Matching number in a string including leading zeros-CodePudding

i am doing this exercise:

Write a function which increments a string, to create a new string. If the string already ends with a number, the number should be incremented by 1. If the string does not end with a number. the number 1 should be appended to the new string.

And this is the part where i am stuck

If the number has leading zeros the amount of digits should be considered.

This is what i did.


sumStrings = (a) => {
    
    if (a.match(/\d /g) == null) {
        console.log(a.match(/[a-z] /g)   "1")
    } else {
        let numbers = a.match(/\d /g).map(Number);
        console.log(a.match(/[a-z] /g)   (Number(numbers.join()) 1))
    }
}

sumStrings("hello002")

returns hello3

Why doesn't match the leading zeros?

CodePudding user response：

By calling map(Number) you cast the string to a Number-type. The number does only contain it's value.

CodePudding user response：

You have to first get the leading zeros from the string, then add them back later once you've incremented the number. Data of Number type will never show leading zeros -- "002" as a number will always be 2 - so you have to manipulate the string first.

This line tests for leading zeros, and will default to an empty string if they're not there.

let leadingz = a.replace(/[a-z] /g, '').match(/^0 /)?.join('') || '';

const sumStrings = a => {
  if (a.match(/\d /g) == null) {
    console.log(a.match(/[a-z] /g)   "1")
  } else {
    let leadingz = a.replace(/[a-z] /g, '').match(/^0 /)?.join('') || '';
    let numbers = a.match(/\d /g).map(Number);
    console.log(a.match(/[a-z] /g)   leadingz   (Number(numbers.join())   1))
  }
}

sumStrings("hello000002")

CodePudding user response：

Here is a solution that preserves the minimum number of digits:

sumStrings = (a) => {
    let num = a.match(/\d /);
    if (num == null) {
        console.log(a.match(/[a-z] /)   "1")
    } else {
        let s = num[0];
        let snum = ""   (Number(s)   1);
        while (snum.length < s.length)
            snum = "0"   snum;
        console.log(a.match(/[a-z] /)   snum);
    }
}

sumStrings("hello002")

Output: hello003

CodePudding user response：

You have to split the letters and the numbers...

About the numbers part... To know how many leading zeros there is, you can compare its length to the "parsed-and-restringified" length.

Then add 1 and redo the check. Maybe you have an extra leading zero now.

Notice the handy repeat() method in the return statement ;)

sumStrings = (a) => {
  
    // Get the letters part
    let letters = a.match(/[a-z] /g)[0];
    let lettersLength = letters.length;

    // Get the numbers part (the remainder)
    let numbers = a.substring(lettersLength);

    // Now parse the numbers as an integer
    let numbers_int = parseInt(numbers);
    
    // Then count how namy leading zeros are in front
    let numbersLength_str = numbers.length
    let numbersLength_int = numbers_int.toString().length
    let leadingZeros = numbersLength_str - numbersLength_int
    
    // Calculate the new number
    let newNumber = numbers_int   1;
  
    // If that new number is not the same length, decrease the leadingZeros count
    if(leadingZeros > 0 && newNumber.toString().length > numbersLength_int){
      leadingZeros--;
    }
      
    return letters   "0".repeat(leadingZeros)   newNumber

  }
  
  console.log(sumStrings("hello002")) // Expecting "hello003"
  
  console.log(sumStrings("hello009")) // Expecting "hello010"
  
  console.log(sumStrings("hello999")) // Expecting "hello1000"