JS Breaking out of nested for loop

Question

writing code for the following algorithmic problem and no idea why it's not working. Following a debugger, I found that the elem variable never iterates beyond 's'. I'm concerned that this could be because of my understanding of how to break out of a parent for loop. I read this question: Best way to break from nested loops in Javascript? but I'm not sure if perhaps I'm doing something wrong.

function firstNonRepeatingLetter(s) {
  //input string
  //return first character that doesn't repeat anywhere else.
  //parent for loop points to the char we are analyzing
  //child for loop iterates over the remainder of the string
  //if child for loop doesnt find a repeat, return the char, else break out of the child for loop and cont

 if (s.length == 1) { return s;}

  parent_loop:
  for (var i = 0; i < s.length - 1; i++){ //parent loop
      var elem = s[i];
      child_loop:
      for (var j = i + 1; j < s.length; j++){
          if (elem == s[j]){
            break child_loop;
          }
      }
    return s[i];
    } 
 return "";
}  

console.log(firstNonRepeatingLetter('stress')); // should output t, getting s.

Rahul Raghavan · Accepted Answer · 2016-09-11 03:10:01Z

1

My suggestion would be to use single for loop instead of using two loops.

for( var i = 0; i<s.length - 1;i++) {
    var lastIndex = s.lastIndexOf(s[i]);
    if ( lastIndex == i) {
       return s[i];
    }
}

answered Sep 11, 2016 at 3:10

Rahul Raghavan

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Comments

Sanka · Accepted Answer · 2016-09-11 07:38:31Z

try with this.

this was the key of my code

if( f!=true)
return s[i];

see the full code

function firstNonRepeatingLetter(s) {
  //input string
  //return first character that doesn't repeat anywhere else.
  //parent for loop points to the char we are analyzing
  //child for loop iterates over the remainder of the string
  //if child for loop doesnt find a repeat, return the char, else break out of the child for loop and cont

 if (s.length == 1) { return s;}


  for (var i = 0; i < s.length - 1; i++){ //parent loop
      var elem = s[i];

    var f=false;
      for (var j = i + 1; j < s.length; j++){

          if (elem == s[j]){

            f=true;
            break;
          }

      }

    if( f!=true)
    return s[i];
    } 

 return "";
}  

console.log(firstNonRepeatingLetter('stress')); // should output t, getting s.

davidhu · Accepted Answer · 2016-09-11 03:07:21Z

I think you are breaking out the loop correctly, the reason why the function always return s is because the counter i never increments.

Let's run through the code

i = 0, element = 's', j = 1, s[j] = 't', which is not equal to 's'
j++, j = 2, s[j] = 'r', not equal to 's'
j++, j = 3, s[j] = 'e', not equal to 's' 
j++, j = 4, s[j] = 's', equal to 's'

so you break out of the child loop.

Now we hit the line return s[i], when i = 0, so naturally the function returns s.

If you change the return s[i] to include an if statement, like so

if(j == s.length) { 
    return s[i];
}

The function now returns 't'. You are checking to see if the child loop ran to its full completion, which means break child_loop; never ran, and you have an unique letter.

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