    } 
    return true; 
}

Note: regular expressions can help out a lot with pattern-matching in strings like this.

E.g. your whole function would have simply become the regex:

function SimpleSymbols(str) {
  return !str.match(/[^+][a-z][^+]/) && str.length > 2;
}

Upvotes: 1

ganesh phirke
ganesh phirke

Reputation: 473

<!-- I think you should change the condition operator -->

function SimpleSymbols(str) { 
    var temp=str.split("")
    if(temp.length==1 || temp.length==2 )
    return false;
    for(i=0;i<temp.length;i++){
        if(temp[i]>='a' && temp[i]<='z'){
            if(i===0 || i===(temp.length-1)|| 
            (temp[i-1]!=='+' || temp[i+1]!=='+')){
                return false;
            }   
        }
    } 
  return true; 
}

Upvotes: 2

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